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ID : ae-9-Quadrilaterals [1] Grade 9 Quadrilaterals For more such worksheets visit www.edugain.com Answer t he quest ions (1) In a quadrilateral ABCD, the angles A, B, C and D are in ratio 2:3:4:6. Find the measure of each angle of the quadrilateral. (2) In the parallelogram ABCD, ∠D is 50°, then the measurement of ∠A is _____. (3) ABCD is a parallelogram with E and F as the mid-points 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. (4) If angles P, Q, R and S of the quadrilateral PQRS taken in order are in ratio 8:12:11:9, then what kind of shape does PQRS represent. (5) ABCD is a parallelogram and the diagonals of it intersect at point O. If ∠DAO = 37 °, ∠BAO = 33° and ∠COD = 97°, f ind ∠ODC. (6) In the parallelogram PQRS, PQ = 10 cm. T he altitude corresponding to the side PQ and PS are 5 and 6 respectively. Find PS. (7) In the parallellogram ABCD, f ind the measurement of ∠BAC and ∠DAC, when ∠BCA is 41° and ∠DCA is 45°. (8) 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. Prove that BC = 4ED. (9) In a parallelogram, prove that the ratio of any side to altitude of adjacent side is same f or all its sides. Choose correct answer(s) f rom given choice (10) In a parallelogram ABCD, f ind ∠CDB if ∠DAB = 54° and ∠DBC = 59°. a. 31° b. 121° c. 54° d. 67° (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ae-9-Quadrilaterals [2] (11) 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 + o c. x + w = l + m d. all of above (12) 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. 2CQ b. (AB + BC)/2 c. 4CQ d. 3CQ (13) T he three angles of a quadrilateral are 64°, 77° and 111° respectively. Find the f ourth angle. a. 252° b. 72° c. 18° d. 108° (14) In a quadrilateral ABCD, O is a point inside the quadrilateral such that AO and BO are the bisectors of ∠A and ∠B respectively. T he value of ∠AOB is a. 1 (∠C + ∠D) 2 b. 1 (∠A + ∠B) 2 c. 180° - (∠A + ∠B) d. ∠C + ∠D (15) ABCD is a parallelogram. T he angle bisectors of ∠A and ∠D meet at O. What is the measure of ∠AOD? a. sum of ∠A and ∠D b. 90° c. 45° d. Can not be determined © 2016 Edugain (www.edugain.com). All Rights Reserved (C) 2016 Edugain (www.Edugain.com) Many more such worksheets can be generated at www.edugain.com Personal use only, commercial use is strictly prohibited ID : ae-9-Quadrilaterals [3] Answers (1) ∠A = 48°, ∠B = 72°, ∠C = 96°, ∠D = 144° Step 1 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 2:3:4:6. T heref ore, ∠A = 2x, ∠B = 3x, ∠C = 4x and ∠D = 6x. Step 2 We know that the sum of all interior angles of a quadrilateral is equal to 360°. T heref ore, ∠A + ∠B + ∠C + ∠D = 360° ⇒ 2x + 3x + 4x + 6x = 360 ⇒ 15x = 360 ⇒x= 360 15 ⇒ x = 24 Step 3 Hence, ∠A = 2x = 2 × 24 = 48°, ∠B = 3x = 3 × 24 = 72°, ∠C = 4x = 4 × 24 = 96° and ∠D = 6x = 6 × 24 = 144°. (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ae-9-Quadrilaterals [4] (2) 130° Step 1 Following f igure shows the parallelogram ABCD, Step 2 According to the question, ∠D = 50°. If we look at the f igure caref ully, we notice that, the ∠A and the ∠D are the consecutive angles, the consecutive angles of a parallelogram are supplementary. T heref ore, ∠A + ∠D = 180° ⇒ ∠A + ∠50° = 180° ⇒ ∠A = 180° - 50° ⇒ ∠A = 130°. Step 3 T hus, the measurement of ∠A is 130°. (3) 9 cm (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ae-9-Quadrilaterals [5] (4) T rapezium Step 1 Following f igure shows the quadrilateral PQRS, Let's assume x is the common ratio of the angles of the quadrilateral PQRS. T heref ore, the ∠P, ∠Q, ∠R and ∠S of the quadrilateral are 8x, 12x, 11x and 9x respectively. Step 2 We know that the sum of all the angles of a quadrilateral is equal to 360°. T heref ore, 8x + 12x + 11x + 9x = 360 ⇒ 40x = 360 ⇒ x = 360/40 ⇒x=9 Step 3 Now, ∠P = 8x = 8 × 9 = 72°, ∠Q = 12x = 12 × 9 = 108°, ∠R = 11x = 11 × 9 = 99° and ∠S = 9x = 9 × 9 = 81°. Step 4 T hus, ∠P + ∠Q = 72° + 108° = 180°, ∠R + ∠S = 99° + 81° = 180° Since, the adjacent angles of the quadrilateral PQRS are supplementary and hence, the PQRS is a T rapezium. (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ae-9-Quadrilaterals [6] (5) 50° Step 1 Following f igure shows the parallelogram ABCD, According to the question, ∠DAO = 37 °, ∠BAO = 33° and ∠COD = 97°. ∠OCD = ∠BAO = 33° [Alternate interior angles] Step 2 In ΔCOD, ∠COD + ∠OCD + ∠ODC = 180° [Since the sum of all the angles of a triangle is 180°] ⇒ 97° + 33° + ∠ODC = 180° ⇒ 130° + ∠ODC = 180° ⇒ ∠ODC = 180° - 130° ⇒ ∠ODC = 50° Step 3 Hence, the value of the ∠ODC is 50°. (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ae-9-Quadrilaterals [7] (6) PS = 8.33 cm Step 1 Following f igure shows the parallelogram PQRS, We know that, the area of a parallelogram = Base × Height According to the question, DS = 5 cm, T he area of the parallelogram PQRS = PQ × DS = 10 × 5 = 50 cm2 Step 2 Similarly, QE = 6 cm T he area of the parallelogram PQRS = PS × QE ⇒ 50 = PS × 6 ⇒ PS = 50 6 ⇒ PS = 8.33 cm2 Step 3 Hence, PS = 8.33 cm2 (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ae-9-Quadrilaterals [8] (7) ∠BAC = 45° ∠DAC = 41° Step 1 In the parallelogram ABCD, AB||DC, T heref ore, ∠BAC = ∠DCA [Alternate interior angles] Similarly, since AD||BC, ∠DAC = ∠BCA [Alternate interior angles] Step 2 According to the question, ∠BCA = 41°, ∠DCA = 45° T heref ore, ∠BAC = ∠DCA = 45°, ∠DAC = ∠BCA = 41° Step 3 Hence, the measurement of the ∠BAC and ∠DAC is 45° and 41° respectively. (8) (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ae-9-Quadrilaterals [9] (9) Step 1 Let ABCD is a parallelogram as shown below. We have extended its sides to show the altitudes as CP and CQ Step 2 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. T heref ore required ratio is, AB CQ Step 3 Similarly these ratios will be as f ollowing f or other sides CD AD , CQ CP , BC CP Step 4 T heref ore we need to prove that, AB = CQ CD CQ = AD CP = BC 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 CP = CD CQ (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ae-9-Quadrilaterals [10] Step 8 Also since AB = CD and BC = AD, AB = CQ CD CQ = AD = CP BC CP (10) d. 67° Step 1 Following f igure shows the parallelogram ABCD, According to the question ∠DAB = 54° and ∠DBC = 59°. ∠A = ∠C = 54° [Since the opposite angles of a parallelogram are congruent.] Step 2 In ΔBCD, ∠DBC + ∠BCD + ∠CDB = 180° [Since the sum of all the angles of a triangle is 180°] ⇒ 59° + 54° + ∠CDB = 180° ⇒ 113° + ∠CDB = 180° ⇒ ∠CDB = 180° - 113° ⇒ ∠CDB = 67° Step 3 Hence, the value of the ∠CDB is 67°. (11) d. all of above Step 1 We know that the sum of the all angles of a parallelogram is 360°.T heref ore, x + y + z + w = 360° and l + m + n + o = 360° T heref ore x + y + z + w = l + m + n + o Step 2 We also know that, the sum of the consecutive angles of a parallelogram is supplementary. T heref ore, x + y = x + w = l + m = l + o = 180° Step 3 As we can see that all options are correct, the correct answer is, 'all of above'. (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ae-9-Quadrilaterals [11] (12) c. 4CQ (13) d. 108° Step 1 According to the question, the three angles of the quadrilateral are 64°, 77° and 111° respectively. Let's assume, the f ourth angle of the quadrilateral be x. Step 2 We know that, the sum of all interior angles of a quadrilateral is 360°. T heref ore, 64° + 77° + 111° + x = 360° ⇒ 252° + x = 360° ⇒ x = 360° - 252° ⇒ x = 108°. Step 3 Hence, the f ourth angle of the quadrilateral is 108°. (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ae-9-Quadrilaterals [12] (14) a. 1 (∠C + ∠D) 2 Step 1 Following f igure shows the quadrilateral ABCD, According to the question, AO and BO are the bisectors of ∠A and ∠B respectively. T heref ore, ∠BAO = ∠A/2, ∠ABO = ∠B/2 Step 2 We know that the sum of all angles of a quadrilateral is equals to 360°. T heref ore, ∠A + ∠B + ∠C + ∠D = 360° ⇒ ∠C + ∠D = 360° - (∠A + ∠B) -----(1) Step 3 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 = 180° - ∠A/2 - ∠B/2 ⇒ ∠AOB = 360° - ∠A - ∠B 2 ⇒ ∠AOB = 360° - (∠A + ∠B) 2 ⇒ ∠AOB = (∠C + ∠D) [From equation (1)] 2 Step 4 Hence, ∠AOB = 1 (∠C + ∠D) 2 (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited ID : ae-9-Quadrilaterals [13] (15) b. 90° Step 1 Following f igure shows the parallelogram ABCD, AO and DO are the bisectors of ∠DAB and ∠ADC respectively. T heref ore, the ∠DAB = 2∠DAO , the ∠ADC = 2∠ADO Step 2 T he ∠DAB and the ∠ADC are consecutive angles of the parallelogram ABCD, we know that, the consecutive angles of a parallelogram are supplementary. T heref ore, ∠DAB + ∠ADC = 180° ⇒ 2∠DAO + 2∠ADO = 180° ⇒ 2(∠DAO + ∠ADO) = 180° ⇒ ∠DAO + ∠ADO = 90° ------(1) Step 3 We know that, the sum of all the angles of a triangle is equal to 180°. In ΔAOD, ∠DAO + ∠ADO + ∠AOD = 180° ⇒ 90° + ∠AOD = 180° [From equation (1), ∠DAO + ∠ADO = 90°] ⇒ ∠AOD = 180° - 90° ⇒ ∠AOD = 90° Step 4 Hence, the measure of ∠AOD is 90°. (C) 2016 Edugain (www.Edugain.com) Personal use only, commercial use is strictly prohibited