BOARD ANSWER PAPER : MARCH 2014

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Ferdinand Stevens
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1 OARD ANSWER PAPER : MARH 04 GEOMETRY. Solve any five sub-questions: i. RP : PK 3 : ----[Given] A( TRP) RP ---- [Ratio of the areas of two triangles having equal heights A( TPK) PK is equal to the ratio of their corresponding bases.] A(ΔTRP) A(ΔTPK) 3 Geometry ii. If two circles touch internally, then the distance between their centres is difference of their radii. The distance between their centres cm iii [Given] sin (60) sin [ sin( ) sin ] sin(60) 3 iv. Let A (, 3) (x, y ), (4, 7) (x, y ) y y Slope of line A x x v. Radius (r) 7 cm ----[Given] ircumference of a circle r cm vi. The longest side is 0 cm. (0) (i) Now, sum of the squares of the other two sides will be (6) + (8) (ii) (0) (6) + (8) ----[From (i) and (ii)] Yes, the given sides form a right angled triangle ----[y converse of Pythagoras theorem]. Solve any four sub-questions: i. In A, ray AD is the bisector of A. A D A D ----[y property of angle bisector of a triangle] 4.8 x x x x 6 cm

2 oard Answer Paper: March 04 ii. QPR [Given] PQM 90 PRM 90 In PQMR, QPR + PQM + QMR + PRM [Tangent is perpendicular to radius] ----[ Sum of the measures of the angles of a quadrilateral is 360] QMR QMR 360 QMR QMR 0 iii. Equation of the line in slope-intercept form is y mx + c Given equation of the line is 3x y 4 0 y 4 3x y 4 3 x y 3 The equation of the line in slope-intercept form is y 3 x. slope m 3 and y-intercept c iv. sin [Given] We know that, sin + cos 3 + cos cos cos 3 4 cos 4 cos ----[Taking square root on both sides] v. Slope of the line PQ (m) and the line passes through point (5, ) (x, y ) Equation of line in slope point form is y y m(x x ) y () (x 5) y + x 0 x y + 0 x y The equation of line PQ is x y.

3 Geometry vi. P 3.5 cm A For drawing the circle of radius 3.5 cm Extend the line passing through A Draw the perpendicular at point A [] 3. Solve any three sub-questions: i. Given: In RST, line l side ST. Line l intersects side RS and side RT in points P and Q respectively, such that RPS and RQT. R To prove : RP PS RQ QT onstruction: Draw seg SQ and seg PT. Proof: In RPQ and SPQ, where RPS, A( RPQ) RP ----(i) A( SPQ) PS In RPQ and TQP, where RQT, A( RPQ) RQ ----(ii) A( TQP) QT A(SPQ) A(TQP) ----(iii) S P [Ratio of the areas of two triangles having equal heights is equal to the ratio of their corresponding bases.] [Ratio of the areas of two triangles having equal heights is equal to the ratio of their corresponding bases.] [Areas of two triangles having equal bases and equal heights are equal.] Q l T A( RPQ) A( SPQ) A( RPQ) A( TQP) ----(iv) [From (i), (ii) and (iii)] RP PS RQ QT ----[From (i), (ii) and (iv)] 3

4 oard Answer Paper: March 04 ii. x x P R y z A y Q z Let P R x ----(i) AP AQ y ----(ii) Q R z ----(iii) [The length of two tangent segments to a circle from an external point are equal.] Perimeter of A [Given] A + A + 40 (P + PA) + (AQ + Q) + (R + R) [AP, R, AQ] x + y + y + z + z + x [From (i), (ii) and (iii)] x + y + z 40 x + y + z 0 x + A [ A y + z] x [ A 0] x 0 P R 0 cm ----[From (i)] The length of the tangent segment from to the circle is 0 cm. iii. R P 3.4 cm M 7. cm Q For drawing the circle of radius 3.4 cm Plot point Q such that PQ 7. cm Draw perpendicular bisector of seg PQ [] Draw circle with centre M Draw tangent QR 4

5 Geometry iv. L.H.S. sec A + cosec A cos A + sin A sin A cos A cos A sin A cos A sin A ----[ sin A + cos A ] [] cos A sin A sec A cosec A R.H.S. sec A + cosec A sec A cosec A v.. Let A (3, 4) (x, y ), Origin O (0, 0) (x, y ) Equation of a line in two point form is x x x x y y y y x ( 3) y x 3 y (x + 3) 3(y 4) 4x + 3y + 4x + 3y + 0 4x + 3y 0 Equation of the line is 4x + 3y 0.. Let P (3, 5) Equation of line parallel to X-axis and passing through P(3, 5) is y y-coordinate of P i.e., y 5 equation of the line is y Equation of X-axis is y 0. Equation of Y-axis is x Solve any two sub-questions: i. Let A represent the height of lighthouse. A 90 m Point represent the position of the ship. Draw ray AP seg. PA is the angle of depression, PA 60 A P m 60 [] 5

6 oard Answer Paper: March 04 Also, A PA ----[Alternate angles] A 60 In right angled A, tan 60 A m The distance of the ship from the lighthouse is 5.9 m. [] ii. Given: AD is a cyclic quadrilateral. A O D To prove: AD + D 80, A + AD 80 Proof: mad m(arc D) ----(i) [Inscribed angle theorem] m D m(arc AD) ----(ii) [Inscribed angle theorem] Adding equations (i) and (ii), we get mad + md m(arc D) + m(arc AD) mad + md [m(arc D) + m(arc AD)] AD + D [Measure of a circle is 360] AD + D (iii) In AD, D +AD +A +AD [Sum of measures of angles of a quadrilateral] 80 + A + AD [From (iii)] A + AD A + AD 80 Hence, the opposite angles of a cyclic quadrilateral are supplementary. 6

7 Geometry iii. Diagonal of a cuboid l b h ut, length of the diagonal cm l b h cm ----(i) l + b + h 38 cm ----(ii) [Given] (l + b + h) (l + b + h) (l + b + h) (l + b + h) (l + b + h ) + (lb + bh + hl) (38) () + (lb + bh + hl) ----[From (i) and (ii)] (lb + bh + hl) (lb + bh + hl) (lb + bh + hl) 960 ut, total surface area of a cuboid (lb + bh + hl) The total surface area of the cuboid 960 cm. 5. Solve any two sub-questions: i. A b D c p a In A, A [Given] seg D hypotenuse A ----[Given] A AD D ----(i) [y theorem on similarity in right angled triangle] []. A AD ----[From (i)] A AD D A A ----[c.s.s.t.] b AD a p c b ----[Substituting the given values] a p c b cp ab ----(ii). A A [y Pythagoras theorem] c b + a ----[Substituting the given values] c b a + a b a b a b ----[Dividing throughout by a b ] c + ab a b ----(iii) c + cp a b ----[From (ii) and (iii)] c + cp a b p a + b 7

8 oard Answer Paper: March 04 ii. A O 6 cm M For drawing A of side 6 cm [] Draw the perpendicular bisectors of side A and side [] Draw incircle with centre O Draw circumcircle with centre Radius of circumcircle (O) 3.4 cm Radius of incircle (OM).7 cm Radius of circumcircle O 3.4 Radius of incircle OM.7 The ratio of radii of cicumcircle and incircle is :. [] iii. Given: For the stair-step (cuboid): length (l) 50 cm, breadth (b) 5 cm height of st step (h ) cm height of nd step (h ) + 4 cm height of 3 rd step (h 3 ) cm For the brick (cuboid): length (l ).5 cm, breadth (b ) 6.5 cm, height (h ) 4 cm volumeof the3stairsteps No of bricks used volumeof brick st nd rd volumeof step volumeof step volumeof 3 step volume of brick lbhlbh lbh3 lbh l b(hh h 3) l b h 8 505( 4 36) bricks The number of bricks used are 88.

BOARD ANSWER PAPER : MARCH 2014

BOARD ANSWER PAPER : MARCH 2014 ORD NSWER ER : MRH 04 GEOMETRY. Solve any five sub-questions: i. R : K :...[Given] ( TR) R...[Ratio of the areas of two triangles having equal heights ( TK) K ( TR) ( TK) is equal to the ratio of their