Area Problems The ratio of the corresponding sides of two similar triangles is 3 : 4, The ratio of their areas is? 9 : 16 4 : 3 3 : 4 ? 3: 2 9 : 16 4 : 3 3 : 4 ? 3: 2 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 Find the area of a rectangle having 15m length and 8m breadth. 120 sq m 111 sq m 125 sq m 115 sq m 120 sq m 111 sq m 125 sq m 115 sq m ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Required area = Length x Breadth= 15 x 8 = 120 sq m
Area Problems If 88 m wire is required to fence a circular plot of land, then the area of the plot is? 616 m2 556 m2 526 m2 None of these 616 m2 556 m2 526 m2 None of these ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Circumference of circular plot= 88 m? 2 x (22/7) x r = 88? r = 88 x (7/22) x (1/2) = 14 m Now area = ?r2=( 22/7) x 14 x 14 m2= 616 m2
Area Problems The radius of a circle is so increased that its circumference increased by 5%. The area of the circle, then increases by 11.25% 10.25% 12.5% 10.5 % 11.25% 10.25% 12.5% 10.5 % ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Increase in circumference of circle = 5%? Increase in radius is also 5%.Now, increase in area of circle = 2a + (a2/100) %Where, a = increase in radius= 2 x 5 + (5 x 5)/100 % = 10.25%
Area Problems Angles of a quadrilateral are in the ratio 3 : 4 : 5 : 8. The smallest angle is 36° 18° 54° 40° 36° 18° 54° 40° ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Let First angle = 3kSecond angle = 4kThird angle = 5k and Fourth angle = 8kWe know 3k + 4k + 5k + 8k = 360°? 20k = 360° ? k = 18 °Measure of smallest angle = 3k =3 x 18 ° = 54 °
Area Problems If the radius of a circle is increased by 6%, find the percentage increase in its area. 15% 17% 12.36% 8.39% 15% 17% 12.36% 8.39% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Given that, a = 6 According to the formula,Percentage increase in area= 2a + [a2/100]%= 2 x 6 + [36/100]%= (12 + 0.36)%= 12.36%
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