9 Exercise 9.1 Question 1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. Solution Given, the ratio of the angles of quadrilateral are 3 : 5 : 9 : 13. Let the angles of the quadrilateral are 3x, 5x, 9x and 13x. We know that, sum of angles of a quadrilateral = 360 3x + 5x + 9x + 13x = 360 30x = 360 x = ngles of the quadrilateral are 3x = 3 1 = 36 5x = 5 1 = 60 9x = 9 1 = 108 and 13x = 13 1 = 156 360 = 1 30 Question. If the diagonals of a parallelogram are equal, then show that it is a rectangle. Solution Let given parallelogram is whose diagonals and are equal. ie.., =. Now, we have to prove that is a rectangle. Proof. In and, we have = (Opposite sides of parallelogram) = (ommon in both triangles) and = (y SSS rule) = (orresponding Part of ongruent Triangle) ut and transversal intersect them. + = 180 (Qoth are interior angles on the same side of the transversal) + = 180 [From Eq. (i)] = 180 = 90 = Thus, is a parallelogram and one of angles is 90. Hence, is a rectangle.
Question 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. Solution Given, a quadrilateral whose diagonals and bisect each other at right angles. O ie.., O = O and O = O and O = O = O = O = 90 To prove, is a rhombus. Proof. In O and O, we have O = O and O = O O = O (Vertically opposite angles) O O (y SS rule) = (orresponding part of congruent triangles) gain, in O and O, we have O = O and O = O and O = O (Vertically opposite angle) O O (y SS rule) = (ii) (orresponding part of congruent triangles) Similarly, we can prove that = = (iii) Hence, from Eqs. (i), (ii) and (iii), we get = = = Hence, is a rhombus. Question 4. Show that the diagonals of a square are equal and bisect each other at right angles. Solution Given square whose diagonals and intersect at O. O To prove iagonals are equal and bisect each other at right angles. ie.., =, O = O, O = O and
Proof In and, we have = (ommon) = (Sides of a square) = = 90 (y SS rule) Hence, = (orresponding Parts of ongruent Triangle) In O and O = (Side of square) O = O (Q and transversal intersect) and O = ( Q and transversal intersect) O O O = O and O = O (orresponding Parts of ongruent Triangle) Now, in O and O, we have O = O (Prove in above) = (Sides of a square) O = O (ommon) O O (y SSS) O = O (y PT) ut O + O = 180 (Linear pair) O = O = 90 Thus, O ie..,. Hence, =, O = O, O = O and Question 5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. Solution Given quadrilateral in which = and such that O = O and O = O. So, is a parallelogram. O To prove is a square. Proof Let and intersect at a point O. In O and O, we have =
O O = O = O (ommon) O = O = 90 O O (y SS) = (y PT) lso, = and = (Opposite sides of parallelogram) = = = gain, in and, we have = (ommon) = = [From Eq. (i)] (y SSS) = (ii) ut + = 180 (Sum of interior angles of a parallelogram) = = 90 [From Eq. (ii)] Thus, = = = and = 90 is a square. Question 6. iagonal of a parallelogram bisects (see figure). Show that (i) it bisects also, (ii) is a rhombus. Solution Given, diagonal of a parallelogram bisects. 1 ie.., = = Here, and is transversal. O = (Pair of alternate angle) (ii) and = (Pair of alternate angle) (iii)
From Eqs. (i), (ii) and (iii), we get = = = Now, = + iagonal also bisects. = + = gain, in O and O, we have O O = O (Qiagonals bisect each other) = O (ommon) O = O = 90 O O (y SS) = (y PT) Now, = and = = = = Hence, is a rhombus. 13 (Opposite sides of parallelogram) Question 7. is a rhombus. Show that diagonal bisects as well as and diagonal bisects as well as. Solution Given is a rhombus. = = = To prove (i) iagonal bisect as well as. (ii) iagonal bisects as well as. Proof (i) Let and are the diagonals of rhombus. In and, = = [From Eq. (i)] and = (ommon) (y SSS rule) = (y PT) and = lso, = and = This shows that bisect as well as. O
(ii) gain, in and, = = = and = lso, = and = This shows that bisect as well as. = (ommon) (SSS rule) Question 8. is a rectangle in which diagonal bisects as well as. Show that (i) is a square (ii) diagonal bisects as well as. Solution Given is a rectangle. O = and = To Prove (i) is a square. ie.., = = = (ii) ioagonal bisects as well as. Proof (i) In and, we have Since, and transversal intersect = = and = (ommon) (y S rule) = (y PT) and = (ii) Hence, from Eqs. (i) and (ii), we get is a square. = = =
(ii) In O and O, we have = (Side of square) O O = O (ommon) = O O O O = O This shows that O or bisect. Similarly, in O and O, we have O and O = O O O O = O This shows that O or bisect. (Qiagonal of square bisect each other) (y SSS rule) = (Side of square) = O (ommon) (Qiagonal of square bisect each other) (y SSS rule) Question 9. In parallelogram, two points P and Q are taken on diagonal such that P = Q (see figure). Show that P Q (i) P Q (ii) P = Q (iii) Q P (iv) Q = P (v) PQ is a parallelogram. Solution Given, is a parallelogram and P and Q are lie on such that P Q P = Q
= (i) We have to show, P Q Now, in P and Q, we have P = Q = (Opposite sides are equal in parallelogram) Q and is a transversal. P = Q (lternate interior angle) P Q (y SS) (ii) Since, P Q P = Q (iii) Here, we have to show, Q P Now, in Q and P, we have Q = P = (Opposite sides of parallelogram) Q and is a transversal. Q = P (lternate interior angle) Q P (iv) Since, Q P Q = P (v) Now, in PQ and PQ, we have Q = P [From part (iv)] P = Q [From part (ii)] PQ = QP (ommon) PQ PQ (y SSS) PQ = PQ and QP = PQ (Vertically opposite) Now, these equal angles form a pair of alternate angle when line segment P and Q are intersected by a transversal PQ. P Q and Q P Now, both pairs of opposite sides of quadrilateral PQ are parallel. Hence, PQ is a parallelogram. Question 10. is a parallelogram and P and Q are perpendiculars from vertices and on diagonal (see figure). Show that P (i) P Q (ii) P = Q. Q
Solution Given, is a parallelogram and P and Q are perpendicular from vertices and on diagonal. Q and is a transversal. = (i) Now, in P and Q, we have = (Sides of parallelogram) Q = P = 90 Q = P [From Eq. (i)] P Q (y S rule) (ii) Q P Q (y PT) P = Q Question 11. In and EF, = E, E, = EF and EF. Vertices, and are joined to vertices, E and F, respectively (see figure). Show that (i) quadrilateral E is a parallelogram (ii) quadrilateral EF is a parallelogram (iii) F and = F (iv) quadrilateral F is a parallelogram (v) = F (vi) EF Solution Given, in and EF, = E, E and = EF, EF (i) Now, in quadrilateral E, E is a parallelogram. (ii) In quadrilateral EF, EF is a parallelogram. E = E and E (Q pair of opposite sides is equal and parallel) = EF and EF F (Q pair of opposite sides is equal and parallel)
(iii) Since, E is a parallelogram. E and = E lso, EF in a parallelogram. F E and F = E (ii) From Eqs. (i) and (ii), we get F and = F (iv) In quadrilateral F, we have F is a parallelogram. (v) Since, F is a parallelogram. F and = F [From part (iii)] = F and F (vi) Now, in and EF, = E = EF and = F [From part (v)] EF (y SSS rule) Question 1. is a trapezium in which and = (see figure). Show that (i) = (ii) = (iii) (iv) diagonal = diagonal E [Hint Extend and draw a line through parallel to intersecting produced at E]. Solution Given, is a trapezium. E and = Now, extend and draw a line through parallel to intersecting produced at E. Now, E is a parallelogram. =
E and = E ut = = = E (i) We know that, + E = 180 (QInterior angles on the same side of the transversal E) E = 180 Since, = E E = E = 180 lso, = 180 E (QE is straight line) = 180 180 + = (ii) Now, + = 180 (QInterior angles on the same side of the transversal ) = 180 = 180 [From Eq. (i)] (ii) lso, + = 180 (QInterior angles on the same side of the transversal ) = 180 (iii) From Eqs. (ii) and (iii), we get = (iii) Now, in and, we have (iv) Since, = = (ommon) = = [From Eq. (i)] (y SS)
9 Exercise 9. Question 1. is a quadrilateral in which P, Q, R and S are mid-points of the sides,, and (see figure). is a diagonal. Show that (i) SR and SR = 1 (ii) PQ = SR (iii) PQRS is a parallelogram. Solution Given, PQR,, and S are mid-points of the sides. P = P, Q = Q (i) In, we have R = R and S = S S is mid-point of and R is mid-point of the. We know that, the line segment joining the mid-points of two sides of a triangle is parallel to the third side. SR lso, SR = 1 S P R Q (ii) (ii) Similarly, in, we have and PQ = 1 Now, from Eqs. (ii) and (iv), we get PQ (iii) (iv) SR = PQ = 1 (v) (iii) Now, from Eqs. (i) and (iii), we get PQ SR and from Eq. (v), PQ = SR Since, a pair of opposite sides of a quadrilateral PQRS is equal and parallel. So, PQRS is a parallelogram.
Question. is a rhombus and P, Q, R and S are the mid-points of the sides,, and, respectively. Show that the quadrilateral PQRS is a rectangle. Solution Given, is a rhombus and PQR,, and S are mid-points of,, and. S F O R E Q P y mid-point theorem, In, SR and SR = 1 In, PQ and PQ = 1 (ii) From Eqs. (i) and (ii), we get PQ SR and PQ = SR = 1 PQRS is a parallelogram. Now, we know that diagonals of a rhombus bisect each other at right angles. EOF = 90 Now, RQ (y mid-point theorem) RE OF lso, SR [From Eq. (i)] FR OE OERF is a parallelogram. So, ERF = EOF = 90 (Opposite angle of a quadrilateral is equal) Thus, PQRS is a parallelogram with R = 90. Hence, PQRS is a rectangle. Question 3. is a rectangle and P, Q, R ans S are mid-points of the sides,, and, respectively. Show that the quadrilateral PQRS is a rhombus. Solution Given, is a rectangle. R = = = = 90 and =, = lso, given PQR,, and S are mid-points of,, and,respectively. PQ and PQ = 1 and SR and SR = 1 S P Q
and SR and SR = 1 In rectangle, = PQ = SR Now, in SP and QP P S SP QP = P = Q = (y SS) SP = PQ (y PT) (ii) Similarly, in RS and RQ, S R RS RQ = Q = R = (y SS) SR = RQ (y PT) (iii) From Eqs. (i), (ii) and (iii), it is clear that quadrilateral PQRS is a rhombus. Question 4. is a trapezium in which, is a diagonal and E is the mid-point of. line is drawn through E parallel to intersecting at F (see figure). Show that F is the mid-point of. E F Solution Given, is a trapezium in which and E is mid-point of and EF. In, we have and E is mid-point of. EP So, by theorem, if a line drawn through the mid-point of one side of a triangle parallel to another side bisect the third side. P is mid-point of. Similarly, in, we have, PF and P is mid-point of. So, by converse of mid-point theorem, F is mid-point of. E P F
Question 5. In a parallelogram, E and F are the mid-points of sides and respectively (see figure). Show that the line segments F and E trisect the diagonal. P F Q Solution Given is a parallelogram and E, F are the mid-points of sides and respectively. To prove Line segments F and E trisect the diagonal. Proof Since, is a parallelogram. and = (Opposite sides of a parallelogram) E F and 1 1 = E F and E = F EF is a parallelogram. F E EQ P and FP Q In P, E is the mid-point of and EQ P, so Q is the mid-point of P. E (y converse of mid-point theorem) Q = PQ gain, in Q, F is the mid-point of and FP Q, so P is the mid-point of Q. (y converse of mid-point theorem) QP = P (ii) From Eqs. (i) and (ii), we get Q = PQ = P Hence, E and F trisect the diagonal. Question 6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. Solution Let is a quadrilateral and PQR,, and S are the mid-points of the sides,, and, respectively. ie.., S = S, P = P, Q = Q and R = R. We have to show that PR and SQ bisect each other ie.., SO = OQ and PO = OR.
= R S E O F Q P Now, in, S and R are mid-points of and. We know that, the line segment joining the mid-points of two sides of a triangle is parallel to the third side. (y mid-point theorem) SR and SR = 1 Similarly, in, P and Q are mid-points of and. PQ and PQ = 1 From Eqs. (i) and (ii), we get PQ SR and PQ = SR = 1 (y mid-point theorem) (ii) Quadrilateral PQRS is a parallelogram whose diagonals are SQ and PR. lso, we know that diagonals of a parallelogram bisect each other. So, SQ and PR bisect each other. Question 7. is a triangle right angled at. line through the mid-point M of hypotenuse and parallel to intersects at. Show that (i) is the mid-point of (ii) M (iii) M = M = 1 Solution Given, is a right angled triangle. = 90 and M is the mid-point of. lso, M (i) In, M and M is mid-point of. is the mid-point of. (y converse of mid-point theorem) M
(ii) Since, M and is transversal. M = (orresponding angles) ut = 90 M = 90 M (iii) Now, in M and M, we have M M M = M (ommon) = (Q is mid point of ) M = M (Each equal to 90 ) (y SS) M = M (y PT) lso, M is mid-point of. M = M = 1 (ii) From Eqs. (i) and (ii), we get M = M = 1