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: Draw a line segment BC = 5cm. Step 2: With Center as B and radius 5cm, with center as C and radius 5cm draw two arcs which intersect each other at point A. Step...
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]...
Solution: (a) Construction: Join BC, and AC then ABCD is a cyclic quadrilateral. Now in ∆DCF Ext. ∠2 = x + z and in ∆CBE Ext. ∠1 = x + y Adding (i) and (ii) x + y + x + z = ∠1 + ∠2 2 x + y + z =...
(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 given that, (i) The area of ∆ABC to the area of ∆PQR if their corresponding sides are in the ratio 1 : 3 Then, ∆ABC ~ ∆PQR area of ∆ABC/area of ∆PQR = BC2/QR2 So,...
Solution:- From the figure, PQ || AC, AP = 4 cm, PB = 6 cm and BC = 8 cm ∠BQP = ∠BCA … [because alternate angles are equal] Also, ∠B = ∠B … [common for both the triangles] Therefore, ∆ABC ~ ∆BPQ...
Solution:- From the given figure it is given that, CM and RN are respectively the medians of ∆ABC and ∆PQR. (i) We have to prove that, ∆AMC ~ ∆PQR Consider the ∆ABC and ∆PQR As ∆ABC ~ ∆PQR ∠A = ∠P,...
Solution:- From the figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC We have to prove that, BE/DE = AC/BC Consider the ∆ABC and ∆DEB, ∠C = 90o ∠A + ∠ABC = 90o [from the figure equation (i)] Now in ∆DEB ∠DBE +...
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...
(v) Let E be the event of getting a king or a queen. There will be 4 cards of king and 4 cards of queen. Number of favourable outcomes = 4+4 = 8 P(E) = 8/52 = 2/13 Hence the probability of getting a...
Solution: Edge of the cube, a = 44 cm Volume of cube = a3 = 443 = 85184 cm3 Diameter of shot = 4 cm So radius of shot, r = 4/2 = 2 cm Volume of a shot = (4/3)r3 = (4/3)×(22/7)×23 = 704/21 cm3 Number...
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: Given radius of the hemisphere, r = 8 cm Volume of the hemisphere, V = (2/3)r3 = (2/3)×83 = (1024/3) cm3 Radius of cone, R = 6 cm Since hemisphere is melted and recasted into a cone, the...
solution; Given height of the tent above the ground = 85 m Height of the cylindrical part, H = 50 m height of the cone, h = 85-50 h = 35 m Diameter of the base, d = 168 m Radius of the base of...
Given edge of the cube, a = 14 cm Radius of the cone, r = 14/2 = 7 cm Height of the cone, h = 14 cm Volume of the cube = a3 = 143 = 14×14×14 = 2744 cm3 Volume of the cone = (1/3)r2h =...
Solution: (i) Given radius of the sphere, r = 14 cm Surface area of the sphere = 4r2 = 4×(22/7)×142 = 4×22×14×2 = 2464 cm3 Hence the surface area of the sphere is 2464 cm2. (ii) Given radius of the...
Solution: Given base area of the cone = 616 cm2 r2 = 616 (22/7)×r2 = 616 r2 = 616×7/22 r2 = 196 r = 14 Given volume of the cone = 9856 cm3 (1/3)r2h = 9856 (1/3)×(22/7)×142 ×h = 9856 h =...
Solution: (i)Given height of the tent, h = 10 m Radius, r = 24 m We know that l2 = h2+r2 l2 = 102+242 l2 = 100+576 l2 = 676 l = √676 l = 26 (ii) Curved surface area = rl = (22/7)×24×26 = 13728/7 m2...
Solution: Given height of a cone, h = 9 cm Volume of the cone = 48 (1/3)r2h = 48 (1/3)r2×9 = 48 3r2 = 48 r2 = 48/3 = 16 r = 4 So diameter = 2×radius = 2×4 = 8 cm Hence the diameter of the cone is 8...
Solution: Given slant height of the cone, l = 10 cm Base radius, r = 7 cm Curved surface area of the cone = rl = (22/7)×7×10 = 220 cm2 Hence the curved surface area of the cone is 220...
Solution: Given line: 3x – 2y = -4 Slope (m1) is given by 2y = 3x + 4 y = (3/2) x + 2 So, m1 = 3/2 Now, the slope of the line parallel to the given line will have the same slope as 3/2 = m And the...
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...