EXERCISE Two angles of a quadrilateral are 70 and 130 and the other two angles are equal. Find the measure of these two angles.

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1 CHPTER 3 UNIT-5 EXERCISE Two angles of a quadrilateral are 70 and 130 and the other two angles are equal. Find the measure of these two angles. ns: Let and be x, C=70 and D = C + D = 360 (Theorem 7) x + x = 360 2x = X = = 80 = 80 and = In the figure suppose P and Q are supplementary angles and R = 125. Find the measures of S. S 125 R P Q ns: P + Q = 180 (Supplementary angles) P + Q + R + S = 360 (Theorem 7) S = 360 S = 360 ( )

2 S = S=55 3. Three angles of a quadrilateral are in the ratio 2:3:5 and the fourth angle is 90. Find the measures of the other three angles. ns: Let the angles be 2x, 3x, 5x. + + C + D = 360 2x + 3x + 5x + 90 = x = x = x = 27 The angles are, = 2x = 2 27 = 54 = 3x = 3 27 = 81 C = 5x = 5 27 = In the adjoining figure, CD is a quadrilateral such that D + C =100. The bisectors of and meet at P. Determine P. D C P ns: P and P are angular bisectors D + C = 100 To find : P Proof : + + C + D = 360 (Theorem 7)

3 = = Multipyling by 1, = a + b =130 In Δ P, a + b + P = 180 P = P = 50 EXERCISE In a trapezium PQRS, PQ RS, and P = 70 and Q = 80. Calculate the measure of S and R. ns : P Q S R P + S = 180 (Supplementary angles) S = = 110 Q + R = 180 (Supplementary angles) R = = 100

4 2. In a trapezium CD with CD. It is given that D is not parallel to C. Is ΔC Δ DC? Give reasons. ns: D C In Δ C and Δ DC, (1) C = C (Common side) (2) C = CD (lternate angles, CD) Δ C Δ DC It cannot be proved with the help of any beam postulated. 3. In the figure, PQRS is an isosceles trapezium; SRP = 30, and PQS=40 Calculate the angles RPQ and RSQ. ns: S R 30 P 40 Data:PQRS is an isosceles trapezium, PS = RQ and P = Q SRP = 30 and PQS = 40 To find : RPQ and RSQ Proof: RPQ = SRP = 30 (lternate angles, PQ SR) RSQ = PRS = 40 (lternate angles, PQ SR) Q

5 4. Proof that the base angles of an isosceles trapezium are qual. ns: D C Data: CD is an isosceles trapezium. To Prove: = Construction: Join the diagonal D and C. Proof : In Δ C and ΔD (1) C = D Isosceles) (2)C = D (3) = (Common side) trapezium) C Δ D (SSS postualate) = (Conguency property) 5. Suppose in a quadrilateral CD, C= D and D = C. Prove that CD is a trapezium. ns: D C Data: CD is aquadrilateral, C = D and D = C To prove: CD is trapezium Proof: In Δ D and Δ C (1) D = C (Data) (2) C = D (Data)

6 (3) = (Common side) Δ D Δ C (SSS postulate) = (Conguency property) C = D (Data), D = C (Data) CD is an isosceles trapezium (lternate method to be given later) EXERCISE The adjacent angles of a parallelogram are in the ratio 2:1 Find the measures of all the angles. ns: Let the angles be 2x and x + = 2x + x = 180 (djecent angleso of percellelograms ae supplementary) 2x + x = 180 3x = 180 X = = 60 = 2x = 60 2 = 120 = 60 C = = 120 (Opposite angles of parallelogram) D = = 60 (Opposite angles of parallelogram)

7 2. field is in the form of a parallelogram, whose perimeter is 450 m and one of its sides is larger than the other by 75m. Find the lengths of all sides ns: K 60 L N M Perimeter = D + DC + C = x + x x + x = 4x = 4x 300 x = x X = 75 Side = 75m(x) Opposite side = x + 75 = = 150m 3. In the figure, CD is a parallelogram. The diaglonals C and D intersect at O; and DC = 40, C = 35, and DOC = 10. DC, C, and CD. D 110 C ns: Data: CD is a parallelogram D C and DC. The diagonals C and D intersect at O DC = 40, C = 35 and DOC = 110 To find : (1) O (2) DC (3) C (4) CD Proof: DC + C = = = 75

8 C = = 75 (Opposite angles of parallelogram are equal) D + = 180 (Supplementary angles) D = = 105 = D = 105 (Opposite angles of pallelgoram are equal) DOC = O = 110 (Vertically opposite angles) In Δ O, + O + = 180 (Sum of all angles of a Δ is 180 ) = 180 = (1) O = 35 (2) DC = 105 (Proved) (3) C = CD = 40 (lternate angles, D C) CD = (4) CD = 70 (1) O = 35 (2) DC = 105 (3) C = 40 and (4) CD = In a parallelogram CD, the side DC is produced to E and CE = 105. Calculate and,, C and D. ns: D C E CD + CE = 180 (Linear pair) CD = = 75

9 In CD, = C = 75 (Opposite angles of parallelogram) C = CE = 105 D = = 105 (lternate angles, DE) (Opposite angles of parallelogram) 5. Prove logically the diagonals of a parallelogram bisect each other. Show conversely that a quadrilateral in which diagonals bisect each other is a parallelogram. ns: D C Data: CD is a parallogram ( ln ) To Prove : O = OC and DO = O Proof: In ΔO and DOC (1) = DC (Opposite sides of angles) (2) O = DOC (V.O.) (3) D = DC (lternate angles ΔO ΔDOC (S portulate) O = OC & DO = 3O (Conguency properties) 6. In a parallelogram KLMN, K = 60. Find the measure of all the angles. ns: K L 60 N M

10 M = K = 180 (Supplementary angles) N = = 120 M = K = 60 Opposite angles of parallelogram L = N = Let CD be a quadrilateral in which = C and = D. Prove that CD is a parallelogram. ns: x y y x D C Data: = C and D = To Prove: CD is a parallelogram Proof: + + C + D = 360 x + y + x + y = 360 2x + 2y = 360 2(x + y) = 360 x + y = 360 = djacent angles are supplementary CD is a parallelogram. 8. In a quadrilateral CD, suppose = CD and D = C. Prove that CD is a parallelogram. ns: D C Data: CD is a quadrilateral, b = DC, D = C To Prove : CD is a parallelogram Construction: Join D

11 Proof: In Δ D and Δ DC (1) D = D (Common side) (2) D = C (Data) (3) = DC (Data) D Δ DC (SSS postulate) D = DC (Congruency property) ut they are alternate angles These we have DC D = DC (C.P) ut they are alternate angles These we have D C CD is a parallelogram. EXERCISE The sides of the rectangle are in the ratio 2:1. The perimeter is 30 cm. Calculate the measure of all the sides. ns: CD is a rectangle Let :C = 2:1 = 2x and C = x Perimeter of the rectangle = 2(1+b) 2(1+b) = 30 D C 2(2x+x) = x = 30 6x = 30 x = 30 6 = 5 2x = 2 x = 2 5 = 10 1x = 1 4 = 1 5 = 5 = 10m C = 5cm CD = 10cm and D = 5cm 2. In the adjacent rectangle CD, OCD = 30. Caclulate OC. What type of triangle is OC?

12 CD is a rectangle Diogonals C and D bisect each other D C t O and C = D. 30 O = OC = O = OD OCD = ODC = 30 On ΔCOD, ODC + OCD + COD = COD = COD = 180 COD = =120 COD + CO = 180 CO = CO = 60 Δ OC is an isosceles triangle 3. ll rectangle are parallelograms, but all parallelograms are not rectangles. Justify this statements. ll rectangles have all the properties of parallelograms but a parallelogram May not have all the properties of a rectangle. 1. In a rectangle all the angles are right angle but in aparallelogram only opposite angles are equal. 2. In a rectangle the diagonals are equal but in a parallelogram diagnolas are not equal. 4. Prove logically that the diagonals of a rectangle are equal. Data : CD is a rectangle. C and D are the diagonals. D C To Prove : C = D Proof : In ΔC and Δ D C = D [90 ] C = D [Opp. Sides] = [Common side] ΔC ΔD [SS] C = D [CPCT] 5. The sides of a rectangualar park are in the ratio 4:3. If the area is 1728 m 2, find the cost of fencing it at the rate of Rs.2.50/m. ns: CD is a rectangualar park let :C = 4:3

13 = 4x and C = 3x D C rea of rectangle = 1 b 4x 3x = x 2 = 1728 X 2 = 144 X = 144 = 12 Length = 4x = 4 12 = 48m readth = 3x = 3 12 = 36m Perimeter of a rectangle = 2(1+b) 2( ) = 2 84 = 168m Cost of fencing = perimeter Rate = = Rs rectangualar yard contains two flower beds in the shape of congruent isosceles right triangle. The remaining portion is a yard of trapezoidal. Shape (see fig) whose parallel sides have legths 15 m and 25 m. What fraction of the yard is occupied by the flower bed? CD is a rectangualar yard 5m E F 5m C EF is a trapezium D EF = 25 m and EF = 15m 5m 5m D = C = 5m rea of rectangle = 1 b = 25 5 = 125m 2 25m rea of each flower bed = 1 2 b h = = 25 2 m2 rea of both flower beds = = 25m2 Fraction of flower beds to yard = 25 = In a rhombus CD C = 70. Find the other angle of the rhombus. CD is a rhombus LC = 70 ut = C = 70 = 70

14 + =180 (adjacent angles) D C 70 + = = = 110 = D = 110 = 70 = 110 D = In a rhombus PQRS, SQR = 40 and PQ = 3 cm. Find SPQ, QSR and the perimeter of the rhombus. PQRS is a rhombus SQR = 40 and pq = 3cm P S PQS = SQR = 40 PQS = 40 (diagonal bisector angles ) 3 cm ut PQS = QSP [PQ = PS] QSP = In Δ PQS, PQS + QSP + SPQ = 180 Q R SPQ = 180 SPQ = = 100 SPQ = 100 In a rhombus PQ = QR = RS = SP = 3cm Perimeter of the Rhombus PQRS = 3 4 = 12cm 9. In a rhombus PQRS, if PQ = 3x -7 and QR = x + 3, find PS. In a rhombus all sides are equal PQ = QR = RS = SP PQ = QR 3x -7 = x + 3 3x -x = x = 10 X = 5 PS = x+3 = 5+3 = 8 cm 10. Let CD be a rhombus and C = 124. Calculate, D and C. In a rhombus, opposite angles are equal = D = =180 Consecutive angles = 180

15 + 124 = 180 = =56 = 56 = 124 and C = Rhombus is a parallelogram: justify. Rhombus has all the properties of parallelogram i.e a) Opposite sides are equal and parallel. b) Opposite angles are equal. c) Diagonals bisect each other Rhombus is a parallelogram. 12. In a given square CD, if the area of triangle D is 36 cm 2, find (i) the area of triangle CD; (ii) the area of the square CD. CD is a square d is the diagonal The diagonal divides the square into two congruent triangles Δ D Δ CD D C rea of Δ D =area of Δ CD rea of Δ D= 36 cm 2 (given) rea of Δ CD = 36 cm 2 rea of the square CD = rea of Δ D + rea of ΔCD =36 cm cm 2 rea of the squre CD =72 cm The side of a square CD is 5 cm and another square PQRS has perimeter equal to 40cm. Find the ratio of the area CD to the area of PQRS. Perimeter of the square CD = 4 side = 4 5 = 20 Ratio of perimeter of CD = 20 = 1 Perimeter of PQRS 40 2 rea of CD = (side) 2 = (5) 2 = 25cm Side of PQRS = perimeter 4 rea of PQRS = (side) 2 = (10) 2 = 100cm 2 Ratio of rea of CD rea of PQRS = = 1 4 = 40 4 = 10cm or 1:4 or 1:2

16 14. square field has side 20m. Find the length of the wire required to fence it foure times. Length of one side of the square = 20 Rs Length of wire required to fence one round = 4 20 Length of wire required to fence four rounds = 80m = 4 80m = 320m 15. List out the differences between square and rhombus. Square Rhombus 1. ll the angles are equal Opposite angles are equal 2. Diagonals are equal Diagonals are unequal 3. rea = side side=(s) 2 rea= 1 2 = 1 2 d 1 d 2 Product of diagonals 16. Four congruent rectangles are placed as shown in the figure. rea of the outer square is 4 times that of the inner square. What is the ratio os length to breadth of the congruent rectangles? Let the length of rectangles be a and breadth be b Side of outer square (a+b) units Side of outer square (a-b) units rea of outer square = 4 times area of inner square rea of CD = 4(rea of PQRS) (a + b) 2 = 4(a-b) D b a + b = 2(a-b) S R a + b = 2a-2b a 2a 2b = a + b 2a-a = b + 2b a = 3b p q b a = 1 b a b 3 Ratio : length: headth = 3:1 taking square of both sides a b C

17 DDITIONL PROLEMS ON QUDRILTERLS I. Complete the following: a. Quadrilateral has.. sides.(four 4) b. Quadrilateral has..diagonals.(2 two) c. Quadrilateral in which one pair of sides are paralled to each other is. (Trapezium) d. In an isosceles trapezium, the base angles are.(equal) e. In a Rhombus, diagonals bisect each other in angles.(right) 2. Let CD be a parallelogram, What special name will you give it. a. If = C : CD will be Rhombus. D C b.if D = 90 : CD will be Rectangle. c. If = D and D = 90 : CD will be a Square. 3. quadrilateral has three acute angles each measuring 70. What is the measure of the fourth angle? : Sum of 4 angles of a Quadrilateral = 360 Three of the four angle = x = 360 X = X= The difference between the two adjacent angles of a parallelogram is 20. Find measures of all the angles of the parallelogram. : Let CD be a 11 grs D C + = 180 [adjacent of a gm is 180 ] and = 20 [given] = 180

18 180 = = = 2 = 80 If =80 = The angles of a Quadrilateral are in the ratio 1:2:3:4. Find all the angles of the quadrilateral.. Let the angles of a Quadrilateral = 1x, 2x, 3x, and 4x. 1x + 2x + 3x + 4x = 360 [Sum of the 4 angles of a Quadrilateral is 360 ] 10x = 360 X = = 36 angles are 36, 2 36, 3 36, , 72, 108, Let PQRS be a parallelogram with PQ = 10cm, and QR = 6 cm. Calculate the measures of the other two sides and the perimeter of PQRS. : PQ = SR S R PS = QR [opposite sides of a 11 gm are equal] PQ = SR = 10cm 6cm PS = QR = 6 cm P 10cm Q Perimeter = PQ + QR + PS + SR = = 32 cm. 7. The Perimeter of a Square is 60 cm. Find its side length. : Perimeter of a Square = 4a = 60 cm. a = 60 4 = 15 cm Length of the side = 15 cm. 8. Let CD be a square and Let C = D = 10 cm. Let C and D intersect at O. Find OC and OD : In a square, diagonals bisect each other. D C O = OC O = OD C = D = 10 cm [given]

19 O = OC = 1 10 = 5 cm 2 O = OD = 1 10 = 5 cm 2 9. Let PQRS be a rhombus with PR = 15 cm and QS = 8 cm. Find the area of the Rhombus. S.: rea of Rhombus = 1 2 d 1d 2 = 1 2 PR QS = P R R = 60 Sq cm. Q 10. Let CD be a parallelogram and suppose the bisectr of and meet at P.Prove that P = 90. : In the gm CD. D + C = 180 (1) s P and P are angular bisectors. DP = P and P = PC (2) In (1) DP + P + P + PC = 180 From (2) P + P + P + P = P + 2 P = 180 P + P = = 90 C D In Δ P, + P + = = P = 180 P = =90 P 90

20 11. Let CD be a Square. Locate paints P,Q,R,S on the sides, C, CD, D respectively such that CR = P = CR= DS. Prove that PQRS is a square. : D R C S Q P Given:- CD is a Square and P = Q = CR = DS = x T.P.T:- PQRS is a Square or PQ = QR = RS = PS & P = Q = R = S = 90 Proof :- CD is a Square = C = CD = D P + P = Q + QC = CR + DR = SD + S x + P = x + QC = x + DR = x + S P = QC = DR = S Now In Δ s SP and Δ PQ P = P S = Q = = 90 Δ SP Δ PQ (SS Postulate) PS = PQ III ly From ΔDSR, DRCO and ΔPQ, QR= RS = SP = PQ. From Δ PQS and Δ RSQ, y SSS Postulate Δ RQS Δ PSQ. R = P III ly. S = Q P = R = S = Q = 90.

21 12. Let CD be a rextangle and let P,Q,R,S be the midpoint of, C,CD, D respectively. Prove PQRS is a Rhombus. : D R C S Q P Given :- CD is a Rectangle. P = P, Q = QC, QR= RC, S = DS T.P.T:- PQRS is a Rhombus. Solution:- In ΔSP and Δ PQ P = P S = Q = = 90 ΔSP ΔPQ PS = PQ III ly From ΔDRS and ΔRCQ, ΔSP and ΔPQ PQ = RQ = RS = PS PQRS is a Rhombus. 13. Let CD be a Quadrilateral in which diagonals intersect at O Perpendicularly. Prove that + C + CD + D> C + D. : Consider the Δ s C O, OC, COD, DO ll are right led Δ s D, C,CD and D are the greatest sides. i.e > O and O C > O, OC

22 CD > CO and OD D > OD and O + C + CD + D > O + O + O + OC + OC + OD + O + C + CD + D > + CD 14. Let CD be a Quadrilateral with diagonals C and D. Prove that (a) + C + CD > D D : Let us join C C In Δ C + C > C In Δ DC D + DC > C + C + D + DC > C + C + CD > D 15. Let CD be a quadrilateral with diagonals C and D. Prove the following statements. a) + C + CD > D b) + C + CD + D > 2C c) + C + CD + D >2D d) + C + CD + D > C + D D C : a) Consder the Δ DC, C + CD > D..(1) (Sum of two sides of a Δ is greater than III side) In Δ D, + D > D..(2)

23 In Δ DC, C +DC > D C > D + C > D + C + CD > D [From (1), (2) and (3)] b) In ΔDC, D + DC > C..(1) In Δ C, + C > C.(2) + C + D + DC > 2C c) In Δ D, + D > D.(1) In Δ DC, C + CD > D..(2) (1) + (2). + C + D + CD >2 D d) In ΔD, + D > D In ΔC, + C > C + C + CD > D > C + D In Δ DC, CD + D > D In ΔDC, D + CD > C b) + C + CD + D > 2C Similar to Previous solution + C > C D + DC > C C + C + D + DC > 2C c) + C + CD + D > 2 D In ΔD, + D > D In Δ DC, C + CD > D

24 + D + C + CD > 2D d) + C + CD + D > C + D Solution of problems (13) 16. Let PQRS be a Kite such that PQ > PS Prove that PQR > PSR. Solution: p Q S R In PSQ PSQ > PQS [angle opposits greater side is greater than the lesser side] PSQ + QSR > PQS + SQR PSR > PQR 17. Let CD be a quadrilateral in which is the Smallest Side and CD is the largest side. Prove that > C and > D. C In Δbc, is Smallest side C is smallest le > and C, > C + + C = 180 or > C In Δ CD and D is smallest side D is smallest le CD is largest side is largest angle > D D

25 18. In a Δ C, let D be the mid point of C. Prove that + C >2D. D C In Δ D, + D > D In Δ DC, C + CD > D + C + C > 2 D + C > 2 D Properly Δ : Sum of any two sides of Δ is greater than the third side. 19. Let CD be a quadrilateral and let P,Q,R and s be the midpoint of, C, CD and D respectively. Prove that PQRS is a gm. D R C S Q P In Δ DC, S and R Midpoints SR C and SR = 1 2 C In Δ C, P and Q Midpoints PQ C, PQ = 1 2 C SR = PQ and SR PQ SRQP is a 11gm [Midpoint Theorem]

2. A diagonal of a parallelogram divides it into two congruent triangles. 5. Diagonals of a rectangle bisect each other and are equal and vice-versa.

2. A diagonal of a parallelogram divides it into two congruent triangles. 5. Diagonals of a rectangle bisect each other and are equal and vice-versa. QURILTERLS 1. Sum of the angles of a quadrilateral is 360. 2. diagonal of a parallelogram divides it into two congruent triangles. 3. In a parallelogram, (i) opposite sides are equal (ii) opposite angles