BRILLIANT PUBLIC SCHOOL, SITAMARHI (Affiliated up to +2 level to C.B.S.E., New Delhi)

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1 BRILLIANT PUBLIC SCHOOL, SITAMARHI (Affiliated up to + level to C.B.S.E., New Delhi) Class X S.A.-II Maths I.I.T.Foundation, N.T.S.E. & Olympiad Study Package Session: 4-5 Office: Rajopatti, Dumra Road, Sitamarhi (Bihar), Pin-84 Ph.66-54, Mobile: , 9969 Website: brilliantpublic@yahoo.com

2 TOPICS PAGES. Circles -5. Constructions 6-. Trigonometry - 4. Heights and Distances Mensuration Statistics Probability CIRCLES 9. CIRCLE A circle is the locus of a points which moves in a plane in such a way that its distance from a fixed point remains constant. 9. SECANT AND TANGENT : Secant to a circle is a line which intersects the circle in two distinct points. A tangent to a circle is a line that intersects the circle in exactly one point. 9. THEOREM : Statement : A tangent to a circle i perpendicular to the radius through the point of contact. Given : A circle C (O, r) and a tangent AB at a point P. To prove : OP AB Construction : Take any points Q, other than P on the tangent AB. Join OQ. Suppose OQ meets the circle at R. Proof: Among all line segments joining the point O to a point on AB, the shorted one is perpendicular to AB. So, to prove that OP AB, it is sufficient to prove that OP is shorter than any other segment joining O to any point of AB. Clearly OP OR Now, OQ OR + RQ OQ > OR OQ > OP ( OP OR) Thus, OP is shorter than any other segment joining O to any point of AB. Hence, OP AB. 9.4 THEORM : Statement : Lengths of two tangents drawn from an external point to a circle are equal.

3 Given: AP and AQ are two tangents drawn from a point A to a circle C (O, r). To prove : AP AQ Construction : Join OP, OQ and OA. Proof : In AOQ and APO OQA OPA [Tangent at any point of a circle is perp. to radius through the point of contact] AO AO OQ OP [Common] [Radius] So, by R.H.S. criterion of congruency AOQ AOP AQ AP [By CPCT] Hence Proved. Result Result : : (i) (i) If If two two tangents tangents are are drawn drawn to to a circle a circle from from an an external external point, point, then then they they subtend subtend equal equal angles angles at at the centre. the centre. OAQ OAQ OAP [By OAP CPCT] [By CPCT] (ii) (ii) If If two two tangents tangents are are drawn drawn to to a a circle circle from from an external an external point, point, they are they equally are equally inclined inclined to the to the segment, segment, joining joining the the centre centre to to that that point point OAQ OAQ OAP OAP [By [By CPCT] CPCT] Ex. If all the sides of a parallelogram touches a circle, show that the parallelogram is a rhombus. Given : Sides AB, BC, CD and DA of a gm ABCD touch a circle at P,Q,R and S respectively. To prove gm ABCD is a rhombus. Proof : AP AS...(i) BP BQ CR CQ...(ii)...(iii) Ex. So. DR DS...(iv) [Tangents drawn from an external point to a circle are equal] Adding (), (), () and (4), we get AP + BP + CR + DR AS + BQ + CQ + DS (AP + BP) + (CR + DR) (AS + DS_ + (BQ + CQ) AB + CD AD + BC AB + AB AD + AD [In a gm ABCD, opposite side are equal] AB AD or AB AD But AB CD AND AD BC [Opposite sides of a gem] Hence, AB BC CD DA gm ABCD is a rhombus. A circle touches the BC of a ABC at P and touches AB and AC when produced at Q and R respectively as shown in figure, Show that (Perimeter of ABC). Given : A circle is touching side BC of ABC at P and touching AB and AC when produced at Q and R respectively.

4 To prove : AQ (perimeter of ABC) Proof : AQ AR...(i) BQ BP...(ii) CP CR...(iii) [Tangents drawn from and external point to a circle are equal] Now, perimeter of ABC AB + BC + CA AB + BP + PC + CA (AB + BQ) + (CR + CA) [From (ii) and (iii)] AQ + AR AQ + AQ [From (i)] AQ (perimeter of ABC). Ex. Prove that the tangents at the extremities of any chord make equal angles with the chord. Let AB be a chord of a circle with centre O, and let AP and BP be the tangents at A and B respectively. Suppose, the tangents meet at point P. Join OP. Suppose OP meets AB at C. We have to prove that PAC PBC In triangles PCA and PCB PA PB [ Tangent from an external point are equal] APC BPC [ PA and PB are equally inclined to OP] And PC PC [Common] So, by SAS criteria of congruence PAC BPC PAC PBC [By CPCT] Ex.4 Prove that the segment joining the points of contact of two parallel tangents passes through the centre. Let PAQ and RBS be two parallel tangents to a circle with centre O. Join OA and OB. Draw OC PQ Now, PA CO PAO + COA 8 [Sum of co-interior angle is 8 ] 9 + COA 8 [ PAO 9] COA 9 Similarly, CON 9 COA + COB Hence, AOB is a straight line passing through O.

5 OBJECTIVE DPP The length of the tangent drawn from a point 8 cm away from the centre of a circle of radius 6 cm is (A) 7 cm (B) 7 cm (C) cm (D) 5 cm. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q, so that OQ cm. Length of PQ is : (A) cm (B) cm (C) 8.5 cm (D) 9 cm. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 8 then POA is equal to (A) 5 (B) 6 (C) 7 (D) 8 4. Two circle touch each other externally at C and AB is a common tangent to the circle. Then ACB (A) 6 (B) 45 (C) (D) 9 5. ABC is a right angled triangle, right angled at B such that BC 6 am and AB 8 cm. A circle with centre O is inscribed in ABC. The radius of the circle is (A) cm (B) cm (C) cm (D) 4 cm SUBJECTIVE DPP ABCD is a quadrilateral such than D 9. A circle C (O, r) touches the sides AB, BC, CD and DA at P, Q, R and S respectively. If BC 8 cm, CD 5 cm and BP 7 cm, find r.. Two concentric circles are of radius 5 cm and cm. Find the length of the chord of the larger circle which touches the smaller circle.. In a circle of radius 5 cm, AB and AC are two chords, such that AB AC 6 cm. Find the length of chord BC. 4. The radius of the incircle of a triangle is 4 cm and the segments into which one side is divided by the point of contact are 6 cm and 8 cm. Determine the other two sides of the triangle. 5. In figure, l and m are two parallel tangents at P and R. The tangent at Q makes an intercept ST between l and m. Prove that SOT 9 6. PQR is a right angled triangle with PQ cm and QR 5 cm. A circle with centre O and radius x is inscribed in PQR. Find the value of x. 7. From an external point P, two tangents PA and PB are drawn to the circle with centre O. Prove that OP is the perpendicular dissector of AB. 8. Two tangent TP and TQ are drawn to a circle with centre O from an external point T. Prove that PTQ OPQ. 4

6 9. A circle touches the sides of a quadrilateral ABCD at P, Q, R, S respectively. Show that the angles subtended at the centre by a pair of opposite sides are supplementary.. In figure, a circle touches all the four sides of a quadrilateral ABCD with AB 6 cm, BC 7 cm and CD 4 cm. Find AD. [CBSE - ]. Prove that the lengths of the tangents drawn from an external point to a circle are equal. Using the above, do the following : In figure, TP and TQ are tangents from T to the circle with centre O and R is any point on the circle. If AB is a tangent to the circle at R, prove that TA + AR TB + BR. [CBSE - 8]. In figure, if ATO 4, find AOB [CBSE - 8]. In figure OP is equal to diameter of the circle. Prove that ABP is an equilateral triangle. [CBSE - 8] ANSWERS (Objective DPP 9.) Qus. 4 5 Ans. B D A D B (Subjective DPP 9.). 4 cm. 8 cm. 9.6 cm 4. cm and 5 cm 6. cm. cm. 5

7 CONSTRUCTION. DIVISION OF A LINE SEGENT : In order to divide a line segment internally is a given ratio m: n, where both m and n are positive integers, we follow the following steps: Step of construction : (i) Draw a line segment AB of given length by using a ruler. (ii) Draw and ray AX making an acute angle with AB. (iii) Along AX mark off (m + n) points A, A,..., A m+n such that AA A A...A m+n+ A m+n. (iv) Join B A m+n (v) Through the point A m draw a line parallel to A m+n B by making an angle equal to AA m+ n B at A m. Suppose this line meets AB at a point P. The point P so obtained is the required point which divides AB internally in the ratio m : n. Ex. Divide a line segment of length cm internally in the ratio :. Following are the steps of construction. Step of construction : (i) Draw a line segment AB cm by using a ruler. (ii) Draw any ray making an acute angle BAX with AB. (iii) Along AX, mark-off 5 ( + ) points A,A,A,A 4 and A 5 such that AA A A A A A A 4 A 4 A 5. (iv) Join BA 5 (v) Through A draw a line A P parallel to A 5 B by making an angle equal to AA5 B at A intersecting AB at a point P. The point P so obtained is the required point, which divides AB internally in the ratio :. 6

8 . ALTERNATIVE METHOD FOR DIVISION OF A LINE SEGMENT INTERNALLY IN A GIVEN RATIO : Use the following steps to divide a given line segment AB internally in a given ration m : n, where m and natural members. Steps of Construction : (i) Draw a line segment AB of given length. (ii) Draw any ray AZ making an acute angle BAX with AB. (iii) Draw a ray BY, on opposite side of AX, parallel to AX making an angle ABY equal to BAX. (iv) Mark off a points A, A,...A m on AX and n points B, B,...B n on BY such that AA A A... A m- A m B B...B n- B n. (v) Join A m B n. Suppose it intersect AB at P. The point P is the required point dividing AB in the ratio m : n. Ex. Decide a line segment of length 6 cm internally in the ratio :4. Follow the following steps : Steps of Construction : (i) Draw a line segment AB of length 6 cm. (ii) Draw any ray AX making an acute angle BAX with AB. (iii) Draw a ray BY parallel to AX by making ABY equal to BAX. (iv) Mark of three point A,A,A on AX and 4 points B, B m B, B 4 on BY such that AA A A A A BB B B B B B B 4. (v) Join A B 4. Suppose it intersects AB at a point P. Then, P is the point dividing AB internally in the ratio :4. 7

9 . CONTRUCTION OF A TRIANGLE SIMILAR TO A GIVEN TRIANGLE : Scale Factor : The ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle is known as their scale factor. Ex. Steps of Construction when m<n : (i) Construct the given triangle ABC by using the given data. (ii) Take any one of the three side of the given triangle as base. Let AB be the base of the given triangle. (iii) At one end, say A, of base AB. Construct an acute angle BAX below the base AB. (iv) Along AX mark of n points A, A,A,...A n such that AA A A... A n- A n. (v) Join A n B. (vi) Draw A m B parallel to A n B which meets AB at B. (vii) From B draw B C CB meeting AC at C. m Triangle AB C is the required triangle each of whose side is of the corresponding side of ABC. n Construction a ABC in which AB 5 cm, BC 6 cm and AC 7 cm. Now, construct a triangle similar to ABC such that each of its side is two-third of the corresponding side of ABC. Steps of Construction (i) Draw a line segment AB 5 cm. (ii) With A as centre and radius AC 7 cm, draw an arc. (iii) With B as centre and BC 6 cm, draw another arc, intersecting the arc draw in step (ii) at C. (iv) Join AC and BC to obtain ABC. (v) Below AB, make an acute angle BAX. (vi) Along AX, mark off three points (greater of and in ) A,A,A such that AA A A A A. th (vi) Join A B. (viii) Draw A B A B, meeting AB at B. (iv) From B, draw B C BC, meeting AC at C. AB C is the required triangle, each of the whose sides is two-third of the corresponding sides of ABC. 8

10 Steps of Construction when m > n: (i) Construct the given triangle by using the given data. (ii) Take any of the three sides of the given triangle and consider it as the base. Let AB be the base of the given triangle. (iii) At one end, say A, of base AB construct an acute angle BAX below base AB i.e. on the composite side of the vertex C. (iv) Along AX, mark-off m (large of m and n) points A, A,...A m on AX such that AA A A... A m- A m. (v) Join A n to B and draw a line through A m parallel to A n B, intersecting the extended line segment AB at B. (vi) Draw a line through B parallel to BC intersecting the extended line segment AC at C. (vii) AB C so obtained is the required triangle. Ex.4 Draw a triangle ABC with side BC 7 cm, B 45, A 5 Construct a triangle whose side are (4/) times the corresponding side of ABC. In order to construct ABC, follow the following steps : (i) Draw BC 7 cm. (ii) At B construct CBX 45 and at C construct BCY 8 - ( ) Suppose BC and CY intersect at A. ABC so obtained is the given triangle. (iii) Construct an acute angle CBZ at B on opposite side of vertex A of ABC. (iv) Mark-off four (greater of 4 and in 4 ) points, B,B,B,B 4 on BZ such that BB - B B V B B B 4. (v) Join B ( the third point) to C and draw a line through B 4 parallel to B C, intersecting the extended line segment BC at C. (vi) Draw a line through C parallel to CA intersecting the extended line segment BA at A Triangle A BC so obtained is the required triangle such that A' B' BC' A' C' 4 AB BC AC 9

11 .4 CONSTRCUTION OF TANGENT TO A CIRCLE :.4 (a)to Draw the Tangent to a Circle at a Given Point on it, When the Centre of the Circle is Known : Given : A circle with centre O and a point P and it. Required : To draw the tangent to the circle at P. Steps of Construction. (i) Join OP. (ii) Draw a line AB perpendicular to OP at the point P. APB is the required tangent at P. Ex.5 Draw a circle of diameter 6 cm with centre O. Draw a diameter AOB. Through A or B draw tangent to the circle. Given : A circle with centre O and a point P on it. Required : To draw tangent to the circle at B or A. Steps of Construction. (i) With O as centre and radius equal to cm ( 6 ) draw a circle. (ii) Draw a diameter AOB. (iii) Draw CD AB. (iv) So. CD is the required tangent..4 (b) To Draw the Tangent to a Circle at a Given Point on it, When the Centre of the Circle is not Known : Given : A circle and a point P on it. Required : To draw the tangent to the circle at P. Steps of Construction (i) Draw any chord PQ and Joint P and Q to a point R in major arc PQ (or minor arc PQ). (ii) Draw QPB equal to PRQ and on opposite side of chord PQ. The line BPA will be a tangent to the circle at P. Ex.6 Draw a circle of radius 4.5 cm. Take a point P on it. Construct a tangent at the point P without using the centre of the circle. Write the steps of construction. Given : To draw a tangent to a circle at P. Steps of Construction (i) Draw a circle of radius 4.5 cm. (ii) Draw a chord PQ, from the given point P on the circle. (iii) Take a point R on the circle and joint PR and QR.

12 (iv) Draw QPB PRQ on the opposite side of the chord PQ. (v) Produce BP to A. Thus, APB is the required tangent..4 (c) To Draw the Tangent to a Circle from a Point Outside it (External Point) When its Centre is known : Given : A circle with centre O and a point P outside it. Required : To construct the tangents to the circle from P. Steps of Construction : (i) Join OP and bisect it. Let M be the mid point of OP. (ii) Taking M as centre and MO as radius, draw a circle to intersect C (O, r) in two points, say A and B (iii) Join PA and PB. These are the required tangents from P to C(O,r) Ex.7 Draw a circle of radius.5 cm. From a point P, 6 cm apart from the centre of a circle, draw two tangents to the circle. Given : A point P is at a distance of 6 cm from the centre of a circle of radius.5 cm Required : To draw two tangents to the circle from the given point P. Steps of Construction : (i) Draw a circle of radius.5 cm. Let it centre be O. (ii) Join OP and bisect it. Let M be mid-point of OP. (iii) Taking M as centre and MO as radius draw a circle to intersect C in two points, say A and B. (iv) Join PA and PB. These are the required tangents from P to C..4 (d) To Draw Tangents to a Circle From a Point Outside it (When its Centre is not Known): Given : P is a point outside the circle. Required : To draw tangents from a point P outside the circle. Steps of Construction : (i) Draw a secant PAB to intersect the circle at A and B. (ii) Produce AP to a point C, such that PA PC. (iii) With BC as a diameter, draw a semicircle. (iv) Draw PD CB, intersecting the semicircle at D. (v) Taking PD as radius and P as centre, draw arcs to intersect the circle at T and T. (iv) Join PT and PT. Then, PT and PT are the required tangents.

13 Ex.8 Draw a circle of radius cm. From a point P, outside the circle draw two tangents to the circle without using the centre of the circle. Given : A point P is outside the circle of radius cm. Required : To draw two tangents to the circle from the point P, without the use of centre. Steps of constructing (i) Draw a circle of radius cm. (ii) Take a point P outside the circle and draw a secant PAB, intersecting the circle at A and B. (iii) Produce AP to C such that AP CP. (iv) Draw a semicircle, wit CB as a diameter. (v) Draw PD AB, intersecting the semi-circle AT D. (vi) With PD as radius and P as centre draw two arcs to intersect the given circle at T and T. (vii) Joint PT and PT. Which are the required tangents. DAILY PRATICE PROBLEMS # SUBEJCTIVE DPP -.. Draw a circle of radius.5 cm. Take a point P on it. Draw a tangent to the circle at the point P.. From a point P on the circle of radius 4 cm, draw a tangent to the circle without using the centre. Also, write steps of construction.. Draw a circle of radius.5 cm. Take a point P on it. Draw a tangent to the circle at the point P, without using the centre of the circle. 4. Draw a circle of radius cm. Take a point P at a distance of 5.6 cm from the centre of the circle. From the point P, draw two tangents to the circle. 5. Draw a circle of radius 4.5 cm. Take point P outside the circle. Without using the centre of the circle, draw two tangents to the circle from the point P. 6. Construct a triangle ABC, similar to a given equilateral triangle PQR with side 5 cm. such that each of its side is 6/7th of the corresponding side of the PQR. 7. Construct a triangle ABC. similar to a given isosceles triangle PQR with QR 5 cm, PR PQ cm, such that each of its side is 5/ of the corresponding sides of the PQR. 8. Draw a line segment AB 7 cm. Divide it externally in the ratio of (i) : 5 (ii) 5 : 9. Draw a ABC with side BC 6 cm, AB 5cm and ABC 6. Construct a AB C similar to ABC such that sides of AB C are 4 of the corresponding sides of ABC. [CBSE - 8]

14 TRIGONOMETRY. TRIGONOMETRY : Trigonometry means, the science which deals with the measurement of triangles.. (a) Trigonometric Ratios : A right angled triangle is shown in Figure. are two other angles i.e. B Is of 9 Side opposite to B be called hypotenuse. There A and C. It we consider C as θ, then opposite side to this angle is called perpendicular and side adjacent to θ is called base. (i) Six Trigonometry Ratio are : Perpenicular P sin θ Hypotenuse H Base B cos θ Hypotenuse H Perpendicular P tan θ Base B AB AC BC AC AB BC Hypoteuse H cos esθ Perpendicular P Hypotenuse H sec θ Base B Base B cot θ Parpendicular P AC BC BC AB AC AB (ii) Interrelationship is Basic Trigonometric Ratio : tan θ cot θ cot θ tan θ cos θ sec θ sec θ cos θ sin θ cos ecθ cos ecθ sin θ We also observe that sin θ tan θ cos θ cos θ cot θ sin θ

15 . (b) Trigonometric Table : θ Sin Cos Tan Cot Not defined Sec Cosec Not defined Not defined Not defined. (c) Trigonometric Identities : (i) sin θ + cos θ (A) sin θ cos θ (ii) (iii) (B) cos θ sin θ + tan θ sec θ (A) sec θ tan θ (B) sec θ tan θ (C) tan θ sec θ + cot θ cos ec θ (A) cos ec θ cot θ (B) cos ec θ cot θ (C) cot θ cos ec θ. (d) Trigonometric Ratio of Complementary Angles : sin( 9 θ) cos θ cos ( 9 θ) sin θ EX. tan ( 9 θ) cot θ cot ( 9 θ) tan θ sec ( 9 θ) cos ec θ cos ec ( 9 θ) sec θ ILLUSTRATIONS : In the given triangle AB cm and AC 5 cm. Find all trigonometric ratios. Using Pythagoras theorem AC AB + BC 5 + p 6 p P cm Here P 4 cm, B cm, H 5 cm P 4 sin θ H 5 B cos θ H 5 P 4 tan θ B B cot θ P 4 H 5 sec θ B H 5 cos ecθ P 4 4

16 m Ex. If tan θ, then find sin θ. n Let P m α and B nα P m tan θ B n H P + B Ex. H m α + n α H α m + n P ma tanθ H a m + n m sin θ m + n If cosec A the prove than tan A-sing A sin 4 A sec A. 5 Hypotenuse We hare coses A 5 Perpendicular So, we draw a right triangle ABC, right angled at C such that hypotenuse AB units and perpendicular BC 5 units B Pythagoras theorem, AB BC + AC () (5) + AC AC AC 44 units BC 5 tan A AC BC 5 sin A AB AB and sec A AC L.H.S. tan A - Sin A R.H.S. sin 4 A sec A ( 69 44) So, L.H.S. R.H.S. Hence Proved. Ex.4 In ABC, right angled at B. AC + AB 9 cm. Determine the value of cot C, cosec C, sec C. In ABC, we have (AC) (AB) + BC (9 - AB) AB + () [ AC + AB 9cm AC 9 - AB] (8 + AB - 8AB AB AB 7 AB 4cm. 8 Now, AC + AB 9 cm AC cm BC AC 5 AC 5 So, cot C, cosec C, sec C. AB 4 AB 4 BC 4 5

17 Ex.5 Given that cos (A - B) cos A cos B + Sin B, find the value of cos 5. Putting A 45 and B We get cos (45 - ) cos 45 cos + sin 45 sin Ex.6 5. cos cos A Rhombus of side of cm has two angles of 6 each. Find the length of diagonals and also find its area. Let ABCD be a rhombus of side cm and BAD BCD 6. Diagonals of parallelogram bisect each other. S, AO OC and BO OD In right triangle AOB OB OA sin cos AB AB OB OA OB 5 cm OA 5 BD (OB) AC (OA) BD (5) AC ( 5 ) BD cm AC cm So, the length of diagonals AC cm & BD cm Area of Rhombus Ex.7 Evaluate : sec cos ec sec AC BD 5 cos ec cot 6 tan cm. cot 6 tan + sin sec ( 9 6 ) cot 6 cos ec ( 9 ) tan 8 + sin sec + sin sec sin 45 5 sin 45 + tan7 tan 6 tan 7 + tan7 tan 6 tan 7 sec ( 9 8 ) sin 45 + tan( 9 7 ) tan 7 tan 6 cos ec 6 cot 6 + sin 8 cos ec cot tan sec tan + sin [ cos ec θ cot θ,sec θ tan θ ] sin 8 tan Ex.8 Prove that : cos ec( 65 + θ) sec( 5 θ) tan( 55 θ) + cot( 5 + θ) cos ec( 65 + θ) cos ec{ 9 ( 5 θ)} sec( 5 θ).(i) cot( 5 + θ) cot{ 9 ( 55 θ)} tan( 55 θ)..(ii) L.H.S. cos ec( 65 + θ) sec( 5 θ) tan( 55 θ) + cot( 5 + θ) sec( 5 θ) sec( 5 θ) tan( 55 θ) + tan( 55 [u sin g(i)& (ii)] R.H.S. θ) 6

18 Ex.9 cos θ Prove that : cot θ tan θ sin θ cos θ L.H.S. cot θ tan θ cos θ sin θ sin θ cos θ cos θ sin sin θ cos θ cos θ ( cos θ) sin θ cos θ cos θ sin θ cot θ, tan θ sin θ cos θ [ sin θ cos θ] cos θ + cos θ cos θ R.H.S. sin θ cos θ sin θ cos θ Ex. Prove that : (coses A sin A) (sec A cos A) (tan A + cot A). L.H.S. (cosec A sin A) (sec A cos A) (tan A + cot A) sin A cos A sin A cos A + sin A cos A cos A sin A sin A cos A sin A + cos A sin A cos A sin A cos A Hence Proved. cos A sin A [ sin A + cos A ] sin A cos A sin A cos A R.H.S. Hence Proved. Ex. If sin θ + cos θ m and sec θ + cos ecθ n, then prove that n (m - ) m. L.H.S. n(m - ) (sec θ + cos ecθ)[(sin θ + cos θ) ] + (sin θ + cos θ + sin θ cos θ ) cos θ sin θ cos θ + sin θ ( + sin θ cos θ ) sin θ cos θ (cos θ + sin θ) ( sin θ cos θ) sin θ cos θ (sin θ + cos θ) m R.H.S. Hence Proved. Ex. If sec θ x +, then prove that sec θ + tan θ x or. 4x x sec θ x +...(i) 4x + tan θ sec θ tan θ sec θ tan θ x + 4x tan θ x + + x 6x 4x tan θ x + + tan θ ± x 6x 4x So, tan θ x 4x tan θ x 4x.(ii) tan θ ± x 4x 7

19 or tan θ x..(iii) 4x Adding equation (i) and (ii) sec θ + tan θ x + + x 4x 4x sec θ + tan θ x Adding equation (i) and (ii) sec θ + tan θ x + x + 4x 4x x Hence, sec θ + tan θ + x or. x Ex. If θ is an acute angle and tan θ + cot θ find the value of We have, tan θ + cot θ tan θ + tan θ tan θ + tan θ tan θ + tan θ tan θ tan θ + (tan θ ) tan θ tan θ tan θ tan 45 θ tan θ + cot θ 9 9 tan 45 + cot 45 9 (tan 45 ) + (cot 45) ( ) + ( ) tan θ + cot θ DAILY PRACTIVE PROBLEMS # OBJECTIVE DPP -. π. If α + β and α, then sin β is (A). If 5 tan θ 4, then value of (A) (B) 5 sin θ 5 sin θ + (B) 6 cos θ is cos θ (C) (C) 5 4 (D) 4 (D) π. If 7 sin α 4 cos α; < α <, then value of 4 tan α 75 cos α 7 sec α is equal to (A) (B) (C) (D) 4 4. Given β + 5 cos α; β 5, then the value of ( cos β 5 sinβ ) is equal to 9 (A) 9 (B) (C) (D) 5 9 8

20 5. If tan θ tan θ 4, then is equal to sin θ + sin θ cos θ cos θ (A) (B) (C) (D) 6. The value of tan 5 tan tan 5... tan 85, is (A) (B) (C) (D) None of these 7. As x increases from to π the value of cos x (A) increases (B) decreases (C) remains constant (D) increases, then decreases π π π cot sec tan π π 8. Find the value of x from the equation x sin cos π π cos ec cos ec 4 6 (A) 4 (B) 6 (C) - (D) 9. The area of a triangle is sq. cm. Two sides are 6 cm and cm. The included angle is (A) cos (B) cos (C) sin (D) sin 6 6. If α + β 9 and α β then cos α + sin β equals to (A) (B) (C) (D) SBJECTIVE DPP -.. Evaluate : sin θ cos θ sin( 9 θ) cos θ sin θ cos( 9 (A) + cos( 9 θ) sin( 9 θ) (B) cos cos cos cos8 θ) sin + cos sin + cos C) sin( 5 + θ) cos( 4 θ) + tan tan tan 7 tan 8 tan (D) (cos sin 45 ) (sin 6 sec 45 ) + cot 4 cos + cos 7 (E) + cos ec 58 cot 58 tan 4 tan sec 5 cot 4. sec θ cos ecθ If cot θ, prove that. 4 sec θ + cos ecθ 7. If A + B 9, prove that : tan A tanb + tan A cot B sin B tan A sin A sec B cos A 4. If A, B, C are the interior angles of a ABC, show that : B + C A B + C A (i) sin cos (ii) cos sin Prove the following (Q, 5 to Q. ) 5. tan θ sin θ tan θ sin θ 6 5 tan 7 tan 45 tan 5 tan (sin θ + cos ecθ) + (cos θ + sec θ) 7 + tan θ + cot θ [CBSE - 8] tan θ cot θ 7. + sec θ cos ecθ + cot θ tan θ 9

21 sin θ 8. sec θ tan θ + sin θ 9.. sin A + cos A sin A cos A + sin A + cos A sin A + cos A sin A cos A cos (sin θ + sec θ) + (cos θ + cos ecθ) ( + sec θ cos ecθ). ( + cot θ cos ecθ)( + tan θ + sec θ) 8 8. (sin θ cos θ) (sin θ cos θ)( sin θ cos θ). tan θ + sec θ + sin θ tan θ sec θ + cos θ 4. If x r sin θ cos φ, y r sin θ sin φ, z r cos θ, then Prove that : x + y + z r. 5. If cot θ + tan θ x and sec θ - cos θ y, then prove that (x y) / (xy ) / 6. If sec θ + tan θ p, then show that 7. Prove that : tan A tan cos B cos A sin A sin B B cos B cos A sin A sin B 8. Prove that : sec x tan x cos x cos x cos x sec x + tan x A p sin θ [CBSE - 4] p + [CBSE - 5] [CBSE - 5] 9. Prove : ( + tan A) [CBSE - 6] tan A sin A sin A. Evaluate : tan 7 tan tan 6 tan 67 tan 8 cot sin sec 7. [CBSE - 7] tan 6. Without using trigonometric tables, evaluate the following : (sin 65 + sin 5 ) + (tan 5 tan 5 tan tan 75 tan 85 ) [CBSE - 8]. If sin θ cos ( θ - 6 ) and θ and θ - 6 are acute, find the value of θ [CBSE - 8]. If sin θ cos θ, find the value of θ. [CBSE - 8] 4. If 7 sin θ + cos θ 4, show that tan θ [CBSE - 8] 5. Prove : sin θ ( + tan θ ) + cos θ ( + cot θ ) sec θ + cosec θ. [CBSE - 8] ANSWERS (Objective DPP -.) Qus Ans. B B B A D A B B D A (Subjective DPP.). (A) (B) (C) (D) 4 (E)

22 HEIGHTS & DISTANCES. ANGLE OF ELEVATION : In order to see an object which is at a higher level compared to the ground level we are to look up. The line joining the object and the eye of the observer is known as the line sight and the angle which this line of sight makes with the horizontal drawn through the eye of the observer is known as the angle of elevation. Therefore, the angle of elevation of an object helps in finding out its height (figure). ANGLE OF DEPRESSION : When the object is at a lower level tan the observer s eyes, he has to look downwards to have a view of the object. It that case, the angle which the line of sight makes with the horizontal thought the observer s eye is known as the angle of depression (Figure). ILLUSTRACTIONS : EX. A man is standing on the deck of a ship, which is 8 m above water level. He observes the angle of elevations of the top of a hill as 6 and the angle of depression of the base of the hill as. Calculation the distance of the hill from the ship and the height of the hill. [CBSE 5] Let x be distance of hill from man and h + 8 be height of hill which is required. is right triangle ACB. AC tan 6 BC h x h x In right triangle BCD. 8 x x 8 Height of hill h + 8. x + 8 ( )( 8 ) + 8 m. Distance of ship from hill x 8 m.

23 Ex. A vertical tower stands on a horizontal plane and is surmounted by vertical flag staff of height 5 meters. At a point on the plane, the angle of elevation of the bottom and the top of the flag staff are respectively and 6 find the height of tower. [CBSE-6] Let AB be the tower of height h metre and BC be the height of flag staff surmounted on the tower, Let the point of the place be D at a distance x meter from the foot of the tower in ABD tan h x x h In ABD tan 6 AC AD 5 + h x 5 + h x From (i) and (ii) 5 h + h h 5 + h h 5 AB AD..(i).(ii) Ex. 5 h. 5m So, the height of tower.5 m The angles of depressions of the top and bottom of 8m tall building from the top of a multistoried building are and 45 respectively. Find the height of multistoried building and the distance between the two buildings. Let AB be the multistoried building of height h and let the distance between two buildings be x meters. XAC ACB 45 [Alternate angles AX DE] XAD ADE [Alternate angles AX BC] In ADE AE tan ED h 8 ( CB DE x) x x (h 8) In ACB h tan 45 x h x x h Form (i) and (ii).(i).(ii) ( h 8) h h 8 h h h 8 h( ) 8 8 ( + ) h +

24 Ex.4 Ex.5 8 ( + ) h h 4 ( + ) h 4 ( + ) metres Form (ii) x h So, x 4( + ) metres Hence, height of multistoried building 4( + ) metres Distance between two building 4( + ) metres The angle of elevation of an aeroplane from a point on the ground is 45. After a flight of 5 sec, the elevation changes to. If the aeroplane is flying at a height of metres, find the speed of the aeroplane. Let the point on the ground is E which is y metres from point B and let after 5 sec flight it covers x metres distance. In AEB. tan 45 AB EB y y m (i) In CED CD tan ED x + y ( AB CD) x + y From equation (i) and (ii) (ii) x + x x ( ) x (. 7 ) x 96 m Speed of Aeroplane Dis tan ce cov ered 96 m / sec m / sec. Tiem taken Km / hr 57.4 Km/hr 5 5 Hence, the speed of aeroplane is 57.4 Km/hr. If the angle of elevation of cloud from a point h metres above a lake is α and the angle of depression of its h sec α reflection in the lake is β, prove that the distance of the cloud from the point of observation is. tan β tan α Let AB be the surface of the lake and let C be a point of observation such that AC- h metres. Let D be the position of the cloud and D be its reflection in the lake. Then BD BD. In DCE DE tan α CE H CE (i) tan α In CED ED' tan β EC h + H + h CE tan β h + H CE (ii) tanβ

25 From (i) & (ii) H h + H tan α tan β Htan β h tan α + H tan α Htan β H tan α + h tan α H(tan β tan α) h tan α h tan α H tan β tan α (iii) In DCE DE Sin α CD DE CD sin α Substituting the value of H from (iii) h tan α CD (tan β tan α) sin α H CD sin α CD (tan sin α h cos α β tan α ) sin h tan α CD tan β tan α h sec α Hence, the distance of the cloud from the point of observation is Hence Proved. tan β tan α Ex.6 A boy is standing on the ground and flying a kite with m of string at an elevation of. Another boy is standing on the roof of a m high building and is flying his kite at an elevation of 45. Both the boys are on opposite sides of both the kites. Find the length of the string that the second boy must have so that the two kites meet. Let the length of second string b x m. In ABC Sin AC AB AC AC 5m In AEF Sin AF AE AC FC x 5 4 [ AC 5 m, FC ED m] x x x 4 m (So the length of string that the second boy must have so that the two kites meet α 4 m.) DAILY PRACTICE PROBLEMS # OBJECTIVE DPP -.. Upper part of a vertical tree which is broken over by the winds just touches the ground and makes an angle of with the ground. If the length of the broken part is metres, then the remaining part of the trees is of length (A) metres (B) metres (C) metres (D) metres. The angle of elevation of the top of a tower as observed from a point on the horizontal ground is x. If we move a distance d towards the foot of the tower, the angle of elevation increases to y, then the height of the tower is d tan x tan y d tan x tan y (A) (B) d(tan y + tan x) (C) d(tan y tan x) (D) tan y tan x tan y + tan x 4

26 . The angle of elevation of the top of a tower, as seen from two points A & B situated in he same line and at distances p and q respectively from the foot of the tower, are complementary, then the height of the tower p is (A) pq (B) (C) pq (D) noen of these q 4. The angle of elevation of the top of a tower at a distance of height of the tower 5 metres from the foot is 6. Find the (A) 5 metres (B) metres (C) -5 metres (D) 5 metres 5. The Shadow of a tower, when the angle of elevation of the sun is, is found to be 5 m longer than when its was 45, then the height of tower in metre is (A) (B) ( ) (C) ( + ) (D) None of these. + SUBJECTIVE DPP -.. From the top a light house, the angles of depression of two ships of the opposite sides of it are observed to be α and β. If the height of the light house be h meters and the line joining the ships passes thought the h(tan α + tan β) foot of the light house. Show that the distance between the ships is meters. tan α tan β. A ladder rests against a wall at angle α to the horizontal. Its foot is pulled away from the previous point through a distance a, so that is slides down a distance b on the wall making an angle β. With the a cos α cos β horizontal show that b sin β sin α. From an aeroplanne vertically above a straight horizontal road, the angle of depression of two consecutive kilometer stone on opposite side of aeroplane are observed to be α and β. Show that the height of tan α tan β aeroplane above the road is kilometer. tan α + tan β 4. A round balloon of radius r subtends an angle θ at the eye of an observer while the angle of elevation of its centre is φ. Prove that the height of the centre of the balloon is r sin φ cosec θ. 5. A window in a building is at a height of m from the ground. The angle of depression of a point P on the ground from the window is. The angle of elevation of the top of the building from the point P is 6. Find the height of the building. 6. A man on a cliff observers a boat at an angle of depression of which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later, the angle of depression of the boat is found to be 6. Find the total time taken by the boat from the initial point to reach the shore. 7. The angles of elevation of the top of a tower two points P and Q at distances of a and b respectively from the base and in the same straight line with it, are complementary. Prove that the height of the tower is ab. [CBSE - 4] 8 Two pillars of equal height are on either side of a road, which s m wide. The angles of elevation of the top the pillars are 6 and at a point on the road between the pillar. Find the position of the pint between the pillars. Also find the height of each pillar, [CBSE - 5] 9 At a point, the angle of elevation of a tower is such that its tangent is 5,On walking 4mnearer the tower, the tangent to the angle of elevation becomes 4, Find the height of the tower. [CBSE - 6] From a window x mtres high above the ground in a street, the angles of elevation and depression of the top and foot of the other hose on the opposite side of the street are α and β respectively, Show that the opposite house is x ( + tan α cot β ) metres. [CBSE - 6] A pole 5m high is fixed on the top of a towel, the angle of elevation of the top of the pole observed from a point A on the ground is 6 an the angle of depression the point ;A; from the top of the tower is 45 Find the height of the tower. [CBSE - 7] 5

27 The angle of elevation of a jet fighter from a point A on the ground is 6 After a flight of 5 seconds, the angle o elevation changes to If the jet is flying at a spies of 7km/fr, find the constant height at which the jet is flying. [use.7] [CBSE - 8] ANSWERS (Objective DPP.) Q 4 5 A. C A C D C 5. m 6. 9 min. (Subjective DPP.) 8. Height 4. m, Position point is 5 m from st end and 75 m from nd end m. 6.8 m. 598 m. MENSURATION. MENSRTION: Figure lying in a plane is called a plane figure. A plane figure made up of lines or curve or both, is said to be a closed figure if it has on free ends. Closed figure in a plane covers some part of the plane, then magnitude o that part of the plane is called the area of that closed figure. the unit of measurement of that part of the plane is called the area of that closed figure. the unit o measurement of area is square unit (i.e. square centimeter, square metre etc.). (a)mensuration of a Triangle: perimeter a + b + c Area Base Height ah Heron s formula: Area s(s a)(s b)(s c) Where s semi perimeter.(b) Menstruation of a Rectangle: a + b + c Perimeter ( l + b) Area l b Length of diagonal l + b.(c) Menstruation of a Square: Perimeter 4 a Area a Length of diagonal a 6

28 .(d) Menstruation of a parallelogram: Perimeter (a + b) Area ah bh.(e)mensuration of a Rhombus: Perimeter 4a Area dd d d +. (f) Mensuration of a Quadrilateral: Let AC d Area d(h + h).(g) Menstruation of a Trapezium: Area h (a + b). AREA RELTED TO CIRCLE: Circle: Circle is a point, which moves so such a manner that its distance from a fixed point id always equal. The fixed point is called center of the circle of the circle and the fixed distance is called radius of the circle. Area of circle (A) π r Circumference (C) πr Diameter (D) r Circle RESULTS: (i) If two If circles two circles touch touch internally. internally. then then distance the distance between between their their centers centers is equal is equal to the to the difference of their radii, of their radii, (ii) If two If two circles circles touch touch externally, externally, then then distance the distance between between their their centers centers is equal is equal to the to the sum sum of their of their radii. radii. (iii) Distance Distance moved moved by a by rotating a rotating wheel wheel in one in revolution one revolution is the is the circumference of the of wheel. the wheel. (iv) Number of revolutions completed by a rotating wheel in one Disminute tan ce moved in one min ute (iv) Number of revolutions completed by a rotating wheel in one minute Dis tan ce moved in one min ute Circumference (v) Angle described Circumference by minute hand is one minute 6. (vi) (v) Angle Angle described described by hour by minute hand in hand one hour is one minute. 6. (vi) Angle described by hour hand in one hour..5(a) Semicircle: Perimeter πr + r ( π + ) r Area (A) πr Semi-Circle 7

29 .(b)sector: Area (A) πr θ 6 Length of arc πrθ (l) 8 Area(A) l r Perimeter l + r.(c)segment : Shaded portion in the figure id called segment of a circle. Sector of a Circle Minor segment Major segment Minor Segment Area of minor segment Area of the sector -Area of triangle OAB πr θ θ θ πr θ r A r sin cos OR A sin θ 6 6 Here, segment ACB is called manor segment while ADB is called major segment.. MENSURATION (SOLID FIGURES) : If any figure such as cuboids, which has three dimensions length, width and height are height are known as three dimensional figures. Where as rectangle has only two dimensional i.e., length and width. Three dimensional figures have volume in addition to areas of surface from which these soils figures are formed. Some of the main solid figures are:. (a) Cuboid: Total Surface Area (T.S.A.) : The area of surface from which cuboid is formed. There are six faces (rectangular), eight vertices and twelve edges n a cuboid. (i)total Surface Area (T.S.A.) [ l 5b + b h + h l] (ii) Lateral Surface Area (L.A.A.) [ b h + h l] (or Area of 4 walls) h [ l + b] (iii) Volume of Cuboid (Area of base) height (iv) Length of diagonal l + b + h. (b) Cube : Cube has six faces. Each face is a square. (i) T.S,A [. x + x. x + x. x] x + x + x ] (x ) 6x (ii) L.S.A. [x + x ] 4x (iii) Volume (Area of base) Height) (x ). x x (iv) Length of altitude x 8

30 . (c) Cylinder : Curved surface area of cylinder (C.S.A.) : It is the area of surface from which the cylinder is formed. When we cut this cylinder, we will find a rectangle with length πr and height h units. (i) C.S.A. of cylinder ( πr) h πrh. (ii) Total Surface Area (T.S.A.) : T.S.A. C.S.A. + circular top & bottom π rh + πr πr (h + r) sq. units. (iii) Volume of cylinder : Volume Area of base height ( π r ) h π r h cubic units. (d) Cone : (i) C.S.A. π rl (II) T.S.A. C.S.A. + Other area π rl πr ( l + r) (iii) Volume π r h Where, h height r radius of base l slant height. (e) Sphere : T.S.A. S.A. 4πr 4 Volume πr. (f) Hemisphere : C.S.A πr T.S.A C.S.A. + other area πr + πr πr Volume πr. (g) Frustum of a Cone : When a cone is cut by a plane parallel to base, a small cone is obtained at top and other part is obtained at bottom. This is known as Frustum of Cone. ABC ~ ADE AC AE AB AD BC DE h l r h h l l r Or h l h l r r r 9

31 Volume of Frustum πr h πr (h h) π[ r h r (h h)] r h π r r r r πh [ r + r + r r ] r h r r Curved Surface Area of Frustum rl πr ( l l) π h r l rl π r r r r r r πl r r ) ( + Total Surface Area of Frustum CSA of frustum + πr Slant height of a Frustum h + (r r ) r r πh r r r l πl r r r r r + πr + πr r r ) + πr r πl ( + l where, h - height of the frustum r radius of larger circular end r radius of smaller circular end ILLUSTRACTION : Ex. A chord of circle 4 cm makes an angle of 6 at the center of the circle. Find : (i) area of minor sector (ii) area of the minor segment (iii) area of the major sector (iv) area of the major segment Given, r 4 cm, θ 6 (i) Area of minor sector OAPB θ 6 πr cm πr θ r (ii) Area of minor segment APB sin θ sin (iv) 7.8 cm (iii) Area of major sector Area of circle - Area of minor sector OAPB π (4) cm Area of major segment AQB Area of circle - Area of minor segment APB cm.

32 Ex. ABCP is a quadrant of a circle of radius 4 cm. With AC as diameter, a semicircle is drawn. Find the area of the shaded portion (figure). In right angled triangle ABC, we have. AC AB + BC AC AC 4 4 cm Now required Area Area APCQA Area ACQA - Area ACPA Area ACQA - (Area ABCPA - Area of ABC) 4 π π( 4) cm Ex. The diameter of cycle wheel is 8 cm. How many revolution will it make in moving. km? Distance traveled by the wheel is one revolution πr 8 88 cm 7 and the total distance covered by the wheel. cm cm Number of revolution made by the wheel Ex. 4 How many balls, each of radius cm, can be made from a solid sphere of lead of radius 8 cm? 4 Volume of the spherical ball of radius 8 cm π 8 cm Ex.5 4 Also, volume of each smaller spherical ball of radius cm π cm. Let n be the number of smaller balls that can be made. Then, the volume of the larger ball is equal to the sum of all the volumes of n smaller balls. 4 4 Hence, π n π 8 n 8 5 Hence, the required number of balls 5. An iron of length m and diameter 4 cm is melted and cast into thin wires of length cm each. If the number of such wires be, find the radius of each thin wire. Let the radius of each thin wire be r cm. The, the sum of the volumes of thin wire will be equal to the volume of the iron rod. Now, the shape of the iron rod and each thin wire is cylindrical. 4 Hence, the volume of the iron rod of radius cm cm is π cm Again, the volume of each thin wire πr Hence, we have π πr 4r 4 r r Hence, the required radius of each thin wire is cm. of. cm. [Taking positive square root only]

33 Ex.6 Ex.7 Ex.8 By melting a solid cylindrical metal, a few conical materials are to be made. If three times the radius of the cone is equal to twice the radius of the cylinder and the ratio of the height of the cylinder and the height of the cone is 4 : find the number of cones which can be made. Let R be the radius and H be the height of the cylinder and let r and h be the radius and height of the cone respectively. Then. r R and H : h 4 :...(i) H 4 h H 4h...(ii) Let be the required number of cones which can be made from the material of the cylinder. The, the volume of the cylinder will be equal to the sum of the volumes of n cones. Hence, we have n π R H πr h R H nr h 9r 4h R H n 4 r h r h r 4h [ From (i) and (ii), R and H ] 9 4 n 4 n 9 Hence, the required number of cones is 9. The base diameter of solid in the form of a cone is 6 cm and the height of the cone is cm. It is melted and recast into spherical balls of diameter cm. Find the number of balls, thus obtained. Let the number of spherical balls be n. Then, the volume of the cone will be equal to the sum of the volumes of the spherical balls. The radius of the base of the cone 6 cm cm and the radius of the sphere cm Now, the volume of the cone π cm πcm 4 π and, the volume of each sphere π cm cm 6 Hence, we have π n π n Hence, the required number of balls 8. A conical empty vessel is to be filled up completely by pouring water into it successively with the help of a cylindrical can of diameter 6 cm and height cm. The radius of the conical vessel if 9 cm and its height is 7 cm. How many times will it required to pour water into the conical vessel to fill it completely, if, in each time, the cylindrical can is filled with water completely? Let n be the required number of times. Then, the volume of the conical vessel will be equal to n times the volume of the cylindrical can. Now, the volume of the conical vessel π 9 7cm 4 8π cm Add the volume of the cylindrical can π cm 9 π cm Hence, 4 8 π 9 π n 4 8 n 8 9 Hence, the required number of times 8.

34 Ex.9 The height of a right circular cylinder is equal to its diameter. It is melted and recast into a sphere of radius equal to the radius of the cylinder, find the part of the material that remained unused. Let n be height of the cylinder. Then, its diameter is h and so its radius is h. Hence, its volume is V h π πh h 4 Again, the radius of the sphere h Hence, the volume of the sphere is V 4 h π πh 6 πh πh πh ( ) πh πh The volume of the unused material V V V Hence, the required volume of the unused material is equal to of the volume of the cylinder. Ex. Water flows at the rate of m per minute through a cylindrical pipe having its diameter as 5 mm. How much time till it take to fill a conical vessel whose diameter of the base is 4 cm and depth 4 cm? Diameter of the pipe 5 mm 5 cm cm. Radius of the pipe cm cm. 4 In minute, the length of the water column in the cylindrical pipe m cm. Volume, of water that flows out of the pipe in minute π cm. 4 4 Also, volume of the cone π 4 cm. Hence, the time needed to fill up this conical vessel π 4 π minutes minutes 56 minutes 5. minutes. 5 Hence, the required time of 5. minutes. Ex. A hemispherical tank of radius is full of water. It is connected with a pipe which empties it at the rate of 4 7 liters per second. How much time will it take to empty the tank completely? 7 7 Radius of the hemisphere m cm 75 cm 4 4 Volume of the hemisphere π cm The cylindrical pipe empties it at the rate of 7 liters i.e., 7 cm of water per second. Hence, the required time to empty the tank s min min min min, nearly.

35 Ex. A well of diameter m is dug 4 m deep. The earth taken out of its is spread evenly all around it to a width of 5 m to from an embankment. Find the height of the embankment. Let n be the required height of the embankment. The shape of the embankment will be like the shape of a cylinder of internal radius m and external radius (5 + ) m 6 m [figure]. The volume of the embankment will be equal to the volume of the earth dug out from the well. Now, the volume of the earth volume of the cylindrical well π 4m 4 π m Also, the volume of the embankment π (6 - ) h cm 5 π h m Hence, we have 5 π h 4 π 4 h Hence, the required height of the embankment.4 m Ex. Water in a canal, dm wide and dm deep, is flowing with a speed of km/hr. How much area will it irrigate in minutes if 8 cm of standing water is required from irrigation. 5 Speed of water in the canal km. h m.6 min m/min. 5 The volume of the water flowing out of the canal in minute m 6 m In min, the amount of water flowing out of the canal (6 ) m 6 m If the required area of the irrigated land is m, then the volume of water to be needed to irrigate the land 8 x x m x 5 m Hence, 8 x Hence, the required area is 5 m. Ex.4 A bucket is 4 cm in diameter at the top and 8 cm in diameter at the bottom. Find the capacity of the bucket in litters, if it is cm deep. Also, find the cost of tin sheet used in making the bucket, if the cost of tin is Rs..5 per sq dm. Given : r cm r 4 cm and h cm π Now, the required capacity (i.e. volume) of bucket h (r r r r + + ) ( ) cm 876 cm 97 cm 97 liters 9.7 liters. 7 Now, I ( r r ) + h ( 4) + cm 6 + cm cm 477 cm. 84 cm. Total surface area of the bucket (which is open at the top) + r ) + r πl ( r π [(r + r ) r ] 6 + π l [( + 4) ] cm Required cost of the tin sheet at the rate of Rs..5 per dm i.e., per cm Rs Rs

36 Ex.5 A cone is divided into two parts by drawing a plane through a point which divides its height in the ratio : starting from the vertex and the place is parallel to the base. Compare the volume of the two parts. Let the plane XY divide the cone ABC in the ratio AE : ED :, where AED is the axis of the cone. Let r and r be the radii of the circular section XY and the base BC of the cone respectively and let h - h and h be their heights [figure]. h Then, h h h h r h And r h h h r r Volume of cone AXY πr (h h) πr ( h h) π r h 6 Volume of frustum XYBC πh(r + r + rr ) πh( 9r + r + r ) π h( r ) πr h Volume of cone AXY So, 6 Volume of frustum XYBC πr h Volume of cone AXY. Volume of frustum XYBC 6 i.e. the ratio between the volume of the cone AXY and the remaining portion BCYX is : 6. DAILY PRACTIVE PROBLEMS # OBJECTIVE DPP -.. If BC passed through the centre of the circle, then the area of the shaded region in the given figure is a (A) ( π) (B) a π (C) a a π ( π ) (D). The perimeter of the following shaded portion of the figure is: (A) 4 m (C) 4.8 m (B) 4.7 m (D) 5 m. If a rectangle of sides 5 cm and 5 cm is be divided into three squared of equal area, then the sides of the squares will be : (A) 4 cm (B) 6 cm (C) 7 cm (D) None 5

37 4. The area of the shaded region in the given figure is : (A) π sq. units (B) π units (B) 4 π sq. units (D) π sq. units 5. The area of the shaded portion in the given figure is : (A) 7.5 π sq. units (C) 5.5 π sq. units (B) 6.5 π sq. units (D) 4.5 π sq. units 6. In the adjoining figure, the radius of the inner circle, if other circles are of radii m, is : (A) ( )m (B) m (C) m (D) m 7. The height of a conical tent of the centre is 5cm. The distance of any point on its circular base from the top of the tent is m. The area of the slant surface is : (A) 44 π sq m (B) π sq m (C) 56 π sq m (D) 69 π sq m 8. The radius of circle is increased by cm, then the ratio of the new circumference to the new diameter is : (A) π + (B) π + (C) π (D) π 9. A hemispherical bowl of internal diameter 6 cm is full of some liquid. This liquid is to be filled in cylindrical bottles of radius cm and height 6 cm., Then no of bottles needed to empty the bowl. (A) 6 (B) 75 (C) 8 (D) 44. There is a cylinder circumscribing the hemisphere such that their bases are common. The ratio of their volume is (A) : (B) : (C) : (D) : 4. A sphere of radius cms is dropped into a cylindrical vessel of radius 4 cms. If the sphere is submerged completely, then the height (in cm) to which the water rises, is (A).5 (B). (C).5 (D).5. If a rectangular sheet of paper 44 cm cm is rolled along its length of form a cylinder, then the volume of cylinder in cm is (A) 694 (B) 8 (C) 88 (D) none of these. Two cones have their heights in the ratio : and the radii of their bases are in the ratio :, then the ratio of their volumes is (A) : (B) 7 : (C) : (D) : 7 4. The total surface area of a cube is numerically equal to the surface area of a sphere then the ratio of their volume is (A) 6 π (B) π 6 π (C) 6 5. A cone is dived into two parts by drawing a plane through the mid point of its axis parallel to its base then the ratio of the volume of two parts is (A) : (B) : 7 (C) : 8 (D) : 9 (D) 6 π 6

38 . A cylindrical container of radius 6 cm and height 5 cm is fulled with ice-cream. The whole ice-cream has to be distributed to children in equal cones with hemispherical tops. If the height of the conical portion is four times the radius of its base, find the radius of the ice-cream cone.. A hemi-spherical depression is cutout from one face of the cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube., Determine the surface are of the remaining solid. 4. In figure there are three semicircles, A,B and C having diameter cm each, and another semicircle E having a circle D with diameter 4.5 cm are shown. Calculate. (i) the area of the shaded region (ii) the cost of painting the shaded region of the 5 paisa per cm, to the nearest rupee. 5. The height of a cone is cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be of the volume of the given cone, at what above the vase is the section made? 7 6. A solid cylinder of diameter 5 cm and height 5 cm is melted and recast into toys in the shape of a right circular cone mounted on a hemisphere. Find the radius of the hemisphere and the total height of the to if height of the conical par is times its radius. [CBSE - 5] 7. if the rail of the ends of bucket, 45 cm high are 8 cm and 7cm, determine the capacity and total surface area of the bucket. [CBSE - 6] 8. A tent is in the form of cylinder of diameter 4. m and height 4 m, surmounted by a cone of equal base and height.8 m. Find the capacity of the tent and the cost of canvas for making the tent at Rs. per sq. m.? [CBSE - 6] 9. Water flows out through a circular pipe whose internal radius is cm, at the rate of 8 cm/second into an empty cylindrical tank, the radius of whose base is 4 cm. By how much will the level of water rise in the tank in half an hour? [CBSE - 7] A hemispherical bowl of internal radius 6 cm is full of liquid. The liquid is to be filled into cylindrical shaped small bottles each of diameter cm and height 6 cm. How many bottles are need to empty the bowl? 7] In figure ABC is a right - angled triangle right-angled at A. Semicircles are drawn on AB, AC and BC as diameters. Find the area of the shaded region. [CBSE - 8]. Find the permetre of figure, where AED is a semi-circle and ABCD is a rectangle. [CBSE - 8] 7

39 . A tent consists of a frustum of a cone, surmounted by a cone. If the diameters of the upper and lower circular ends of the frustum b 4 m and 6 m respectively, the height of the frustum be 8 m and the slant height of the surmounted conical portion be m, find the area of canvas required to make the tent. (Assume that the radii of the upper circular end of the frustum and the base of surmounted conical portion are equal) [CBSE - 8] ANSWERS (Objective DPP -.) Qus Ans. D C D A D A C C B C C C C B B. 7.7 cm. cm (Subjective DPP -.) cm cm 6. : cm, 8.8 cm 8..5 cm 9. hrs cm. 4 cm. cm l. ( 4 + π) cm, Rs. 5. cm 6. radius cm and height 9 cm cm, 56 cm m, Rs cm sq.. 76 cm m 8

40 STATISTICS 4. INTRODUCTION : The branch of science known as statistics has been used in India from ancient times. Statistics deals with collection of numerical facts. i.e., data, their classification & tabulation and their interpretation. 4. MEASURES OF CENTRAL TENDANCY : The commonly used measure of central tendency (or averages) are : (i) Arithmetic Mean (AM) or Simply Mean (ii) Median (jjj) Mode 4. ARITHMETIC MEAN : Arithmetic mean of a set of observations is equal to their sum divided by the total number of observations. Mean of raw data : x, x, x,..., x n are the n values (or observations) the, A.M. (Arithmetic mean) is x x + x xn n i n n x i n xi i n x - Sum of observations n i.e. product of mean & no. of items gives sum of observation. Ex. The mean of marks scored by students was found to be 4. Later on its was discovered that a score of 56 was misread as 8. Find the correct mean. n, x 4 x ( xi) 4 ( ) n xi Incorrect value of x i 4. i Now, Correct value of x correct value of xi 97 Correct mean 9. 7 n So, the correct mean is 9.7 Method for Mean of Ungrouped Data x i f i F x x f f x x f f x x... f... i fx.. f f x 9

41 Grouped Frequency Distribution (Grouped) (i) Direct method : for finding mean mean x Ex. Find the missing value of P for the following distribution whose mean is.58 x 5 8 P 5 y Given x.58 Calculation of Mean : x i f i f i x i x fi x f i i P P ; 7P 5 ; P 5. Ex. Find the mean for the following distribution : f x P 7 7P i f u f i 5 fixu P Marks Frequency i Marks Mid Values x i No. of students f i f i x i f i 4 f i 4 i x x f x i f i i

42 (ii) Deviation Method : (Assumed Mean Method) x A + fd i f i i where, A Assumed mean d i Deviation from mean (x i - A) Find the mean for the following distribution by using deviation method : x i Frequency x i f i Let A 5 f i d i d i x i i x A + i f f 77 i d f d i f i i fu i i (iii) Step - Deviation Method : x A + h f i xi A where, A Assumed mean ui, h Width of class interval h Ex.5 Find the mean of following distribution with step - deviation method : Class Frequency Calculation of Mean : Class x i f i Let A 7.5 xi 7.5 ui 5 f i u i i i f 4 f i u 4

43 Ex. 6 fu x A i i + h x f i 4 The mean of the following frequency distribution is 6.8 and the sum of all frequencies is 5. Compute the missing frequency f and f Class Frequency 5 f f 7 8 Let, h Class x i f i xi A ui h f i u i f f + f f i + f + f f iu i 58 + f Given + f + f 5 f + f fu i i x A + h fi...(i) 58 + f (58 + f ) f f 4f f 48 f Now, f f f + f 8 So, the missing frequencies are f 8 and f. 4

44 Ex.7 Find the mean marks from the following data : Marks No. of Students Below 5 Below 9 Below 7 Below 4 9 Below 5 45 Below 6 6 Below 7 7 Below 8 78 Below 9 8 Below 85 Charging less than type frequency distribution in general frequency distribution. Marks x i f i A 45, h f i u i xi A ui h i According to step deviation formula for mean fu i i x A + h fi 9 x x x 48.4 So, the mean marks is PROPERTIES OF MEAN : i f 85 f i u 9 4

45 (i) Sum of deviations from mean is zero. i.e. (x x n i (ii) If a constant real number a is added to each of the observation than new mean will be x + a. (iii) If a constant real number a is subtracted from each of the observation then new mean will be x a. (iv) If constant real number a is multiplied with each of the observation then new mean will be a x. x (v) If each of the observation is divided by a constant no a, then new mean will be. a 4.5 MERITS OF ARITHETIC MEAN : (i) It is rigidly defined, simple, easy to understand and easy to calculate. (ii) It is based upon all the observations. (iii) Its value being unique, we can use it to compare different sets of data. (iv) It is least affected by sampling fluctuations. (v) Mathematical analysis of mean is possible. So, It is relatively reliable. 4.6 DEMERITS OF ARITHMETCI MEAN : (i) It can not be determined by inspection nor it can be located graphically. (ii) Arithmetic mean cannot be used for qualities characteristics such as intelligence, honesty, beauty etc. (iii) It cannot be obtained if a single observation is missing. (iv) It is affected very much by extreme values. In case of extreme items, A.M. gives a distorted picture of the distribution and no longer remains representative of the distribution. (v) It may lead to wrong conclusions if the details of the data from which it is computed are not given. (vi) It can not be calculated if the extreme class is open, e.g. below or above 9. (vii) It cannot be used in the study of rations, rates etc. 4.7 USES OF ARITHMETIC MEAN : (i) It is used for calculating average marks obtained by a student. (ii) It is extensively used in practical statistics and to obtain estimates. (iii) It is used by businessman to find out profit per unit article, output per machine, average monthly income and expenditure etc. 4.8 MEDIAN : Median is the middle value of the distribution. It is the value of variable such that the number of observations above it is equal to the number of observations below it. Median of raw data (i) Arrange the data in ascending order. (ii) Count the no. of observation (Let there be n observation) n + (A) if n be odd then median value of th i ) observation. n (B) if n is even then median is the arithmetic mean of Median of class - interval data (Grouped) N C Median l + h f th n observation and + l lower limit of median class, N total no of observation C cumulative frequency of the class preceding the median class h size of the median class f frequency of the median class. What is median class : N The class in which th item lie is median class. th observation. 44

46 Ex.8. Following are the lives in hours of 5 pieces of the components of air craft engine. Fin the median : 75, 74, 75, 7, 79, 745, 649, 699, 696, 7, 74, 78, 76, 75, 79 Arranging the data in ascending order , 75, 7, 7, 75, 76, 79, 74, 75, 78, 79, 74, 745 N 5 N + So, Median th th observation 5 + observation 76. Ex. 9 The daily wages (in rupees) of workers in a factory are given below : Daily wages (in Rs.) No. of workers Find the median wage of a worker for the above date. Daily wages (in Rs.) No. of workers Cumulative frequency N (even) Median Median N th th N observation + + observation 5 observation + 5 observation th th Median wage of a workers in the factory is Rs

47 Ex. Calculate the median for the following distribution class : Class Frequency N (i) First we find which lies in -. th th 55 value i.e. 7.5 th Class f c.f class in median class here l - 5 N 7.5, C 5, f, h median Median 6.5 Ex. in the median of the following frequency distribution is 46,find the missing frequencies : Variable Total Frequency? 65? Class Interval Frequency C.F f 4 + f f 5-6 f 7 + f + f f + f f + f Let the frequency of the class - 4 be f and that of the class 5-6 be f. The total frequency is f f f + f 79 It is given that median is 46., clearly, 46 lies in the class 4-5. So, 4-5 is the median class l 4, h, f 65 and C 4 + f, N 9 N C Median l + f h 9 ( 4 + f ) f f 6 f f. 5 or

48 Since, f + f 79 f 45 Hence, f 4 and f 45. Merits of Median : (i) It is rigidly defined, easily, understood and calculate. (ii) It is not all affected by extreme values. (iii) It can be located graphically, even if the class - intervals are unequal. (iv) It can be determined even by inspection is some cases. Demerits of Median : (i) In case of even numbers of observations median cannot be determined exactly. (ii) It is not based on all the observations. (iii) It is not subject to algebraic treatment. (iv) It is much affected by fluctuations of sampling. Uses of Median : (i) Median is the only average to be used while dealing with qualitative data which cannot be measured quantitatively but can be arranged in ascending or descending order of magnitude. (ii) It is used for determining the typical value in problems concerning wages, distribution of wealth etc. 4.9 MODE: Mode or modal value of the distribution is that value of variable for which the frequency is maximum. Mode of ungrouped data : - (By inspection only) Arrange the data in an array and then count the frequencies of each variate. The variate having maximum frequency is the mode. Mode of continuous frequency distribution f + f Mode l + h f f f Where l lower limit of the modal class f frequency of the class i.e. the largest frequency. f frequency of the class preceding the modal class. f frequency of the class succeeding the modal class. h width of the modal class Ex.. Fin the mode of the following data : 5, 6, 9, 48, 9,, 4, 5, 9,,, 4, 9, 6,, 6, 8,, 6, 9. Frequency table for the given data as given below : Value x i Frequency f i has the maximum frequency of 5. So, Mode 9. Ex.. The following table shows the age distribution of cases of a certain disease admitted during a year in a particular hospital. Age (in Years) No. of Cases Here class intervals are not is inclusive form. So, Converting the above frequency table in inclusive form. Age (in Years) No. of Cases Class has maximum frequency. So it is the modal class. l 4.5, h, f, f and f 4. f f Mode l + h f f f Mode Ans. 47

49 Ex.4 Find the mode of following distribution : Daily Wages No. of workers Daily Wages No. of workers Daily wages No of workers Modal class frequency is l 4.5 f f, f 5, h 6 Mode ( ) 5 Mode 46. Merits of Mode (i) It can be easily understood and is easy to calculate. (ii) It is not affected by extreme values and can be found by inspection is some cases. (iii) It can be measured even if open - end classes and can be represented graphically. Demerits of Mode : (i) It is ill - fined. It is not always possible to find a clearly defined mode. (ii) It is not based upon all the observation. (iii) It is not capable of further mathematical treatment. it is after indeterminate. (iv) It is affected to a greater extent by fluctuations of sampling. Uses of Mode : Mode is the average to be used to find the ideal size, e.g., in business forecasting, in manufacture of readymade garments, shoes etc. Relation between Mode, Median & Mean : Mode median - mean. 4. CUMULATIVE FREQUENCY CURVE OR OGIVE : In a cumulative frequency polygon or curves, the cumulative frequencies are plotted against the lower and upper limits of class intervals depending upon the manner in which the series has been cumulated. There are two methods of constructing a frequency polygon or an Ogive. (i) Less than method (ii) More than method In ungrouped frequency distribution : Ex.5 The marks obtained by 4 students in medical entrance exam are given in the following table. Marks Obtained No. of Examinees (i) Draw Ogive by less than method. (ii) Draw Ogive by more than method. (iii) Find the number of examinees, who have obtained the marks less than 65. (iv) Find the number of examinees, who have obtained 65 and more than marks. (i) Cumulative frequency table for less than Ogive method is as following. 48

50 Marks Obtained No. of Examinees Less than 45 Less than 5 75 Less than 55 5 Less than 6 87 Less than 65 4 Less than 7 8 Less than 75 5 Less than 8 4 Following are the Ogive for the above cumulative frequency table by applying the given method and the assumed scale. (ii) Cumulative frequency table for more than Ogive method is as following : - Marks Obtained No. of Examinees 4 and more 4 45 and more 7 5 and more 5 55 and more 65 6 and more 65 and more 59 7 and more 9 75 and more 47 49

51 Following are the Ogive for the above cumulative frequency table. (iii) So, the number of examinees, scoring marks less than 65 are approximately. (iv) So, the number of examinees, scoring marks 65 and more will be approximately 9. Ex.6 Draw on O-give for the following frequency distribution by less than method and also find its median from the graph. Marks Number of students Converting the frequency distribution into less than cumulative frequency distribution. According to graph median 4 marks. Marks No. of Students Less than 7 Less than 7 Less than 4 Less than 4 9 Less than 5 97 Less than 6. 5

52 OBJECTIVE DPP - 4. DAILY PRACTIVE PROBLEMS # 4. The median of following series if 5,, 4, 9, 5, 8,, 5, 8 (A) (B) 5 (C) 9 (D) 5. If the arithmetic mean of 5, 7, 9 x is9 then the value of x is (A) (B) 5 (C) 8 (D) 6. The mode of the distribution, 5, 7, 4,,, 4,, 4 is (A) 7 (B) 4 (C) (D) 4. If the first five elements of the set x, x,...x are replaced by x i + 5, i,,, 4, 5 and next five elements are replaced by x j - 5, j 6, 7,... then the mean will change by n + (A) (B) (C) (D) 5 5. If the mean and median of a set of numbers are 8.9 and 9 respectively, then the mode will be (A) 7. (B) 8. (C) 9. (D). SUBJECTIVE DPP Find the value of p, if the mean of the following distribution whose mean is x p f 4 5p 6. Find the mean of following distribution by step deviation method : - Class interval No. of workers The mean of the following frequency distribution is 6.8 and the sum of all the frequencies is 5. Compute the Class Frequency 5 f f 7 8 missing frequency f and f. 4. Calculate the median from the following data : Rent (in Rs.) No. of House Find the missing frequencies and the median for the following distribution if the mean is.46. No. of accidents 4 5 Total Frequency (No. of 46 f f 5 5 days) 6. If the median of the following frequency distribution is 8.5 find the missing frequencies : Class interval : Total Frequency 5 f 5 f The marks is science of 8 students of class X are given below : Find the mode of the marks obtained by the students in science. 5

53 Class interval : Frequency Find the mode of following distribution : Class interval Frequency During the medical check - up of 5 students of a class, their weights were recorded as follows : Weight (in kg) Number of students Less than 8 Less than 4 Less than 4 5 Less than 44 9 Less than 46 4 Less than 48 8 Less than 5 Less than 5 5 Draw a less than type ogive for the given data. Hence, obtain median weight from the graph and verify the result by using the formula.. The following table gives the height of trees : Height Less than 7 Les than 4 Less than Less than 8 Less than 5 Less than 4 Less than 49 Less than 56 No. of trees Draw less than ogive and more than ogive. 5

54 . If the mean of the following data is 8.75, find the value of p: [CBSE - 5] x 5 p 5 f Find the mean of following frequency distribution [CBSE - 6] Classes Frequency Find the median class of the following data : [CBSE - 8] Marks obtained Frequency Find the mean, mode and median of the following data : [CBSE - 8] Classes Frequency ANSWERS (Objective DPP- 4.) Que. 4 5 Ans. C B B A C (Subjective DPP 4.) ]. p... f 8, f f 76, f 8, and median 6. f 8, f kg Mean 5.6, Median 5.67 and mode