Question

3. a) It is given that DEC = LEDB = x', and LABD = LACE = x - 20". Let ZBOC = 120. (0) Find the value of x. (W) Find the measure of ZEFD. (1) Show that EB-DC. D b) A circle of center A and radius AC = b is shown below, with BC = a, and AB = c. a+b2-c? (i) Prove that cos C = 2ab (1) Find ED. B D с c) In the figure below the two circles have the same center and the sides of the square ABCD are tangent to the smaller circle. If OA = 1 cm, find the area of the blue shaded area, between the square and the smaller circle.

Answer 1

S- irt 8 3. a) Given: LDEC = LEDB= x² and LABD = LACE = x-20. Let <BOC = 1200 A (i) find the value of a of x. E we know that F X- 20 LBAC= 2 Вос 2 J2-20 -X T20 С LBAC = 12 x 120 B CBAC = 60 Now, LBCE oc same [:LB D E = x (Given) angle subtended by arc (BE). 3 [ angle subtended by same are OC), LDEC= x (Given) LDBC=X In SABC, LA = 60°, LB = x + x-go <= x +X-20 Lat & Bt Lc = 180° Go tx+x-20 + x + x -20 = 180 180 - 20 160 - Answer x = 40 V

(ii) Find the measure In D EFD, Lave. of <EFD we LE = ycº, <D = x s LEFD = no In D EFD, we have. <EFD + x + x = 180 from above, solution, oc=40 Hence, LEFD = 180 2x 180-2(40) = 180 - 80 LEFD = 100 »Answer SA show that EB = DC. Here LDECE LEDB = x = 40 (Given) ,دکی we can say E B = DC Because, If arcs are same then two angles subtended two equal sides both equal (hence, Both Both chord EB & DC are quals.) EB=DC arcs be

b) Radius Ac=b and BC = a, AB=C ed ish i prove that E ח 2 Cos C = atb Qab B с Taking D ABC с A b a h G A-x-P с. b b- x from s ABP, c²= x + h2 a ² = (6-2)² th 2 from a CBP Also Cosc こ 3 subtract (2 we have from 0 22 (b-x) c²_a² 2 2 1 2 2 8 - 6² - x² + abx 2 c²_a² = ² - a c² - a² +6² 2 bx - 62 b (2x- b) ar

c - a² +6² bx + buc c - a² +6² a + 6² - 6x = bx 6² - 6x = 2 bx tas 2 2. 2 b(bax) bxtat_c² 1 2 b-x bxt a ²_c² ab Goose 2 (6x + a²_c²) 2 ab 2 bx + 2a² _ 20² 2 ab Cos c = Qbx ta gab a ² - c² + a²_c² Cos c = + a²-c2 Qab [ 26x + a ² cl 2 So Cos ca b²+ a²c2 gab