Class 9 Quadrilaterals


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1 ID : in9quadrilaterals [1] Class 9 Quadrilaterals For more such worksheets visit Answer t he quest ions (1) The diameter of circumcircle of a rectangle is 13 cm and rectangle's width is 5 cm. Find length of the rectangle. (2) In a quadrilateral ABCD, the angles A, B, C and D are in ratio 3:4:5:6. Find the measure of each angle of the quadrilateral. (3) In a parallelogram, prove that the ratio of any side to altitude of adjacent side is same f or all its sides. (4) In a rhombus, one of the diagonals is equal to a side of the rhombus. Find the angles of the rhombus. (5) ABCD is a parallelogram with E and F as the midpoints of side AB and CD. Line segments CE and AF intersect the diagonal BD at point P and Q respectively. If QD = 3 cm and BC = 8 cm, f ind the length of diagonal BD. (6) ABCD is a rectangle and point P is such that PB = 3 2 cm, PC = 4 cm and PD = 3 cm, f ind the value of PA. Choose correct answer(s) f rom given choice (7) ABCD is a parallelogram. L and M are points on AB and DC respectively such that AL = CM. The diagonal BD intersects LM at point O. If AB = 15 cm, AD = 5 cm and LO = 3 cm, f ind the length of OM. a. 7 cm b. 3 cm c. 6 cm d. can not be determined (8) The perimeter of a parallelogram is 180 cm. If one side exceeds the other side by 12 cm, what are the sides of the parallelogram. a. 39 cm, 51 cm b. 41 cm, 53 cm c. 39 cm, 27 cm d. 45 cm, 57 cm
2 ID : in9quadrilaterals [2] (9) In a triangle ABC, AD is the median. F is the point on AC such that line BF bisect AD at E. The measure of AC = a. 3AF b. 3AE c. 2AF d. 3EF (10) The length and breadth of a rectangle is 8 cm and 15 cm. Find the radius of circumcircle of this rectangle. a. 8.5 cm b. 17 cm c cm d cm (11) ABCD is a parallelogram and E is the midpoint of side BC. When DE and AB are extended, they meet at point F. If AB = 10 cm and AD = 7 cm, f ind the measure of AF. a. 15 cm b cm c. 30 cm d. 20 cm (12) In a parallelogram ABCD, f ind CDB if DAB = 51 and DBC = 45. a. 84 b. 45 c. 51 d. 39 (13) In the quadrilateral ABCD, AO and DO are the bisectors of A and D respectively. If B = 115 and C = 97 than AOD is: a. 119 b. 96 c. 106 d. 212 (14) ABCD is a quadrilateral and A = B = C = D = 90. Then ABCD can be called as a. Parallelogram b. Square c. Rectangle d. Both rectangle and parallelogram (15) In a parallelogram ABCD, the diagonals bisect at O. What kind of a triangle AOB is? a. a right angled but not an isosceles triangle b. an isosceles but not right angled triangle c. neither isosceles nor a right angled triangle d. an isosceles right angled triangle 2016 Edugain ( All Rights Reserved Many more such worksheets can be generated at
3 Answers ID : in9quadrilaterals [3] (1) 12 cm Following f igure shows the rectangle ABCD with it's circumcircle, AB, BC and AC are the length, breadth and diameter of circumcircle of the rectangle ABCD. According to the question, AC = 13 cm, BC = AD = 5 cm. Now, in right angled triangle ABC, AB 2 = AC 2  BC 2 AB = [ AC 2  BC 2 ] = [ ] = 12 cm Hence, the length of the rectangle is 12 cm.
4 (2) A = 60, B = 80, C = 100, D = 120 ID : in9quadrilaterals [4] Let's assume x is the common f actor of the angles of the quadrilateral. According to the question, the angles A, B, C and D are in ratio 3:4:5:6. Theref ore, A = 3x, B = 4x, C = 5x and D = 6x. We know that the sum of all interior angles of a quadrilateral is equal to 360. Theref ore, A + B + C + D = 360 3x + 4x + 5x + 6x = x = 360 x = x = 20 Hence, A = 3x = 3 20 = 60, B = 4x = 4 20 = 80, C = 5x = 5 20 = 100 and D = 6x = 6 20 = 120.
5 (3) ID : in9quadrilaterals [5] Let ABCD is a parallelogram as shown below. We have extended its sides to show the altitudes as CP and CQ Lets f irst consider side AB. Its adjacent sides are AD and BC. From the picture we can see that altitude f or AD and BC is CQ. Theref ore required ratio is, AB CQ Similarly these ratios will be as f ollowing f or other sides CD, AD, BC CQ CP CP Step 4 Theref ore we need to prove that, AB = CD = AD = BC CQ CQ CP CP Step 5 If we compare triangles ΔBPC and ΔDQC, we observe f ollowing,  DQC = BPC... (Both are right angle)  DCQ = BCP... ( DCQ = 90  BCD and BCP = 90  BCD) Step 6 Since two angles are equal, triangles ΔBPC and ΔDQC are similar triangles Step 7 Since ratio of corresponding sides of similar triangles is same, BC = CD CP CQ
6 Step 8 ID : in9quadrilaterals [6] Also since AB = CD and BC = AD, AB CQ = CD CQ = AD CP = BC CP (4) 60, 120 Following f igure shows the rhombus ABCD, We know that, all sides of a rhombus have equal length. According to the question, the diagonal AC of the rhombus ABCD is equal to the side of the rhombus. Theref ore, AB = BC = CD = DA = AC In ΔABC, AB = BC = CA, Hence, the ΔABC is a equilateral triangle. Step 4 Similarly, the ΔACD is a equilateral triangle. Step 5 All angles of a equilateral triangle is equal to 60. Theref ore, B = D = 60, A = BAC + CAD = = 120, Similarly, the C = 120 Step 6 Hence, the angles of the rhombus are 60 and 120. (5) 9 cm (6) 5 cm (7) b. 3 cm
7 (8) a. 39 cm, 51 cm ID : in9quadrilaterals [7] Let's assume a and b are the sides of the parallelogram. The perimeter of the parallelogram = 2(a + b) According to the question, one side of the parallelogram exceeds the other side by 12 cm. Theref ore, a = b + 12 Since, the perimeter of the parallelogram = 180 cm Theref ore, 2(a + b) = 180 2(b b) = 180 (2b + 12) = 180/2 2b = b = 78 2 b = 39 cm, a = b + 12 = = 51 cm Step 4 Hence, the sides of the parallelogram are 39 cm and 51 cm.
8 (9) a. 3AF ID : in9quadrilaterals [8] Following picture shows triangle ΔABC with median AD at E, and another line BF, which bisects AD. We have also drawn another line DG which is parallel to BF From proportionality theorem we know that if a line is drawn parallel to one side of a triangle, it divides other two sides in the same ration For triangle ΔBFC, DG is parallel to BF and using proportionality theorem we can f ind that, FG = BD GC DC Step 4 Since we know that BD = DC, we can f ind that GC = FG... (1) Step 5 Similarly we can use proportionality theorem f or triangle ΔADG, AF = AE FG ED Step 6 But since AE = ED, FG = AF... (2) Step 7 Theref ore, AC = AF + FG + GC Using relation (1) and (2), AC = AF + AF + AF, AC = 3 AF
9 (10) a. 8.5 cm ID : in9quadrilaterals [9] Following f igure shows the rectangle ABCD with it's circumcircle, AB, BC and AC are the length, breadth and diameter of circumcircle of the rectangle ABCD. According to the question, AB = 15 cm, BC = 8 cm. Now, in right angled triangle ABC, AC 2 = AB 2 + BC 2 AC = [ AB 2 + BC 2 ] = [ ] = 17 cm The diameter of circumcircle of the rectangle = 17 cm. The radius of circumcircle of the rectangle = 17/2 = 8.5 cm. (11) d. 20 cm
10 (12) a. 84 ID : in9quadrilaterals [10] Following f igure shows the parallelogram ABCD, According to the question DAB = 51 and DBC = 45. A = C = 51 [Since the opposite angles of a parallelogram are congruent.] In ΔBCD, DBC + BCD + CDB = 180 [Since the sum of all the angles of a triangle is 180 ] CDB = CDB = 180 CDB = CDB = 84 Hence, the value of the CDB is 84.
11 (13) c. 106 ID : in9quadrilaterals [11] Following f igure shows the quadrilateral ABCD, According to the question, AO and DO are the bisectors of A and D respectively. Theref ore, DAO = A/2, ADO = D/2, B = 115, C = 97 We know that the sum of all angles of a quadrilateral is equals to 360. Theref ore, A + B + C + D = 360 A D = 360 A + D = 360 A + D = A + D = (1) In ΔAOD, DAO + ADO + AOD = [Since, we know that the sum of all three angles of a triangle is equals to 180 ] A/2 + D/2 + AOD = 180 AOD = A/2  D/ A  D AOD = 2 AOD = ( A + D) AOD = 2 AOD = [From equation (1)] Step 4 Hence, the value of AOD is 106.
12 (14) d. Both rectangle and parallelogram ID : in9quadrilaterals [12] Following f igure shows the quadrilateral ABCD where all f our angles are 90 A quadrilateral with all f our angles of 90 is a rectangle. We also know that all rectangles are parallelogram since opposite sides of rectangles are parallel and equal to each other. T heref ore, the correct answer is 'Both rectangle and parallelogram'. (15) c. neither isosceles nor a right angled triangle Following f igure shows the parallelogram ABCD with diagonals. We know that diagonals of a parallelogram bisects each other but they are not perpendicular, theref ore AOC 90, and AO = AC/2 and OB = BD/2 Also since diagonals of parallelogram are not equal, AC BD AC/2 BD/2 AO OB Step 4 Since AO OB and AOC 90, triangle ΔAOB is neither isosceles nor a right angled triangle.
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