Grade 9 Geometry-Overall

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1 ID : au-9-geometry-overall [1] Grade 9 Geometry-Overall For more such worksheets visit Answer t he quest ions (1) A chord of a circle is equal to its radius. Find the angle subtended by this chord at center. (2) If DAB = 81 and ABD = 69, f ind the value of ACB. (3) If O is center of the circle and ACB = 50, f ind the value of OAB (4) Two circles with radii of 7 and 11 are drawn with the same center. The smaller inner circle is painted orange, and the part outside the smaller circle and inside the larger circle is painted red. What is the ratio of the areas painted red to the area painted orange?

2 (5) If two horizontal lines are parallel, f ind the value of angle x. ID : au-9-geometry-overall [2] Choose correct answer(s) f rom given choice (6) If ADC = 115º and chord BC = chord BE. Find CBE. a. 130º b. 125 c. 135 d. 120 (7) AB is diameter of the circle. If A and B are connected to E, circle is intersected at C and D respectively. If AB = 28 cm and CD = 14 cm, f ind AEB. a. 75 b. 60 c. 45 d. 65 (8) WXYZ is a quadrilateral whose diagonals intersect each other at the point O such that OW = OX = OZ. If OWX = 30, then f ind the measures of OZW. a. 65 b. 60 c. 50 d. 70

3 ID : au-9-geometry-overall [3] Fill in the blanks (9) The chords AB and CD of a circle are perpendicular to each other. If radius of the circle is 21 cm, and length of the arc AQD is 54 cm, the length of arc BPC = cm (Assume π = 22/7). (10) Value of angle a is (11) T he bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q. If radius of the circle is 10 cm, the distance between points P and Q = cm. (12) AC is diameter of the semicircle and BD is perpendicular to AC. If AB = 18 cm and BC = 8 cm, f ind the length of BD.

4 (13) ABCD is a square. Two arcs are drawn with A and B as centers, and radius equal to the side of square. If arcs intersects at point E, the angle ACE =. ID : au-9-geometry-overall [4] Check True/False (14) The angles subtended by a chord at any two points of a circle are equal. True False (15) The angle subtended by an arc at the centre is half the angle subtended by it at any point on the remaining part of the circle, True False 2016 Edugain ( All Rights Reserved Many more such worksheets can be generated at

5 Answers ID : au-9-geometry-overall [5] (1) 60 Take a look at the image below The chord AB has a length equal to the radius of the circle This means that the triangle OAB is an equilateral triangle (all three sides are of the same length) AOB = 60 (2) 30 The angle subtended by a chord to two points on the circumf erence are equal, if the two points are on the same side of the chord This means that ACB = BDA Now, ABC and ABD are triangles, so the sum of the internal angles is 180 Step 4 For triangle ABD, this means DAB + ABD + BDA = BDA = 180 BDA = 180 ( ) BDA = 30

6 (3) 40 ID : au-9-geometry-overall [6] The key point to note here is that AB is a chord of the circle C is a point on the circumf erence, and O is the center Since AB is a chord, OAB = OBA Since AOC is a triangle, this means that OAB + OBA + AOB = 180 In this case, 2x( OAB) + AOB = 180 AOB = (180-2x( OAB) Step 4 We also know that the angle subtended by a chord at the centre of a circle is double the size of the angle subtended by the same chord at the circumf erence of the circle. This means AOB = 2 x ACB Step 5 So we have 2 equations 1) AOB = (180-2x( OAB) 2) AOB = 2 x ACB Step 6 ACB = 50 Using these two equations we calculate that OAB = 40

7 (4) 72:49 ID : au-9-geometry-overall [7] Following f igure shows the circles with radii 7 and 11 are drawn with the same center, We know that the area of a circle = π(r) 2 According to the question, the smaller inner circle is painted orange, and the part outside the smaller circle and inside the larger circle is painted red. The area painted orange = The area of the smaller inner circle = π(7) 2 = 49π The area painted red = The area of the larger circle - The area of the smaller inner circle = π(11) 2 - π(7) 2 = π( ) = π(121-49) = 72π Step 4 Thus, the ratio of the areas painted red to the area painted orange = 72π 72:49. 49π = =

8 (5) 70 ID : au-9-geometry-overall [8] We know that the angle made by a straight line is 180. Theref ore, we can write, y = 180 or y = = 70 When a straight line cuts any two parallel lines, its Corresponding Angles are equal. Since angles x and y are Corresponding angles. Theref ore, x = y = 70...[Corresponding angles of two parallel lines are equal] (6) a. 130º ABCD is a cyclic quadrilateral since all 4 points A, B, C and D lie on the circumf erence The opposite angles of a cyclic quadrilateral add up to 180º From this, we f ind ADC + CBA = 180º CBA = 180º - ADC CBA = 180º - 115º CBA = 65º We know that chord BC = chord BE We have also learnt that the center of the circle lies on the bisector of CBE This means that BA is the bisector of CBE Theref ore CBE = 2 CBA = 2 x 65º = 130º (7) b. 60 (8) b. 60

9 (9) 12 ID : au-9-geometry-overall [9] Consider the representative image below We have drawn a diameter RS parallel to AB Now RS is a diameter and perpendicular to CD (whence RS is parallel to AB and AB is perpendicular to CD) Theref ore RS bisects arc CD Arc RC = arc RD Since RS is a diameter Arc RS = Arc RC + Arc CS = 180 Arc RD + Arc CS = 180 This implies that the remaining arcs should add up to the remaining 180 Arc RC + Arc SD = 180 Step 4 Now looking at the question We know length of arc AQD is 54 cm, and need to f ind length of arc BPC From the previous analysis, we know that arc AQD and arc BPC cover a semi-circle, so the total length of the two arcs is half the circumf erence Circumf erence of the circle = 2 x 22 x 21 = length of arc AQD + length of arc BPC = 0.5 x 132 = 66 length of arc BPC = 66 - length of arc AQD = = 12 cm

10 (10) 47 ID : au-9-geometry-overall [10] If you look at the angles 90 and a + 43, these are opposite angles We know that opposite angles are equal. Theref ore a + 43 = 90 a = a = 47

11 (11) 20 ID : au-9-geometry-overall [11] Take a look at the image below We consider the cyclic quadrilateral ABCD. Bisecting the angle DAB, we get the bisector which intersects the circle at P Bisecting the opposite angle DCB, we get the bisector intersecting the circle at Q Given the radius of the circle, we need to f ind the length of the segment PQ Since ABCD is a cyclic quadrilateral, DAB + BCD = ( DAB + BCD) = 90 2 PAD + QCD = 90 (since AP bisects DAB and QC bisects BCD) Now consider triangles QCD And QAD We see that QCD = QAD (angles subtended by the same segment - QD in this case - to points on the circumf erence are equal) PAD + QAD = 90 PAQ= 90 This means that PAQ is the angle in a semi-circle PQ is the diameter of the circle Step 4 As PQ is the diameter of the circle, it is twice the radius PQ = 2 x 10 = 20 cm

12 (12) 12 ID : au-9-geometry-overall [12] Following f igure shows the semicircle with the diameter AC, According to the question, AB = 18 cm and BC = 8 cm. The diameter(ac) of the semicircle = AB + BC = = 26 cm Let's draw a line OD f rom the center of the semicircle, as shown in the f ollowing f igure, The line OD, OA and OC are the radius of the semicircle. Theref ore, OD = OA = OC = AC/2 = 26/2 = 13 cm Now, OB = OC - BC = 13-8 = 5 cm In ΔOBD, OBD = 90 BD 2 = OD 2 - OB 2 [By the Pythagorean theorem] BD 2 = BD 2 = 144 BD 2 = 12 2 BD = 12 cm Step 4 Hence, the length of the BD is 12 cm. (13) 30 (14) False

13 (15) False ID : au-9-geometry-overall [13]

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