GOMTRY RLLL LINS Theorems Theorem 1: If a pair of parallel lines is cut by a transversal, then corresponding angles are equal. Theorem 2: If a pair of parallel lines is cut by a transversal, then the alternate angles are equal. Theorem 3: If a pair of alternate angles formed by a transversal on two lines is equal angles, then the straight lines are parallel. Theorem 4: If a transversal cuts two parallel lines, then the allied angles are supplementary. Theorem 5: If a pair of allied angles formed by a transversal on two lines is supplementary then the lines are parallel. Theorem 6: If two lines are each parallel to a third line, then the lines are themselves parallel. onstruction 1. To draw a line through a given point and parallel to a given straight line. Given : straight line and a point. To construct: line through and parallel to. Steps: 1. Join with any point Q in. 2. t, construct angle RQ equal to angle Q. R is parallel to. 2 To draw a line parallel to a given line and at a given distance from it. Given: straight line. To construct: line parallel to and at a particular distance (say, 3 cm) from. Steps: 1. With any point in as centre and radius equal to the given distance (3 cm), draw an arc. 2. With some other point Q in as centre and with the same radius, draw one more arc on the same side of. 3. raw a line touching both the arcs drawn. is the required parallel line. xercise 1: rom each of the following diagrams, find x: (i) (ii) x 60 0 60 0 x 150 0 140 0
(iii) 50 0 60 0 xercise 2: (i) 42 0 ind x and y x y (ii) x 3x + 10 y 63 0 xercise 3: ind x: (i) 100 0 x (ii) 20 0 128 0 x (iii) 25 0 (iv) x 70 0 25 0 62 0 x xercise 4: In the adjoining figure, prove that is parallel to. 62 0 30 0 148 0 32 0 xercise 5: raw angle = 60 0. ind a point in which is at a distance of 2.5 cm from. xercise 6: onstruct angle QR = 75 0. raw the bisector of angle QR. ind a point O on this angle bisector which is at a distance of 2 cm from Q.
TRINGLS: Some important terms. Median: The Median of a triangle, corresponding to any side, is the line joining the mid-point of that side with the opposite vertex. triangle has three medians and all the three medians are always concurrent i.e. any two medians intersect each other at one point only. The point of concurrence of the medians is called centroid of the triangle. In the figure; G is the centroid of. lso, the centroid of a triangle divides each median in the ratio 2 : 1. i.e. in the given figure: G : G = 2 : 1 ; G : G = 2 : 1 G : G = 2 : 1. G ltitude: n altitude of a triangle, corresponding to any side, is the length of perpendicular drawn from opposite vertex to that side. O triangle has three altitudes and all the three altitudes are always concurrent i.e. any two altitudes intersect each other at one point only. The point of concurrence of all altitudes of a triangle is called orthocenter. In the given figure, O is the orthocenter of triangle. efinition: Incentre: The point of concurrence of the angle bisectors of a triangle is called the incentre of the triangle. In figure, I is the incentre of. I Inradius and Incircle: The distance of the incentre from any side of a triangle is called as its inradius. The circle with centre at the incentre of the triangle and radius equal to its inradius is known as the incircle of the triangle. ircumcentre: The point of concurrence of the perpendicular bisectors of a triangle is called the circumcentre of the triangle. To determine the O circumcentre.
ircumraidus and ircumcirle: The distance of the circumcentre from any vertex of a triangle is called its circumraidus. The circle with centre at the circumcentre of the triangle and radius equal to its circumradius is known as the circumcircle of the triangle. The circumcirle of a triangle passes through the three vertices of the triangle. urther, the circumcentre may lie in the interior or exterior or on the triangle. Theorem 7: The angle bisectors of a triangle pass through the same point. I Theorem 8: The perpendicular bisectors of the sides of a triangle pass through the same point. (or) The perpendicular bisectors of the sides of a triangle are concurrent. In a triangle, the three perpendicular bisectors of the sides, and meet at a point. O Theorem 9: The three altitudes of a triangle pass through the same point. or The altitude of a triangle are concurrent. H In, the three altitudes drawn from the vertices,, upon the sides,, respectively of a meet at a point. Theorem 10: The medians of a triangle pass through the same point which divides each median in the ratio 2 : 1. Given : triangle, in which the medians and intersect in G. Join G. roduce G to meet at (figure) Theorem 11: The sum of the angles of a triangle is equal to two right angles. G H Theorem 12: If the one side of a triangle is produced, the exterior angle so formed is equal to the sum of two interior opposite angles. Illustration 1: rove that the angle between the internal bisector of one base angle and the external bisector of Solution: the other is equal to one half of the vertical angle. Given whose side is produced to. The bisectors and of and respectively, meet at. To rove = ½. roof: Side of is produced to. = + ½ = ½ + ½ = ½ +... (i) gain, side of is produced to.
= + = ½ +... (ii) rom (i) and (ii), we get: ½ + = ½ + ½ [each equal to ] = ½. xercise 7: The angles of a triangles are (2x + 5) 0, (x + 20) 0 and (2x 10 0 ) 0. ind the value of x. lso find the measure of each angle of the triangle. xercise 8: In triangle, 3 = 4 = 6. alculate each angle of the triangle. xercise 9: In the figure given below, is perpendicular to ; //, = 138 0. ind the number of degree in (i) (ii) 138 0 xercise 10: Use the following figure to prove that a + b + = 4 rt. angles. a b c xercise 11: isector of angles and of a quadrilateral meet at. rove that = 1 2 [ + ] xercise 12: rove that the sum of the angles of a quadrilateral is equal to 4 right angles. xercise 13: In the figure given below, is trapezium. is parallel to and. ind the measures of the angles marked x, y and z. x 0 y 0 z 0 120 0 50 0 xercise 14: Use the diagram below to find the value of x in terms of a, b and c. a 0 x 0 b 0 c 0
xercise 15: rom the figure given below, find the value of x in terms of a, b and c. a 0 0 b0 x 0 xercise 16: In the figure given below // and // ; find b and d. d 50 0 65 0 xercise 17: rove that The medians of a triangle pass through the same point which divides each median in the ratio 2 : 1. olygons: onvex, Regular efinition: or n ( 3), let 1, 2,, n be distinct points in a plane. If the n line segments 1 2, 2 3,., n 1 are such that (i) no two line-segments intersect except at their end points, (ii) no two line-segments with a common end point are collinear, then the union of the n line-segments 1 2, 2 3,., n 1 is called a polygon and the points 1, 2,., n, are called its vertices. It is denoted by olygon 1 2 n. ach of the n line-segments forming the polygon is called its side. The angle determined by two sides meeting at a vertex is called an angle of the polygon. (i) (ii) ig: 9.47 Remark : or = 3, the polygon has the specific name triangle and similarly for n = 4, the polygon has the special name quadrilateral [ig 9.47 (i) and (ii) some other special names are entagon (n = 5) [ig. 9.47 (iii)] Hexagon (n = 6) [ig. 9.47 (iv)] efinition: polygon 1 2 n is called convex if for each side of the polygon, the line containing that side has all the other vertices on the same side of it. In igure. 9.47, (i), (iii) and (iv) are convex if each side of the polygon, whereas (ii) is non-convex. Note: In what follows, we shall be dealing with the convex polygons only and as such, a polygon will mean a convex polygon. (iii) (iv)
efinition: polygon is called a regular polygon if all its sides are equal and all its angles are equal. In the figure (iv) is a regular polygon. ngle Sum roperty of a Quadrilateral Theorem: The sum of the four angles of a quadrilateral is 360 0. Sum of interior angles of a convex polygon is (2n 4) right angles. O Take any point O in the interior of polygon and join to vertices n triangles are formed. Sum of interior angles of n triangles = 2n right angles. ut this sum includes angles at point O = 4 right angles. Sum of Interior angles of a polygon of n sides is (2n 4) right angles. Sum of exterior angles of a convex polygon = 4 right angles. xercise 18: In a quadrilateral, the angles,, and are in the ratio 1 : 2 : 3 : 4. ind the measure of each angle of the quadrilateral. xercise 19: How many sides has a regular polygon, each angle of which is of measure 108 0? xercise 20: measure of 72 ind the number of sides of a regular polygon, when each of its exterior angles has a xercise 21: The sides and of quad. are produced as shown in igure. 9.51. rove that a + b = x + y Y 0 a 0 b 0 X 0 ongruency of Triangles xiom: SS (Side ngle Side) ongruence xiom: Two triangles are congruent if two sides and the included angle of one are equal to the corresponding sides and the included angle of the other. Note: This axiom refers to two sides and the angle between them so it is known as (side angle side) or the SS ongruence xiom or the SS riterion.
Theorem 13: ngles opposite to two equal sides of a triangle are equal. Given: In, = (figure). To rove: = onstruction: We draw the bisector of which meets in. roof: In s and, = = = Hence, = (Given) (construction) (common side) (SS ong. xiom) SS (Side ngle side) ongruence Theorem 14: If two triangle have two sides and the included angle of the one equal to the corresponding sides and the included angle of the other, then the triangle are congruent. Given and in which =, = and =. To rove =. roof lace over such that falls on and falls along. Since =, so falls on. Since =, so will fall along. ut, =. will fall on. Thus, will coincide with nd, Therefore, will coincide with. coincides with. Hence,. S RITRI OR ONGRUN O TRINGLS Theorem 15: If two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle, then the two triangles are congruent. Given and in which =, = and =.
To prove. roof ase 1 Let =. In this case, =, = and =. (SS criteria) ase II If possible, let. Then, construct =. Join. Now, in and, we have =, = and =. (SS criteria) = (c.p.c.t) ut, = (given) =. This is possible only when and coincide.. orollary (S ongruence riterion) If any two angels and a non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent. Theorem 16: If two angles of a triangle are equal, then the side opposite to them are also equal. SSS (Side-side-side) ongruence Theorem 17: Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle. G GIVN and in which =, = and =. To rove =. onstruction Suppose is the longest side of. raw G and G such that G = and G =. Join G. roof In and G, we have: = (given), = G (construction) and G (construction) G (S criteria) = G, = G and = G (c.p.c.t) = G, = G and = G [ = and = ]. Now, in G, = G G = G nd, in G, = G G = G. G + G = G + G G = = [ G = ] =.
Now, in and, we have: =, = and = (SS criteria). RHS (Right ngle-hypotenuse-side) ongruence Theorem 18: Two right triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the side of the other triangle. Given Two right angled triangles and. onstruction roduce to G such that G =. Join G. G roof In and G, we have: = G (construction) = (given) = G = 90 G [SS criteria] = G and = G [c.p.c.t.]. Now, = G and = G = G = = [ G = ] Now, =, = 3 rd = 3 rd. Thus, in and, we have: =, = and =. = (SS criteria). Illustration 2: Solution: In a right angled triangle, one acute angle is double the other, rove that the hypotenuse is double the smallest side. Given in which = 90 and = 2. To prove = 2. onstruction roduce to such that =. Join. roof In and, We have: = (by construction) = (common) = = 90 (SS criteria) = and = = x (say) [c.p.c.t.]. = 2x = =. Now, in, we have: = = 2x. = = = 2 [ = and = 2]. Hence, = 2. x 2x
xercise 22: In the figure explain how one can find the breadth of the river without crossing it O M N xercise 23: In. and are, respectively, the bisectors of and. rove that. xercise 24: In the figure = and =. rove that =. xercise 25: is an isosceles triangle with =. Side is produced to such that =. rove that is a right angle. xercise 26: In igure 10.25, =, = and =. rove that triangles and are congruent, and hence =. Inequality Relation in a triangle Theorem 19: If two sides of a triangle are unequal, the longer side has greater angle opposite to it. Statement Reason 1. = onstruction 2. = Theorem 4 1 2
3. xterior > ny exterior angle is greater than each of its interior opposite angles. 4. > Statements (2), (3) 5. > s is a part of 6. > s = Thus >. Theorem 20: In a triangle, the greater angle has the longer side opposite to it. Given: In, > To prove: > roof: (y ontradiction ) Statement Reason 1. If >, then either (a) = or (b) < The sign > means is greater than and < means is less than. 2. If (a) is true, then is an isosceles triangle. Two sides are equal. 3. Thus = ngles opposite to equal sides 4. This is not true Given that <. 5. If (b) is true, then > Theorem 5 6. This is not true. Given < 7. Hence neither (a) not (b) is true Statements (4) & (7) Thus, by contradiction, >. Theorem 21: The sum of any two sides of a triangle is greater than the third side. ig. 10.28 Theorem 22: Of all the line segments that can be drawn to a given line, from a point not lying on it, the perpendicular line-segment is the shortest. M orollary: The difference of the lengths of any two sides of a triangle is always less then the length of the third side.
Illustration 3: Solution: In the quadrilateral, is th shortest side and is the longest side, Show >. We join and, In, > as is the longest side of the quadrilateral. Thus >..(i) Similarly, from, >..(ii) rom (i) and (ii), we get + > + i.e., > i.e., > xercise 27: Show that the sum of the three altitudes of a triangle is less than the sum of the three sides of the triangle. xercise 28: In figure, QR is a triangle and S is any point in its interior. Show that SQ + SR < Q + R (Hint : roduce QS to meet R in T.) T S Q R Q S R xercise 29: In QR, S is any point on the side QR. Show that Q + QR + R > 2 S. xercise 30: In figure., QRS is a quadrilateral in which diagonals R and QS intersect in O. Show that (i) Q + QR + RS + S > R + QS (ii) Q + QR + RS + S < 2 (R + QS) R S O Q onstruction of triangles onstruction of triangles when:- 1. Length of three sides are given 2. Length of two sides and the included angle are given 3. Two angles and length of included side are given 4. In a right angled triangle length of hypothesis and one side are given 5. onstruction of isosceles triangle when altitude and vertex angle are given 6. onstruction of equilateral triangle when its altitude is given Special constructions: onstruction 7: To construct a triangle when lengths of its three medians are given. Let the lengths of three medians be 6.0 cm, 7.5 cm and 9.0 cm. Steps:
1.onstruct a triangle with = 2 of 6.0 cm = 4.0 cm 3 = 2 of 7.5 cm = 5.0 cm and 3 = 2 of 9.0 cm = 6 cm 3 2. roduce upto such that : = = 6.0 cm 3. ind, the mid point of. 4. Join and produce upto such that =. 5. Join and. is the required triangle. lternative Method: Let in triangle, median = 6.0 cm, median = 7.5 cm and and medium = 9.0 cm Steps: 1.onstruct triangle such that = 6.0 cm = = 7.5 cm and = = 9.0 cm 2. raw medians M and N of triangle and let these medians intersect at point G. 3. rom M produced, cut M = MG. 4.Join and and produce it upto point so that =. 5.Join and. ` G N M is the required triangle onstruction 8: To construct a triangle when the lengths of its two sides and the medium between them are given. Let the lengths of two sides be 4.5 cm and 5.4 cm; and the length of median between them be 3.8 cm. Steps: 1. raw = 2 x medium = 2 x 3.8 cm = 7.6 cm 2. raw two arcs = 4.5 cm and = 5.4 cm, intersecting each other at. 3. raw two more arc = 5.4 cm and = 4.5 cm, intersecting each other at. 4. Join, and. is the required triangle. 4.5 cm 5.4 cm 5.4 cm 4.5 cm
onstruction 9: To construct a triangle when the sum of lengths of two sides, the remaining side and the angle between them are given. i.e. the sum of lengths of two sides, the base and one base angle are given. Let the sum of two sides be 10.0 cm, the base be 6.0 cm and one base angle to be 60 o. Steps: 1. raw = 6.0 cm 2. raw, making angle 60 o with. 3. rom, cut = 10.0cm 4. Join and draw its perpendicular bisector which cuts at. 5. Join and, is the required triangle. 60 6 cm onstruction 10: To construct a triangle when difference between the lengths of its two sides, the third side and angle between them are given. i.e., the difference the two sides, the base one base. Let the difference of two sides = 2.5 cm. the base 6.7 cm and one base angle = 45 o Steps: 1. raw = 6.7 cm 2. raw making angle 45 o with, 3. rom, cut = 2.5 cm. 4. Join and. 5. raw perpendicular bisector of which cuts at. 6. Joint and. is the required triangle. 45 6.7cm onstruction 11: To construct a triangle when its perimeter (i.e. the sum of lengths of its three sides) and both the base angles are given. Let its perimeter = 10.0 cm and both the base angles be 45 o and 60 o. Steps: 1. raw = 10.0cm 2. raw and Q such that = 45 0 and Q = 60 o. 3. raw and, the bisectors of angles and Q respectively, intersecting each other at. 4. raw perpendicular bisectors of and, interesting Q at points and respectively. 5. Join and. is the required triangle. xercise 31: onstruct a triangle whose two sides are 4.2 cm and 4.4 cm, and the median between them is 3.6 cm xercise 32: onstruct a triangle whose base is 5.0 cm, median bisecting the base is 5.0 cm, median bisecting the base is 4.6 cm and height of the triangle is 3.2 cm. xercise 33: onstruct a triangle is, when (i) = 4.8 cm, + = 10.6 cm and = 45 o (ii) = 6.4 cm, + = 10.3 cm and = 75 o (iii) = 6.0 cm, = 1.5 cm and = 45 o (iv) = 6.0 cm, = 1.2 cm and = 60 o
xercise 34: onstruct a triangle such that + + = 10.6 cm, = 60 o and = 90 o. xercise 35: onstruct a triangle whose perimeter is 12cm and the ratio between the length of its sides = 2: 3 : 4. xercise 36: onstruct a right angle triangle in which sum of lengths of hypotenuse and side is 9.0 cm and base is 4.5 cm. Measure the length of. xercise 37: Using ruler and compasses only, construct an isosceles triangle of height 2 cm and perimeter 8 cm. Measure the base of the triangle. Quadrilaterals close plane figure bounded by four straight lines is called a quadrilateral. 1. Trapezium: trapezium is a quadrilateral in which one pair of opposite sides is parallel but the pair of opposite sides is non parallel. 2. arallelogram: parallelogram is a quadrilateral in which opposite sides are parallel. 3. Rectangle: rectangle is a parallelogram in which one angle is 90 o roperties of a Rectangle: i) iagonals are equal, i.e. = ; ii) iagonals bisect each other, i.e. O = O and O = O; iii) ach angle is a right angle, i.e. = = = = 90 o. 4. Rhombus rhombus is a parallelogram which adjacent sides are equal. roperties of Rhombus: i) ll the sides are equal, i.e. = = =. ii) iagonals bisect each other at 90 o. 5. Square: Square is a parallelogram in which one angel is 90 and adjacent sides are equal roperties of a square i) ll the sides are equal; ii) ach angel is 90 o iii) iagonals are equal; iv) iagonals bisect each other at right angle. v) ach diagonal bisects angles at the vertices. arallelograms: Theorems on parallelograms: In a parallelogram, The opposite sides are equal In a parallelogram, The opposite angles are equal. The diagonals of a parallelogram bisect each other. Theorem 23: If a pair of opposite sides of a quadrilateral is equal and parallel, it is a parallelogram. Given: quadrilateral in which = and //.
To rove: is a parallelogram. onstruction: Join and. roof: Statement Reason In triangles and : 1. 1 = 2 2. = 3. = = 3 = 4 ut 3 and 4 are alternate angles // is a parallelogram Hence roved Theorem 24: The opposite sides and angels of a parallelogram are equal. Given : parallelogram To rove: = ; = ; = and =. onstruction: Join and. roof : Statement Reason In triangles and :1 1. 1 = 2 2. 3 = 4 3. = = =, = and = lso = lternate angles, since cuts parallel sides and Given ommon S..S. orresponding parts of congruent s are congruent. If alternate angels are equal, line are parallel. Opposite sides are parallel lternate angles, since cuts parallel sides and lternate angels ; cuts parallel sides and. ommon.s.. orresponding parts of congruent triangles are congruent. dding 1 and 2. Theorem 25: ach diagonal of a parallelogram bisects the parallelogram. Theorem 26: The diagonals of a parallelogram bisect each other. Theorems on arallel Line and Triangle 1. In a triangle, the line segment joining the mid points of any two sides of a triangle is parallel to the third side and is half of it. 2. The line drawn through the mid point of one side of a triangle paralleled to another bisects the third side. 3. If there are Three or more paralleled lines and the intercepts made them on are transversal are equal, then the corresponding intercepts on any other transversal are also equal.
xercise 38: rove the following 1. If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram. 2. If the opposite sides of a quadrilateral are equal, the quadrilateral is a parallelogram. 3. If the opposite angles of a quadrilateral are equal, the quadrilateral is a parallelogram. 4. If the diagonals of a parallelogram are equal, the parallelogram is a rectangle. 5. If the diagonals of a parallelogram are equal and intersect each other at right angle, the parallelogram is a square. 6. If the diagonals of a rectangle intersect each other at right angle, the rectangle is a square. 7. diagonal of a square makes an angle of 45 0 with the sides of the square. 8. diagonal of a rhombus bisects the angles at vertices. 9. The diagonals of a rhombus intersect each other at right angle. 10. If the diagonals of a parallelogram intersect each other at right angle, the parallelogram is a rhombus. 11. If the diagonals of a rhombus are equal, the rhombus js square. xercise 39: transversal cuts two parallel lines at and. The two interior angels at are bisected and so are the two interior angles at ; the four bisectors form a quadrilateral. rove that (i) is a rectangle (ii) is parallel to the original parallel line. 1. roved that the bisectors of opposite angles of a parallelogram are parallel. 2. rove that the bisectors of interior angles of a parallelogram form a rectangle. 3. rove that the bisector of the interior angels of a rectangle form a square. 4. In parallelogram, the bisector of angel meets in and = 2. rove that: a. bisects angle b. ngle = 90 0 xercise: 40: In a parallelogram, the bisector of also bisects the side at L. Show that = 2. xercise: 41: is a parallelogram, and Q are the points on the diagonal such that = Q, rove that the Q is parallelogram. xercise: 42: In figure, point H is the mid-point of and Q. Q K is the mid-point of and R. rove that QR =. [Hint: Join H and K.] H K R xercise: 43: In, is the median through and is the mid-point of. is produced to meet in. rove that = 1 3. [Hint: raw G. onsider G and get = G. Similarly, by considering, get G = G.] G
xercise: 44: oint, are the same side of a line l. perpendicular l and perpendicular l, meet l in and respectively. is the mid-point of. (ig. 16.72) rove that =. [Hint: raw M perpendicular l. Now M and is a transversal such that =. M l M = M Show that M M =. ] xercise: 45: In the figure 16.73, = and and is the bisector of exterior of. rove that (i) = (ii) is a gm. 1 2 onstruction 19: xercise: 46: In figure. 16.74, is parallelogram in which is the mid- point of and Q is a point on such that Q = 1. If 4 Q produced meets in R, rove that R is the mid-point of. O Q R xercise 47: is a parallelogram. If and Q are the points on opposite sides and respectively such that 1 3 Q and Q = 1 3, prove that Q is a parallelogram. xercise 48: rove that the straight line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides and is equal to half their difference. M N Theorems on reas rea of a parallelogram Theorem 27: The diagonal of a parallelogram divide it into two triangles of equal area. Theorem 28: arallelograms on the same box and between the same parallels are equal in area. The area of a parallelogram is the product of its base and the corresponding altitude.
xercise 49: In parallelogram, = 10 cm. The altitudes corresponding to side and are N 7 cm respectively 7 cm and 8 cm (ig. 13.10). ind 8 cm. M 10 cm xercise 50: Show that the line segment joining the mid points of pair of opposite sides of a parallelogram divides it into two equal parallelograms. rea of a Triangle Theorem 29: Triangles on the same base and between the same parallels are equal in area. xercise 51: rea of a triangle is half the product of any of its sides and the corresponding altitude. 1. The diagonals of a quadrilateral bisect each other at right angles. Show that the quadrilateral is a rhombus. (Hint: If be the quadrilateral and its diagonals and bisect each other at right angles, then lies on the perpendicular bisector of and hence =. Similarly, = ) 2. If a triangle and a parallelogram are on the same base and between the same parallel, the area of the triangle is equal to half that of the parallelogram. 3. The area of a trapezium is half the product of its height and the sum of the parallel sides. xercise 52: Show that a median of a triangle divides it into two triangles of equal area. L \ xercise 53: If,, are the mid points of the sides, and, respectively of. Show also that R ( ) = 1 R ( ). 4 \ xercise 54: XY is a line parallel to side of. and meet XY in and, respectively. Show that area ( ) = area ( ). xercise 55:,, G, H are, respectively, the mid points of the sides,,, and of gm. Show that the quad. GH is a parallelogram and that its area is half the area of the parallelogram. xercise 56: Given two points and, and positive real number x. ind the locus of a point, such that area ( ) = x.
Loci and oncurrency Theorems Locus is a set of points. The locus of a point is the path traced by the point under certain given conditions. OR The locus is the set of all points, which satisfy some given conditions. To prove Theorems on locus we have to show that 1. very point satisfying the given condition is a point of the locus and 2. very point of the locus must satisfy the conditions. Theorem 30: The locus of a point, equidistant from two given points, is the perpendicular bisector of the line segment joining the two points. 4. l is the perpendicular bisector of line segment Q and R is a point on the same side of l as. The line segment QR intersects l at X (ig. 12.10). rove that X + XR = QR. X R l Theorem 31: The locus of a point equidistant from two intersecting lines is the part of lines bisecting the angles formed by the given lines. Q xercise 57: In, the bisector X of intersects at X. XL and XM are drawn (ig. 12.5). Is XL = XM? Why or why not? L M X xercise 58: Given, determine the locus of a point which lies in the interior of and is equidistant from the two lines and. Similarity igures which are of the same shape, irrespective of their size are called similar figures. ondition for similarity 1. The corresponding angles are equal to each other. 2. The corresponding sides are in the same ratio. Theorems on Similar Triangles Theorem 32: Two triangles are similar if they have a pair of corresponding angles congruent and the sides including them proportional. Theorem 33: Two triangles are similar if they have two pairs of corresponding angles congruent.
5 cm Theorem 34: Two triangles are similar if their corresponding sides are proportional. Theorem 35: line drawn parallel to one side of a triangle divides that other two sides proportionally. Theorem 36: perpendicular drawn from the vertex of the right angled triangle to its hypotenuse divides the triangle into two triangles similar to each other and also to the given triangle. Theorem 37: The ratio of the areas of similar triangles is equal to the ratio of the squares on their corresponding sides. xercise 59: In the following figure prove that QR OQ and hence calculate R if Q = 5 cm., QO = 6 cm and QR = 8 cm. Q 6 cm 8 cm R xercise 60: In figure = 90 o and. rove that 2 = X. xercise 61: In triangle, segment is drawn perpendicular to side. If = prove that is a right angle triangle. xercise 62: rove that the ratio of corresponding altitudes of two similar triangles is equal to the ratio of their corresponding sides. xercise 63: rove that the ratio of the areas of two similar triangles is the same as the ratio of the squares of their corresponding medians. xercise 64: is a rhombus and RS is a straight line such that R = = S. rove that R and S when produced meet at right angle.
xercise 65: In the figure given below, is a parallelogram in which bisects and bisects. Show =. xercise 66: In the figure given below, is a parallelogram and bisects. Show that = G = +. G xercise 67: is a square of side 6cm. is a point on such that the area of is one third the area of the square. ind the length of. onstruction of Quadrilaterals onstruction of Quadrilaterals given 1. our sides and one diagonal 2. our sides and one angle 3. Three sides and two consecutive angles 4. Three sides and two diagonals. onstruction of arallelogram 1. Given two sides and an angle 2. Given two sides and a diagonal 3. Given one side and both diagonals 4. Given both diagonals and an angle included between them 5. Given two sides and the height onstruction of Trapeziums 1. Given four sides onstruction of Rectangle 1. Given two sides 2. Given one side and diagonal
onstruction of rhombous 1. Given two diagonals 2. Given one side and one angle 3. Given one side and one diagonal onstruction of Squares given 1. Given a diagonal 2. Given difference between a diagonal and a side. onstruction of Regular Hexagon 1. Given a side xercise 68: onstruct a trapezium QRS where Q = 4cm, QR = 2.1cm, RS = 2.2cm, S = 1.6cm and Q SR. xercise 69: onstruct a rhombus KLMN in which KL = 3.2cm and KLM = 112 o. ind a point X such that KXL = 90 o and XM = XN. xercise 70: onstruct a rhombus whose one side is 2.5cm and height is 1.3cm. xercise 71: onstruct a parallelogram where = 5cm, = 120 o and the between and = 3.2cm. xercise 72: onstant a quadrilateral with the following measurements. lso construct a triangle equal in area to the given quadrilateral: (i) = 5 cm, = 1250, = 4.4 cm, = 3 cm and = 900. (ii) = 3.8 cm, = 6.2 cm, = 900, = 8 cm and. xercise 73: raw the parallelogram with the measurements given below. onstant a triangle equal in area to the given parallelogram. (i) = 7 cm, = 5 cm and = 600. (ii) = 8 cm, = 4 cm and = 500. (iii) = 5 cm, = 3.9 cm and = 900. xercise 74: raw a rectangle with the measurements given below. onstruct a triangle equal in area to the given rectangle. (i) = 6 cm and = 5 cm. (ii) QR = 7 cm and Q = 5.6 cm.
SSIGNMNTS SUJTIV LVL I 1. The side and of a square are produced to and Q respectively. such that = Q. rove that, Q and are perpendicular to each other. [Hint: Q. = Q]. 2. In the interior (figure) of a square QRS, there is a point O. OMT is also a square. rove that QO = SM. [Hint: QO MS.] M O O Q T S R. 3. In the ig. is square. Sides SR and RQ are produced to T and M respectively, such that RT = QM. rove that SMR = TS and MNT = 90 0. [Hint: TS = SMR.] M Q N O S R T 4. In, > and and are angle bisectors of and respectively. rove that >. 5. In the given igure T is a point on side QR of QR and S is a point such that TR = TS. rove that Q + R > QS. S T Q R 6. In the figure, is a parallelogram. is the mid point of, and, when produced meet at, rove that = 2. 7. In an isosceles triangle ( = ), a line through and parallel to and another line which is the angle bisector of the exterior angle meet at. rove that is a parallelogram gm.
n 8. In a parallelogram QRS, S =6 cm, MS = 2 cm, NR = 4 cm. rove that MSNQ is a parallelogram. M N Q S R 9. In the, = and G = G and, If = 6 cm, find, giving reasons. G 10. and Q are the points of trisection of the diagonal of the parallelogram gm. rove that Q and bisects Q.
LVL II 1. is isosceles with =. is the mid point of. Show that the circumcentre, the in centre, the orthocenter and the centroid all lie on the line.. 2. H is the orthocenter of a, X, Y and Z are respectively the mid points of H, H and H. Show that, H is also the orthocenter of XYZ. 3. Show that the sum of two medians of a triangle is greater than the third median. 4.,, G, H are respectively, the mid points of the sides,, and of a quad.. Show that the quad. GH is a parallelogram and its area is half the area of quad.. 5. In a, is the mid point of. is any point on. Q meets at Q. Show that ( Q) = 1 2 ( ).
OJTIVS LVL I 1. rom the adjoining figure the value of x is () 60 0 () 75 0 () 90 0 () 120 0 25 0 y 0 x 0 35 0 60 0 2. In the figure, is the bisector of of then () > () = () < () none 1 2 4 3 3. In the given figure, is a trapezium in which = 7 cm, = = 5 cm, = x cm and the distance between and is 4 cm. Then the value of x is () 13 cm () 16 cm () 19 cm () cannot be determined L 5 cm x cm M 4 cm 5 cm 7 cm 4. In parallelogram, = 12 cm. The altitude corresponding to the sides and are respectively 9 cm and 11 cm. ind. () 108 11 () 99 10 cm () 108 10 cm cm () 108 17 cm N M 9 cm 12 cm 11 cm 5. If the bases of two similar triangles are in the ratio of 2 : 3 then their areas are in the ratio of () 2 : 3 () 3 : 2 () 4 : 9 () 9 : 4 6. The interior angles of a pentagon are a 0, (a + 20) 0, (a + 40) 0, (a + 60) 0 and (a + 80) 0. The smallest angle of the pentagon is () 58 0 () 68 0 () 78 0 () 88 0 7. In the adjoining figure, and are angle bisectors of and which meet at of the quadrilateral abcd. Then 2 = () + () + () + () + 1 2 3 4 c
8. In the given figure the value of x 0 is () 60 0 () 70 0 () 80 0 () 90 0 30 0 y 0 x 0 40 0 20 0 9. In the adjoining figure O, O are angle bisectors of external angles of. Then O is () 90 + 1 2 () 900 + 1 2 () 180 0-1 2 ()1800 1 2 1 4 2 3 O Q 10. The locus of points equidistant from the three sides of triangle is () circumcircle () incircle () excircle () none of these LVL II 1. The quadrilateral formed by joining the midpoints of consecutive sides of a rectangle is a () rhombus () square () rectangle () none of these 2. In a quadrilateral, the sides and diagonals are related as () + + + < + () + + + > + () + + + = + () none of these 3. The orthocenter of right angled triangle lies () inside the triangle () on the vertex whose angle is 90 0 () outside the triangle () none of these 4. In a, = 70 0 and = 60 0. The bisectors of angles and meet in I. Then I = () 130 0 () 90 0 () 115 0 () none of these 5. In, is the median through and is the midpoint of. produced meets in. Then = () 1 2 () 1 3 () 1 () none of these 3 6. In which of the following figures, is one diagonal the perpendicular bisector of the other diagonal? () arallelogram () Rectangle () Rhombus () none of these
7. rom the given figure here, the value of x = () 18 0 () 120 0 () 36 0 () none of these O 5x 2x Q 8. ach of the base angles of a triangle is half its vertical angle. The triangle () isosceles () right-angled () isosceles right-angled () none of these 9. is a tetrahedron in which = = = 12 metres. = = = 8 metres. is perpendicular from on the plane. Then is equal to () 5 3 m () 3 5 m () 4 m () none of these 10. is a regular pentagon. The bisector of of the pentagon meets the side in M. M is a/an () acute angle () right angle () obtuse angle () none of these NSWRS SUJTIV LVL I 9. 2 cm OJTIV LVL I 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. LVL II 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.