Solution: (i) The point O where the angle bisectors meet is called the incenter of the triangle. (ii) The perpendicular drawn from point O to AB and CA are equal. i.e., OR and OQ. (iii) ∠ACO = ∠BCO....
Solution: Steps to construct: Step 1: Mark a point O. Step 2: With center O and radius 4cm and 6cm, draw two concentric circles. Step 3: Join OA and mark its mid-point as M. Step 4: With center M...
Solution : (a) Given: O is the center of the circle. To Prove : ∠AOC = 2 (∠ACB + ∠BAC). Proof: In ∆ABC, ∠ACB + ∠BAC + ∠ABC = 180° (Angles of a triangle) ∠ABC = 180o – (∠ACB + ∠BAC)….(i) In the...
Solution: (a) triangle ABC is an equilateral triangle Each angle = 60o ∠A = 60o But ∠A = ∠D (Angles in the same segment) ∠D = 600 Now ABEC is a cyclic quadrilateral, ∠A = ∠E = 180o 60o + ∠E = 180o...
Solution: Three circles with centers A, B and C touch each other externally at P, Q and R respectively and the radii of these circles are 2 cm, 3 cm and 4 cm. By joining the centers of triangle ABC...
Solution: (a) Let the circle touch the sides BC, CA and AB of the right triangle ABC at points D, E and F respectively, where BC = a, CA = b and AB = c (as showing in the given figure). As the...
Solution: (i) Join OB ∠OBA = 90° (Radius through the point of contact is perpendicular to the tangent) OB2 = OA2 – AB2 r2 = (r + 7.5)2 – 152 r2 = r2 + 56.25 + 15r – 225 15r = 168.75 r = 11.25 Hence,...
Solution: (a) From A, AP and AS are the tangents to the circle ∴AS = AP = 6 From B, BP and BQ are the tangents ∴BQ = BP = 5 From C, CQ and CR are the tangents CR = CQ From D, DS and DR are the...
Solution: Radii of the circles are 5 cm and 2.8 cm. i.e. OP = 5 cm and CP = 2.8 cm. (i) When the circles touch externally, then the distance between their centers = OC = 5 + 2.8 = 7.8 cm. (ii) When...
Solution: CT is the radius CP = 13 cm and tangent PT = 12 cm CT is the radius and TP is the tangent CT is perpendicular TP Now in right angled triangle CPT, CP2 = CT2 + PT2 [using Pythagoras axiom]...
(b) In the figure given below, if ∠ACE = 43° and ∠CAF = 62°, find the values of a, b and c (2007) Solution: (a) In ∆PQR, ∠PRQ = 90° (Angle in a semi-circle) and ∠PQR = 58° ∠RPQ = 90° – ∠PQR = 90° –...
Solution: (a) ADFE is a cyclic quadrilateral Ext. ∠FEB = ∠ADF ⇒ ∠ADF = 80° ABCD is a parallelogram ∠B = ∠D = ∠ADF = 80° or ∠ABC = 80° (b)In trapezium ABCD, AD || BC (i) ∠B + ∠A = 180° ⇒ 70° + ∠A =...
Solution:- From the question it is given that, In a ∆ABC, D and E are points on the sides AB and AC respectively. AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and AC = 8.8cm Consider the ∆ABC, EC = AC – AE =...
Solution:- From the given figure, AB = AC. If PM ⊥ AB and PN ⊥ AC We have to show that, PM x PC = PN x PB Consider the ∆ABC, AB = AC … [given] ∠B = ∠C Then, consider ∆CPN and ∆BPM ∠N = ∠M … [both...
Solution:- From the question it is give that, AP = 2PB, CP = 2PD (i) We have to prove that, ∆ACP is similar to ∆BDP and AC || BD AP = 2PB AP/PB = 2/1 Then, CP = 2PD CP/PD = 2/1 ∠APC = ∠BPD [from...
Solution: For same material, density will be same. Density = mass/Volume Mass of the smaller sphere, m1 = 1 kg Mass of the bigger sphere, m2 = 7 kg The spheres are melted to form a new sphere. So...
Solution: Let the co-ordinates of P(x, y) divides AB in the ratio m:n. A(-3,3) and B(12,-7) are the given points. Given m:n = 2:3 x1 = -3 , y1 = 3 , x2 = 12 , y2 = -7 , m = 2 and n = 3 By Section...