Grade 9 Quadrilaterals

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1 ID : pk-9-quadrilaterals [1] Grade 9 Quadrilaterals For more such worksheets visit Answer t he quest ions (1) 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) In a square ABCD, the diagonals bisect at O. What kind of a triangle AOB is? (3) Prove that the line segment joining the midpoints of the diagonals of a trapezium is parallel to each of the parallel sides ans is equal to half of the dif f erence of these sides. (4) ABCD is a parallelogram. The angle bisectors of A and D meet at O. What is the measure of AOD? Choose correct answer(s) f rom given choice (5) In the parallellogram ABCD, the sum of angle bisectors of two adjacent angles is. a. 45 b. 30 c. 115 d. 90 (6) The a parallelogram ABCD, the bisector of A also bisects the side BC. If AB = 7 cm, f ind the length of side AD. a. 14 cm b. 7 cm c. 17 cm d. can not be determined

2 (7) In the rectangle below, AB is 8 m and BC is 15 m. If O is the midpoint of BC, then what is the area of the shaded region? ID : pk-9-quadrilaterals [2] a. 63 m 2 b. 59 m 2 c. 60 m 2 d. 54 m 2 (8) The quadrilateral f ormed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle if a. Diagonals of PQRS are equal b. Diagonals of PQRS are perpendicular c. PQRS is a rectangle d. Can not be determined (9) In a triangle, ABC, D is a point on AB such that AB = 4AD and E is a point on AC such that AC = 4AE. Find BC. a. 5ED/2 b. 3ED c. AD + AE d. 4ED (10) In a square ABCD, E, F, G, and H are the mid points of the f our side, what kind of shape is represented by EFGH. a. T rapezium b. Square c. Rectangle d. Can not be determined (11) In a parallelogram ABCD, f ind CDB if DAB = 64 and DBC = 77. a. 103 b. 39 c. 77 d. 64 (12) 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 = 14 cm and AD = 5 cm, f ind the measure of AF. a. 21 cm b. 28 cm c cm d. 42 cm (13) ABCD is a quadrilateral and A = B = C = D = 90. Then ABCD can be called as a. Rectangle b. Parallelogram c. Square d. Both rectangle and parallelogram (14) The quadrilateral f ormed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rhombus if a. PQRS is a rectangle b. Diagonals of PQRS are equal c. PQRS is a parallelogram d. Can not be determined

3 (15) In the parallellogram ABCD, f ind the measurement of BAC and DAC, when BCA is 45 and DCA is 38. ID : pk-9-quadrilaterals [3] a. 38, 45 b. 45, 45 c. 45, 52 d. 45, Edugain ( All Rights Reserved Many more such worksheets can be generated at

4 Answers ID : pk-9-quadrilaterals [4] (1) 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)

5 (2) an isosceles right angled triangle ID : pk-9-quadrilaterals [5] Following f igure shows the square ABCD with diagonals. We know that diagonals of a square are equal and bisects each other perpendicularly, theref ore AC = BD AC/2 = BD/2 AO = OB Also, AOC = (since diagonals are perpendicular) Since AO = OB and AOC = 90, triangle ΔAOB is an isosceles right angled triangle. (3)

6 (4) 90 ID : pk-9-quadrilaterals [6] 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.

7 (5) d. 90 ID : pk-9-quadrilaterals [7] Following f igure shows the parallelogram ABCD, Let's assume, AO and DO are the angle bisectors of the adjacent angles A and D respectively. Theref ore, DAO = A/2, ADO = D/2. We know that the adjacent angles in a parallelogram are supplementary as they are f ormed by a straight line (e.g. AD) intersecting two paralle lines (e.g. AB and CD). Theref ore sum of the adjacent angles equals to 180. A + D = (1) Now, the sum of angle bisectors of the adjacent angles A and D = DAO + ADO = A/2 + D/2 = ( A + D)/2 = 180/2 = 90 Step 4 Hence, the sum of angle bisectors of two adjacent angles is 90. (6) a. 14 cm

(7) c. 60 m 2 ID : pk-9-quadrilaterals [8] Following f igure shows the rectangle ABCD, If we look at the f igure caref ully, we notice that the shaded region makes a ΔAOD.

8 (7) c. 60 m 2 ID : pk-9-quadrilaterals [8] Following f igure shows the rectangle ABCD, If we look at the f igure caref ully, we notice that the shaded region makes a ΔAOD. According to the question, AB = 8 m, AD = 15 m. In ΔAOD, The height of the ΔAOD = AB = 8 m, The base of the ΔAOD = AD = 15 m, The area of the ΔAOD = (AB AD)/2 = (8 15)/2 = 60 m 2 Hence, the area of the shaded region is 60 m 2.

9 (8) b. Diagonals of PQRS are perpendicular ID : pk-9-quadrilaterals [9] Following f igure shows the quadrilateral PQRS, with it's mid-points ABCD connected to f orm another quadrilateral ABCD. Lets draw diagonals of the quadrilateral PQRS, It is given that ABCD is a rectangle, DAB = ABC = BCA = CDA = 90 Step 4 In triangle ΔSRQ, since B and C connects mid-points, BC will be parallel to SQ. Step 5 Now parallel lines BC and SQ are intersected by AB, theref ore x = B = 90 Similarly y = 90 Step 6 In quadrilateral BxOy, B = x = y = 90 O = = 90 Which means PR and SQ are perpendicular to each other.

10 Step 7 ID : pk-9-quadrilaterals [10] Theref ore, f or quadrilateral ABCD to be rectangle, it is required that digoanls of PQRS are perpendicular to each other

11 (10) b. Square ID : pk-9-quadrilaterals [11] 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.

12 ID : pk-9-quadrilaterals [12] (11) b. 39 Following f igure shows the parallelogram ABCD, According to the question DAB = 64 and DBC = 77. A = C = 64 [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 = 39 Hence, the value of the CDB is 39. (12) b. 28 cm (13) d. Both rectangle and parallelogram 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'.

13 (15) a. 38, 45 ID : pk-9-quadrilaterals [13] In the parallelogram ABCD, AB DC, T heref ore, BAC = DCA [Alternate interior angles] Similarly, since AD BC, DAC = BCA [Alternate interior angles] According to the question, BCA = 45, DCA = 38 Theref ore, BAC = DCA = 38, DAC = BCA = 45 Hence, the measurement of the BAC and DAC is 38 and 45 respectively.

Class 9 Full Year 9th Grade Review

ID : in-9-full-year-9th-grade-review [1] Class 9 Full Year 9th Grade Review For more such worksheets visit www.edugain.com Answer the questions (1) In the graph of the linear equation 5x + 2y = 110, there