2 Rectilinear Figures Introduction If we put the sharp tip of a pencil on a sheet of paper and move from one point to the other, without lifting the pencil, then the shapes so formed are called plane curves. Some plane curves are shown below: (i) (ii) (iii) (iv) (v) (vi) (vii) The (plane) curves which have different beginning and end points are called open curves and the curves which have same beginning and end points are called closed curves. In the above figure, (i), (iv) and (vi) are open curves where as (ii), (iii), (v) and (vii) are closed curves. curve which does not cross itself at any point is called a simple curve. In the above figure, (i), (ii), (iv), (v) and (vii) are all simple curves. Note that (ii), (v) and (vii) are simple closed curves. simple closed plane curve made up entirely of line segments is called a polygon. In the above figure, (v) and (vii) are polygons. In this chapter, we shall study about different kinds of polygons (parallelograms, rectangles, rhombuses, squares, kites) and their various properties. We shall also construct some quadrilaterals and regular hexagons by using ruler and compass.
2. Rectilinear Figures plane figure made up entirely of line segments is called a rectilinear figure. Look at the following plane figures: (i) (ii) (iii) (iv) (v) (vi) ll the six figures are made up entirely of line segments, so these are all rectilinear figures. Polygon polygon is a simple closed rectilinear figure i.e. a polygon is a simple closed plane figure made up entirely of line segments. The figures (shown above) (i), (ii), (iv) and (vi) are polygons where as figures (iii) and (v) are not polygons since figure (iii) is not simple and figure (v) is not closed. Thus, every polygon is a rectilinear figure but every rectilinear figure is not a polygon. The line segments are called its sides and the points of intersection of consecutive sides are called its vertices. n angle formed by two consecutive sides of a polygon is called an interior angle or simply an angle of the polygon. polygon is named according to the number of sides it has. No. of sides 3 4 5 6 7 8 0 Name Triangle Quadrilateral Pentagon Hexagon Heptagon ctagon ecagon iagonal of a polygon. Line segment joining any two nonconsecutive vertices of a polygon is called its diagonal. onvex polygon. If all the (interior) angles of a polygon are less than 80, it is called a convex polygon. In the adjoining figure, EF is a convex polygon. In fact, it is a convex hexagon. oncave polygon. If one or more of the (interior) angles of a polygon is greater than 80 i.e. reflex, it is called concave (or re-entrant) polygon. In the adjoining figure, EFG is a concave polygon. In fact, it is a concave pentagon. However, we shall be dealing with convex polygons only. E F E onvex polygon oncave polygon Rectilinear figures 223
Exterior angle of convex polygon If we produce a side of a polygon, the angle it makes with the next side is called an exterior angle. In the adjoining figure, E is a pentagon. Its side has been produced to P, then P is an exterior angle. Note that corresponding to each interior angle, there is an exterior angle. lso, as an exterior angle and its adjacent interior angle make a line, so we have : an exterior angle + adjacent interior angle = 80. E Interior angle Exterior angle Regular polygon. polygon is called regular polygon if all its sides have equal length and all its angles have equal size. Thus, in a regular polygon: (i) all sides are equal in length (ii) all interior angles are equal in size (iii) all exterior angles are equal in size. ll regular polygons are convex. ll equilateral triangles and all squares are regular polygons. P 2.2 Quadrilaterals simple closed plane figure bounded by four line segments is called a quadrilateral. In the adjoining figure, is a quadrilateral. It has four sides,, and four (interior) angles,, and four vertices,, and two diagonals and. In quadrilateral, sides, ;, ;, ;, are pairs of adjacent sides. Sides, ;, are pairs of opposite sides. angles, ;, ;, ;, are pairs of adjacent angles. ngles, ;, are pairs of opposite angles. ngle sum property of a quadrilateral Sum of (interior) angles of a quadrilateral is 360. In an adjoining figure, is any quadrilateral. iagonal divides it inot two triangles. We know that the sum of angles of a triangle is 80, in Δ, + + 2 = 80 (i) in Δ, 4 + + 3 = 80 (ii) n adding (i) and (ii), we get + 4 + + + 2 + 3 = 360 + + + = 360 (from figure) Hence, the sum of (interior) angles of a quadrilateral is 360. 4 3 2 224 Understanding ISE mathematics Ix
Types of quadrilaterals. Trapezium quadrilateral in which one pair of opposite sides is parallel is called a trapezium (abbreviated trap.) The parallel sides are called bases of the trapezium. The line segment joining mid-points of non-parallel sides is called its median. In the adjoining quadrilateral, whereas and ase are non-parallel, so is a trapezium, and are Median E F its bases, and EF is its median where E, F are mid-points of the sides, respectively. ase Isosceles trapezium If non-parallel sides of a trapezium are equal, then it is called an isosceles trapezium. Here, and are non-parallel and =, so is an isosceles trapezium. Trapezium Isosceles trapezium 2. Parallelogram quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram. It is usually written as gm. In the adjoining quadrilateral, and, so is a parallelogram. Parallelogram 3. Rectangle If one of the angles of a parallelogram is a right angle, then it is called a rectangle. In the adjoining parallelogram, = 90, so is a rectangle. f course, the remaining angles will also be right angles. Rectangle 4. Rhombus If all the sides of a quadrilateral are equal, then it is called a rhombus. In the adjoining quadrilateral, = = =, so is a rhombus. [Every rhombus is a parallelogram, see corollary to theorem 2.3] 5. Square If two adjacent sides of a rectangle are equal, then it is called a square. lternatively, if one angle of a rhombus is a right angle, it is called a square. In the adjoining rectangle, =, so is a square. f course, the remaining sides are also equal. Square Rectilinear figures 225
6. Kite quadrilateral in which two pairs of adjacent sides are equal is called a kite (or diamond). In the adjoining quadrilateral, = and =, so is a kite. Kite Remark From the above definitions it follows that parallelograms include rectangles, squares and rhombi (plural of rhombus), therefore, any result which is true for a parallelogram is certainly true for all these figures. 2.2. Properties of parallelograms Theorem 2. diagonal of a parallelogram divides it into two congruent triangles. Given. parallelogram and diagonal divides it into two triangles Δ and Δ. To prove. Δ Δ. In Δ and Δ. =. alt. s, and is transversal. 2. = 2. lt. s, and is transversal. 3. = 3. ommon 4. Δ Δ 4. sa rule of congruency. Theorem 2.2 In a parallelogram, opposite sides are equal. Given. parallelogram. To prove. = and =. onstruction. Join. In Δ and Δ. =. lt. s, and is transversal. 2. = 2. lt. s, and is transversal. 3. = 3. ommon 4. Δ Δ 4. S rule of congruency. 5. = and = 5. c.p.c.t. 226 Understanding ISE mathematics Ix
The converse of the above theorem is also true. In fact we have: Theorem 2.3 If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram. Given. quadrilateral in which = and =. To prove. is a parallelogram. onstruction. Join. In Δ and Δ. =. Given. 2. = 2. Given. 3. = 3. ommon. 4. Δ Δ 4. SSS rule of congruency 5. = 6. = 5. c.p.c.t. alt. s are equal formed by lines, and transversal. 6. c.p.c.t. alt. s are equal formed by lines, and transversal. Hence, is a parallelogram. orollary. Every rhombus is a parallelogram. [In a rhombus, all sides are equal, so opposite sides are equal. Therefore, every rhombus is a parallelogram.] Theorem 2.4 In a parallelogram, opposite angles are equal. Given. parallelogram. To prove. = and =.. + = 80 2. + = 80 3. + = + = Similarly, =.. ad and is a transversal, sum of cointerior angles = 80 2. ab and is a transversal, sum of cointerior angles = 80 3. From and 2 The converse of the above theorem is also true. In fact, we have: Theorem 2.5 If each pair of opposite angles of a quadrilateral is equal, then it is a parallelogram. Given. quadrilateral in which = and =. Rectilinear figures 227
To prove. is a parallelogram.. =. given 2. = 2. given 3. + = + 3. dding and 2 4. + + + = 360 4. Sum of angles of a quadrilateral. 5. 2( + ) = 360 + = 80 Similarly, Hence, is a parallelogram. 5. Using 3 Sum of co-interior angles = 80, formed by lines, and transversal. Theorem 2.6 If a pair of opposite sides of a quadrilateral is equal and parallel, then it is a parallelogram. Given. quadrilateral in which and =. To prove. is a parallelogram. onstruction. Join. In Δ and Δ. =. lt. s, and is a transversal. 2. = 2. Given 3. = 3. ommon 4. Δ Δ 4. SS rule of congruency. 5. = Hence, is a parallelogram. Theorem 2.7 The diagonals of a parallelogram bisect each other. Given. parallelogram whose diagonals and intersect at. To prove. = and =. 5. c.p.c.t. lt. s are equal formed by lines, and transversal. In Δ and Δ. =. lt. s, and is a transversal. 228 Understanding ISE mathematics Ix
2. = 2. lt. s, and is a transversal. 3. = 3. pp. sides of a parallelogram are equal. 4. Δ Δ = and = 4. S rule of congruency c.p.c.t. The converse of the above theorem is also true. In fact, we have: Theorem 2.8 If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. Given. quadrilateral whose diagonals and intersect at such that = and =. To prove. is a parallelogram. In Δ and Δ. =. Given 2. = 2. Given 3. = 3. Vert. opp. s 4. Δ Δ 4. SS rule of congruency 5. = = 5. c.p.c.t. alt. s are equal formed by lines, and transversal. 6. = 6. c.p.c.t. 7. is a parallelogram. 7. In quadrilateral, and = (Theorem 2.6) Properties of a rectangle since every rectangle is a parallelogram, therefore, it has all the properties of a parallelogram. dditional properties of a rectangle are: ll the (interior) angles of a rectangle are right angles. In the adjoining figure, = = = = 90. The diagonals of a rectangle are equal. In the adjoining figure, =. Properties of a rhombus since every rhombus is a parallelogram, therefore, it has all the properties of a parallelogram. dditional properties of a rhombus are: ll the sides of a rhombus are equal. In the adjoining figure, = = =. The diagonals of a rhombus intersect at right angles. In the adjoining figure,. Rectilinear figures 229
The diagonals bisect the angles of a rhombus. In the adjoining figure, diagonal bisects as well as and diagonal bisects as well as. Properties of a square since every square is a parallelogram, therefore, it has all the properties of a parallelogram. dditional properties of a square are: ll the interior angles of a square are right angles. In the adjoining figure, = = = = 90. ll the sides of a square are equal. In the adjoining figure, = = =. The diagonals of a square are equal. In the adjoining figure, =. The diagonals of a square intersect at right angles. In the adjoining figure, ^. The diagonals bisect the angles of a square. In the adjoining figure, diagonal bisects as well as and diagonal bisects as well as. In fact, a square is a rectangle as well as a rhombus, so it has all the properties of a rectangle as well as that of a rhombus. Illustrative Examples Example. If angles,, and of a quadrilateral, taken in order, are in the ratio 3 : 7 : 6 : 4 then is a trapezium. Is this statement true? Give reason for your answer. Solution. s the angles are in the ratio 3 : 7 : 6 : 4, let these angles be 3x, 7x, 6x and 4x. Since sum of angles of a quadrilateral is 360, 3x + 7x + 6x + 4x = 360 26 08 20x = 360 x = 8. The angles are: = 3 8º = 54, 54 72 = 7 8 = 26, = 6 8 = 08 and = 4 8 = 72. We note that + = 54 + 26 = 80. Thus, the sum of co-interior angles is 80 formed by lines, and transversal, therefore,. So, is a trapezium. Hence, the given statement is true. Example 2. Three angles of a quadrilateral are equal. Is it a parallelogram? Why or why not? Solution. It need not be a parallelogram; because we may have = = = 80, then = 360 3 80 = 20, so (opposite angles are not equal). Example 3. In a quadrilateral, and are the bisectors of and respectively. Prove that = ( + ). 2 Solution. Given is a quadrilateral, and are the bisectors of and respectively. Mark the angles as shown in the figure given below. 230 Understanding ISE mathematics Ix
s and are bisectors of and respectively, = 2 and 2 = 2 (i) + + 2 = 80 (sum of angles in ) = 80 ( + 2) = 80 2 ( + ) (using (i)) 2 = 80 2 (360 ( + )) [ sum of angles in a quadrilateral is 360, so + + + = 360 + = 360 ( + )] = 2 ( + ). Example 4. iagonals of a quadrilateral bisect each other. If = 45, determine. Solution. Since the diagonals and of quadrilateral bisect each other, is a parallelogram. and is a transversal, + = 80 (sum of co-interior angle = 80 ) 45 + = 80 45 = 35. Example 5. The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60. Find the angles of the parallelogram. Solution. Let be parallelogram such that dm, N and MN = 60. In quadrilateral MN, 60 MN + M + N + = 360 60 + 90 + 90 + = 360 N = 20. and is a transversal, + = 80 M + 20 = 80 = 60. = and = (opp. s of a gm are equal) = 60 and = 20. Hence, the angles of gm are 60, 20, 60, 20. Example 6. In the adjoining figure, is parallelogram. Find the values of x, y and z. Solution. Given is a parallelogram, 50 3x = 2x + 2 (opp. sides are equal) z 02 x = 3. (3x ) cm = = 02 (opp. s are equal) For, y = 50 + ( ext. = sum of two int. opp. s) y = 50 + 02 = 52 + 02 = 80 (, sum of co-int. s = 80 ) = 80 02 = 78. From figure, z = = 78 50 = 28. Rectilinear figures (2x + 2) cm y 23
Example 7. In the adjoining figure, is a parallelogram. Find the ratio of :. ll measurements are in centimetres. Solution. Given is a parallelogram, 3x 4 = y + 5 (opp. sides are equal) 3x y 9 = 0 (i) and 2x + 5 = y (opp. sides are equal) 2x y + 6 = 0 (ii) n subtracting (ii) from (i), we get x 5 = 0 x = 5. n substituting this value of x in (i), we get 3 5 y 9 = 0 36 y = 0 y = 36. ab = 3x 4 = 3 5 4 = 4 and b = y = 36 = 35. Hence, : = 4 : 35. 2x + 5 y + 5 3x 4 y Example 8. In a rectangle, diagonals intersect at. If = 30, find (i) (ii) (iii) (iv). Solution. (i) = 90 (each angle of a rectangle = 90 ) + 30 + 90 = 80 (sum of angles in ) = 80 30 90 = 60. (ii) a = (diagonals are equal) 2 = 2 (diagonals bisect each other) 30 = = (angles opp. equal sides in ) = 30 ( = 30 given) (iii) + 30 + 30 = 80 (sum of angles in ) = 80 30 30 = 20. but = (vert. opp. s) = 20. (iv) + 20 = 80 = 80 20 = 60. (linear pair) Example 9. In the adjoining figure, is a square and E is an equilateral triangle. Find (i) E (ii) E (iii) reflex E. Solution. (i) From figure, E = 90 60 ( each angle in a square = 90 and each angle in an equilateral triangle = 60 ) E E = 30. ed = (sides of equilateral triangle) ad = (sides of square) E = E = E (angles opp. equal sides in E) but E + E + E = 80 (sum of angles in E) 2 E + 30 = 80 80 30 E = = 75. 2 232 Understanding ISE mathematics Ix
(ii) E = 90 75 = 5 ( E = E = 75 ) (iii) Reflex E = 360 75 60 = 225. Example 0. E is an equilateral triangle in the square. Find the value of x in the figure. Solution. Since is a square and is a diagonal, = 45. s E is an equilateral triangle, E = 60. In F, x + 45 + 60 = 80 x = 80 45 60 = 75. Example. In the adjoining figure, is a rhombus and E is an equilateral triangle. E and lie on opposite sides of. If = 78, calculate E and E. Solution. Since is a rhombus, = = 78. s E is an equilateral triangle, E = 60. From figure, E = + E = 78 + 60 = 38. lso = E (Since E is equilateral triangle) and ba = ( is a rhombus) E = E = E ( angles opp. equal sides in E) E = (80 38 ) = 42 = 2. 2 2 78 E E x F In, = =. = (80 78 ) = (02 ) = 5. 2 2 ut = = 5. From figure, E = E = 5 2 = 30. ( is a rhombus) (, alt. s are equal) Example 2. In parallelogram, = 0 cm and = 6 cm. The bisector of meets in E. E and produced meet at F. Find the length of F. Solution. Mark the angles as shown in the figure. s E is bisector of, = 2 Since is a parallelogram, i.e. F = 3 (i) (alt. s are equal) 2 = 3 (using (i)) In ΔF, 2 = 3 bf = b + F = 0 cm (sides opp. equal angles are equal) ad + F = 0 cm ( =, opp. sides of a gm) 6 cm + F = 0 cm F = 4 cm. 2 E 3 F Rectilinear figures 233
Example 3. is a rhombus in which altitude from to side bisects. Find the angles of the rhombus. Solution. is a rhombus in which M such that M is mid-point of. Join. In ΔM and ΔM, M = M (each = 90 ) am = M dm = M (M is mid-point of ) ΔM ΔM (SS rule of congruency) ad = lso, ad = ( is a rhombus) ad = = Δ is an equilateral triangle = 60. Now, and is a transversal, + = 80 (sum of co-interior s) 60 + = 80 = 20 = 60 and = 20. = and = Hence, the angles of the rhombus are 60, 20, 60, 20. M (opp. s in a gm are equal) Example 4. Prove that the diagonals of a rectangle are equal. Solution. Let be a rectangle. We need to prove that =. In Δ and Δ, b = (opp. sides of a rectangle are equal) ab = = (each angle of a rectangle = 90 ) Δ Δ (SS rule of congruency) = (c.p.c.t.) Example 5. If the diagonals of a parallelogram are equal, then prove that it is a rectangle. Solution. Let be a parallelogram in which =. We need to prove that = 90. In Δ and Δ, b = ab = a = (opp. sides of a gm) (given, diagonals are equal) Δ Δ (SSS rule of congruency) = (c.p.c.t.) s and is a transversal, + = 80 (sum of co-int. s) + = 80 2 = 80 = 90. is a rectangle. Example 6. Show that the diagonals of a rhombus bisect each other at right angles. Solution. Let be a rhombus and let its diagonals and meet at. 234 Understanding ISE mathematics Ix
We need to prove that =, = and = 90. s every rhombus is a gm, therefore, the diagonals bisect each other i.e. = and =. Thus, the diagonals of a rhombus bisect each other. In Δ and Δ, = (proved above) = ab = (sides of a rhombus) Δ Δ (SSS rule of congruency) = (c.p.c.t.) ut + = 80 (linear pair) + = 80 = 90. Hence, the diagonals of a rhombus bisect each other at right angles. Example 7. If the diagonals of a quadrilateral bisect each other at right angles, then prove that the quadrilateral is a rhombus. Solution. Let be a quadrilateral in which the diagonals and bisect each other at and are at right angles i.e. =, = and. We need to prove that is a rhombus. s the diagonals of the quadrilateral bisect each other, is a parallelogram. In Δ and Δ, = (from given) = = (each = 90, because given) Δ Δ (by SS rule of congruency) ab = (c.p.c.t.) Thus, is a parallelogram in which two adjacent sides are equal, therefore, is a rhombus. Example 8. Prove that the diagonals of a square are equal and bisect each other at right angles. Solution. Let is a square and let its diagonals and meet at. We need to prove that =, =, = 90 and =. s is a square, so it is a parallelogram. Therefore, its diagonals bisect each other i.e. = and =. Thus, the diagonals of a square bisect each other. In Δ and Δ, = (proved above) ab = (sides of a square) = Δ Δ (by SSS rule of congruency) = (c.p.c.t.) ut + = 80 (linear pair) + = 80 = 90 s is a square, so one of its angle is 90. Let = 90. Rectilinear figures 235
Now and is a transversal + = 80 (sum of co-int. s) 90 + = 80 = 90. In Δ and Δ, b = (sides of a square) ab = = (each = 90 ) Δ Δ a = (c.p.c.t.) Hence, the diagonals of a square are equal and bisect each other at right angles. Example 9. In the adjoining figure, is a parallelogram and P, Q are perpendiculars from the vertices, respectively on P the diagonal. Show that (i) ΔP ΔQ (ii) P = Q. Q Solution. (i) In ΔP and ΔQ, ab = (opp. sides of a gm) P = Q (each = 90 ) P = Q (alt. s, and is a transversal) ΔP ΔQ (by S rule of congruency) (ii) P = Q (c.p.c.t.) Example 20. If E and F are points on diagonal of a parallelogram such that E = F, then show that FE is a parallelogram. Solution. s is a parallelogram, its diagonals bisect each other i.e. = and =. F Given ae = F E = F E = F E Thus, in quadrilateral FE, E = F and = i.e. its diagonals EF and bisect each other, therefore, FE is a parallelogram. Example 2. is a parallelogram. If the bisectors of and meet the diagonal at P and Q respectively, prove that the quadrilateral PQ is a parallelogram. Solution. Given. is a gm, P bisects and Q bisects. To prove. P Q and P Q. onstruction. Join. P Q. P = 2. P is bisector of. 2. Q = 2 2. Q is bisector of. 3. P = Q 3. =, since is a gm. 236 Understanding ISE mathematics Ix
4. = 4. lt. s, since. 5. P = Q 5. Subtracting 4 from 3. 6. P = Q 6. From figure. 7. P Q Similarly, P Q. Hence, PQ is a parallelogram. Q.E. 7. lt. s are equal. Example 22. Show that the bisectors of the angles of a parallelogram form a rectangle. Solution. Let be a parallelogram and let P, Q, R and S be the points of intersection of the bisectors of and, and, and, and and respectively. We need to show that PQRS is a rectangle. s is a gm, and is a transversal. + = 80 (sum of co-int. s = 80 ) S R P Q 2 + = 90 2 P + P = 90 ( P is bisector of and P is bisector of ) In ΔP, P + P + P = 80 (sum of angles in a Δ) P + 90 = 80 P = 90 SPQ = 90. Similarly, PQR = 90, QRS = 90 and RSP = 90. So, PQRS is a quadrilateral in which each angle is 90. Now, SPQ = QRS (each = 90 ) and PQR = RSP (each = 90 ) Thus, PQRS is a quadrilateral in which both pairs of opposite angles are equal, therefore, PQRS is a parallelogram. lso, in this parallelogram one angle (in fact all angles) is 90. Therefore, PQRS is a rectangle. Example 23. In the adjoining figure, is a parallelogram. If = 2 and P is mid-point of, prove that P = 90. Solution. Given P is mid-point of ap = P = 2. lso = 2 = 2. ap =. In P, P = P = P (angles opp. equal sides) ut + P + P = 80 (sum of angles in a = 80 ) + P + P = 80 ( P = P) 2 P = 80 P = 80 a 2 s P = P and = P =. P (i) (opp. sides of gm ) Rectilinear figures 237
In P, P = P = P. ut + P + P = 80 + P + P = 80 2 P = 80 P = 80 b 2 dding (i) and (ii), we get P + P = 80 2 ( + ) (ii) = 80 (80 ) ( is a gm,, so + = 80 ) 2 P + P = 90 (iii) ut P + P + P = 80 ( P is a line) ( P + P) + P = 80 90 + P = 80 P = 90 (using (iii)) Example 24. In the parallelogram, M is mid-point of, and X, Y are points on and respectively such that X = Y. Prove that (i) triangle XM is congruent to triangle YM. (ii) XMY is a straight line. Y Given. is a gm, M is mid-point of, X, Y are points on, such that X = Y. M To prove. (i) XM YM (ii) XMY is a straight line. onstruction. Join XM and MY. X In s XM and YM. X = Y. Given. 2. M = M 2. M is mid-point of. 3. XM = MY 3. (i) XM YM 4. MY = MX 4. c.p.c.t. lt. s, since. SS rule of congruency. 5. XM = XM + XM 5. Ext. = sum of two int. opp. s. 6. MY + XM 6. dding 4 and 5. = MX + XM + XM 7. MY + XM = 80 (ii) XMY is a straight line Q.E.. 7. Sum of s of a = 80. Sum of adj. s = 80. Example 25. In the adjoining figure, is a kite in which = and =. Prove that: (i) is a bisector of and of. (ii) is perpendicular bisector of. Solution. (i) In and, ab = 238 Understanding ISE mathematics Ix (given)
b = = Rectilinear figures (given) (SSS rule of congruency) = and =. (c.p.c.t.) Hence, is bisector of and of. (ii) In and, b = = = (given) (proved above) (SS rule of congruency) = and = (c.p.c.t.) but + = 80 2 = 80 = 90. Hence, is perpendicular bisector of. Example 26. In the adjoining kite, diagonals intersect at. If = 25 and = 40, find (i) (ii) (iii). Solution. (i) Since the diagonal bisects, = 2 = 2 25 = 50. (ii) = 90 (diagonals intersect at right angles) + 40 + 90 = 80 = 80 40 90 = 50. since the diagonal bisects, (sum of angles in ) = 2 = 2 50 = 00. (iii) Since the diagonal bisects, = = 50. + 50 + 25 = 80 = 80 50 25 = 05. Example 27. In the adjoining figure, is an isosceles trapezium and its diagonals meet at. Prove that: (i) = and =. (ii) =. (iii) = and =. (linear pair) (sum of angles in ) Solution. (i) From and, draw perpendiculars N and M on respectively. In M and N, ad = M = N (given) ( M and N, by construction) md = N (distance between parallel lines) M N (RHS rule of congruency) = (c.p.c.t.) also + = 80 and + = 80 ) (, sum of co-int. s = 80 ) + = + M 25 40 N 239
= ( =, proved above) (ii) In and, = ad = ab = (proved above) (given) (SS rule of congruency) = (c.p.c.t.) (iii) In and, ad = (given) = (vert. opp. s) = (, so = ) (S rule of congruency) = and =. (c.p.c.t.) Example 28. In the adjoining figure, is a trapezium. If = 26 and P = Q = 52, find the values of x and y. Solution. Produce P and Q to meet at R. In R, R = R (each angle = 52 ) dr = R (sides opp. equal angles are equal) R = R Similarly, R = R. R = R ar = R. ad = R R = R R = ab is an isosceles trapezium. (corres. s, ). = (Example 27) =. 80 26 = = 27. 2 = = 52 27 = 25 x = 25. + + = 80 y + 27 + 52 = 80 y = 80 27 52 y = 0. x x P P 52 52 26 R 26 Q 52 52 Q (sum of angle in ) y y Exercise 2.. If two angles of a quadrilateral are 40 and 0 and the other two are in the ratio 3 : 4, find these angles. 2. If the angles of a quadrilateral, taken in order, are in the ratio : 2 : 3 : 4, prove that it is a trapezium. 3. If an angle of a parallelogram is two-thirds of its adjacent angle, find the angles of the parallelogram. 4. (a) In figure () given below, is a parallelogram in which = 70, = 80. alculate angles and. 240 Understanding ISE mathematics Ix