Grade 9 Quadrilaterals

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1 ID : ww-9-quadrilaterals [1] Grade 9 Quadrilaterals For more such worksheets visit Answer t he quest ions (1) 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. (2) In a square ABCD, E, F, G and H are the mid points of the f our sides, what kind of shape is represented by EFGH. (3) If angles P, Q, R and S of the quadrilateral PQRS taken in order are in ratio 12:8:6:14, then what kind of shape does PQRS represent. (4) ABCD is a parallelogram. The angle bisectors of A and D meet at O. What is the measure of AOD? (5) In the parallelogram PQRS, f ind the measurement of PSQ and PQS when QPS is 43 and QSR is 77. (6) The length and breadth of a rectangle is 3 cm and 4 cm. Find the radius of circumcircle of this rectangle. (7) In a quadrilateral ABCD, O is a point inside the quadrilateral such that AO and BO are the bisectors of A and B respectively. Prove that AOB = 1 2 ( C + D).

2 ID : ww-9-quadrilaterals [2] Choose correct answer(s) f rom given choice (8) ABCD and PQRS are parallelograms as shown in the f igure below: a. x + y + z + w = l + m + n + o b. x + y = l + m c. x + y = l + o d. all of above (9) In a triangle ABC, AD is the median. F is the point on AC such that line BF bisect AD at E. If AD = 9 cm and AF = 3 cm, f ind the measure of AC. a cm b. 9 cm c. 7.5 cm d. 6 cm (10) ABCD is a parallelogram and P and R are the midpoints of side DC and BC respectively. If line PR intersect diagonal AC at Q, than AC =. a. 4CQ b. 3CQ c. 2CQ d. (AB + BC)/2 (11) ABCD is a parallelogram and the diagonals of it intersect at point O. If DAO = 30, BAO = 36 and COD = 93, f ind ODC. a. 51 b. 30 c. 60 d. 57 (12) The quadrilateral f ormed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rhombus if a. Diagonals of PQRS are perpendicular b. PQRS is a parallelogram c. Diagonals of PQRS are equal d. Can not be determined (13) In a parallelogram ABCD, f ind CDB if DAB = 63 and DBC = 74. a. 16 b. 43 c. 74 d. 27 (14) The diagonals of a parallelogram ABCD intersect at O. A line through O intersect AB at X and DC at Y. if AB = 15 cm, AD = 5 cm, and OX = 3 cm, f ind the length of OY. a. 3 cm b. 2 cm c. 5 cm d. can not be determined

3 ID : ww-9-quadrilaterals [3] (15) In a parallelogram the ratio of any side to altitude of adjacent side is a. twice the ratio of bigger side and its altitude b. equal to the ratio of other sides and their corresponding altitudes c. not same f or all its sides and corresponding altitudes d. equal to the ratio of other sides and their corresponding adjacent side's altitudes 2016 Edugain ( All Rights Reserved Many more such worksheets can be generated at

4 Answers ID : ww-9-quadrilaterals [4] (1) 5 cm

5 (2) Square ID : ww-9-quadrilaterals [5] Following f igure shows the square ABCD, Let's assume the side of the square be a. In ΔGDH, DG = DH = a/2 [Since, G and H are the midpoints of the sides CD and DA respectively.] D = 90 [Since, ABCD is a square] GH 2 = DG 2 + DH 2 [By the pythagorean theorem] GH 2 = DG 2 + DG 2 [Since GH = GD] GH 2 = 2DG 2 GH 2 = (2a/2) 2 GH 2 = a 2 GH = a Similarly, HE = EF = FG = a and hence, HE = EF = FG = GH The ΔGDH is an isosceles triangle. [Since, DG = DH] In ΔGDH, D = 90, Theref ore, DHG = DGH = 45 [Since, the sum of all the angles of a triangle is equals to 180 ], Similarly, AHE = 45 Step 4 Now, DHG + AHE + GHE = 180 [Since, the angles on one side of a straight line will always add to 180 degrees.] GHE = GHE = 180 GHE = GHE = 90, Similarly, HEF = EFG = FGH = 90 and hence, HEF = EFG = FGH = GHE = 90 Step 5 Thus, EF = FG = GH = HE and HEF = EFG = FGH = GHE = 90. We know that quadrilateral with f our equal sides and f our right angles is a square. Theref ore, EFGH is a Square.

6 ID : ww-9-quadrilaterals [6]

7 (4) 90 ID : ww-9-quadrilaterals [7] Following f igure shows the parallelogram ABCD, AO and DO are the bisectors of DAB and ADC respectively. Theref ore, the DAB = 2 DAO, the ADC = 2 ADO T he DAB and the ADC are consecutive angles of the parallelogram ABCD, we know that, the consecutive angles of a parallelogram are supplementary. Theref ore, DAB + ADC = DAO + 2 ADO = 180 2( DAO + ADO) = 180 DAO + ADO = (1) We know that, the sum of all the angles of a triangle is equal to 180. In ΔAOD, DAO + ADO + AOD = AOD = 180 [From equation (1), DAO + ADO = 90 ] AOD = AOD = 90 Step 4 Hence, the measure of AOD is 90.

8 (5) PSQ = 60 PQS = 77 ID : ww-9-quadrilaterals [8] According to the question, QPS = 43, QSR = 77 T he QPS and the PSR are consecutive angles of the parallelogram ABCD, we know that, the consecutive angles of a parallelogram are supplementary. Theref ore, QPS + PSR = 180 QPS + PSQ + QSR = 180 [Since PSR = PSQ + QSR] 43 + PSQ + 77 = 180 PSQ = 180 PSQ = PSQ = 60 PQS = QSR [Alternate interior angles] PQS = 77 Step 4 Hence, the measurement of the PSQ and PQS are 60 and 77 respectively.

9 (6) 2.5 cm ID : ww-9-quadrilaterals [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 = 4 cm, BC = 3 cm. Now, in right angled triangle ABC, AC 2 = AB 2 + BC 2 AC = [ AB 2 + BC 2 ] = [ ] = 5 cm The diameter of circumcircle of the rectangle = 5 cm. The radius of circumcircle of the rectangle = 5/2 = 2.5 cm.

10 (7) ID : ww-9-quadrilaterals [10] Following f igure shows the quadrilateral ABCD, According to the question, AO and BO are the bisectors of A and B respectively. Theref ore, BAO = A/2, ABO = B/2 We know that the sum of all angles of a quadrilateral is equals to 360. Theref ore, A + B + C + D = 360 C + D = ( A + B) -----(1) In ΔAOB, BAO + ABO + AOB = 180 [Since, we know that the sum of all three angles of a triangle is equals to 180 ] A/2 + B/2 + AOB = 180 AOB = A/2 - B/ A - B AOB = 2 AOB = ( A + B) 2 AOB = Step 4 ( C + D) 2 [From equation (1)] Hence, AOB = 1 2 ( C + D)

11 (8) d. all of above ID : ww-9-quadrilaterals [11] We know that the sum of the all angles of a parallelogram is 360.Theref ore, x + y + z + w = 360 and l + m + n + o = 360 Theref ore x + y + z + w = l + m + n + o We also know that, the sum of the consecutive angles of a parallelogram is supplementary. Theref ore, x + y = x + w = l + m = l + o = 180 As we can see that all options are correct, the correct answer is, 'all of above'. (9) b. 9 cm (10) a. 4CQ (11) a. 51 Following f igure shows the parallelogram ABCD, According to the question, DAO = 30, BAO = 36 and COD = 93. OCD = BAO = 36 [Alternate interior angles] In ΔCOD, COD + OCD + ODC = 180 [Since the sum of all the angles of a triangle is 180 ] ODC = ODC = 180 ODC = ODC = 51 Hence, the value of the ODC is 51. (12) c. Diagonals of PQRS are equal

12 (13) b. 43 ID : ww-9-quadrilaterals [12] Following f igure shows the parallelogram ABCD, According to the question DAB = 63 and DBC = 74. A = C = 63 [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 = 43 Hence, the value of the CDB is 43. (14) a. 3 cm

13 ID : ww-9-quadrilaterals [13] (15) d. equal to the ratio of other sides and their corresponding adjacent side's altitudes 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

14 Step 8 ID : ww-9-quadrilaterals [14] Also since AB = CD and BC = AD, AB CQ = CD CQ = AD CP = BC CP

Class 9 Quadrilaterals

ID : in-9-quadrilaterals [1] Class 9 Quadrilaterals For more such worksheets visit www.edugain.com Answer t he quest ions (1) The diameter of circumcircle of a rectangle is 13 cm and rectangle's width