Area Problems The ratio of the corresponding sides of two similar triangles is 3 : 4, The ratio of their areas is? 9 : 16 3 : 4 ? 3: 2 4 : 3 9 : 16 3 : 4 ? 3: 2 4 : 3 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Ratio of similar triangle = Ratio of the square of corresponding sides = (3x)2 / (4x)2 = 9x2 / 16x2 = 9/16 = 9 : 16
Area Problems A wire can be bent in the form of a circle of radius 56cm. If it is bent in the form of a square, then its area will be 8844 5544 4444 7744 8844 5544 4444 7744 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP length of wire = 2 ?r = 2 x (22/7 ) x 56 = 352 cmside of the square = 352/4 = 88cmarea of the square = 88 x 88 = 7744sq cm
Area Problems The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres? 40 120 Data inadequate None of these 40 120 Data inadequate None of these ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Let breadth = x metres. Then, length = (x + 20) metres. Perimeter = ❨ 5300 ❩ m = 200 m. 26.50 ∴ 2[(x + 20) + x] = 200 ⟹ 2x + 20 = 100 ⟹ 2x = 80 ⟹ x = 40. Hence, length = x + 20 = 60 m
Area Problems The circumference of a circle is 25 cm. Find the side of the square incribed in the circle. 25/??2 cm 29/??3 cm 23/??2 21/??3 cm 25/??2 cm 29/??3 cm 23/??2 21/??3 cm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Circumference of the circle = 2?r = 25? r = 25 / (2?)According to the formula,Side of inscribed square = r?2 = 25 / (2?r x ?2) = 25 / (??2 ) cm
Area Problems The radius of the wheel of a vehicle is 70 cm. The wheel makes 10 revolutions in 5 seconds. The speed of the vehicle is? 32.72 km/hr. 36.25 km/hr. 31.68 km/hr. 29.46 km/hr. 32.72 km/hr. 36.25 km/hr. 31.68 km/hr. 29.46 km/hr. ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Circumference = 2?r = 2 x (22 / 7) x 70 cm = 440 cm Distance travelled in 10 revolutions = 440 x 10 cm = 4400 cm = 44 m? Speed = distance / time= 44 / 5 m/sec= (44 / 5) x (18 / 5) km/hr = 31.68 km/hr
Area Problems The ratio of the area of the circumcircle and the incircle of a square is 2 : 1 ?2 : 1 1 : ?2 1 : 2 2 : 1 ?2 : 1 1 : ?2 1 : 2 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Ratio of the areas of the circumcircle and incircle of a square = [(Diagonal)2?] / [(Side)2?]= [(Side x ?2)2] / (Side)2 = 2/1 or 2 : 1
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