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 +...
(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: 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...