Area Problems
The sides of a triangle area in the ratio of 1/3 : 1/4 : 1/5 and its perimeter is 94 cm. Find the length of the smallest side of the triangle .
Given ratio = 1/3 : 1/4 : 1/5 = 20 : 15 : 12Let length of the sides be 20k, 15k and 12k.Then, according to the question,20k + 15k + 12k = 94? 47k = 94? k = 94/47 = 2Smallest side = 12k = 12 x 2 = 24 cm
We know that area of triangle = ( base x height ) / 2So area of triangle = (BC x AD) / 2 = (AC x BE) / 2 ? (10 x 4.4) / 2 = (11 x h) / 2? h= (10 x 4.4)/11 = 4 cm
Area to be plastered = [2(l + b) x h] + (l x b) = {[2(25 + 12) x 6] + (25 x 12)} m2 = (444 + 300) m2 = 744 m2. ∴ Cost of plastering = Rs. ❨ 744 x 75 ❩ = Rs. 558
Let the width of the room be x membersThen, its area = (4x) m2Area of each new square room = (2x)m2Let the side of each new room = y metersThen, y2 = 2xClearly, 2x is a complete square when x=2? y2 = 4? y = 2 m .
Let lateral side = (5y) cm and base = (4y) cm ? perimeter = 5y + 5y + 4y = 14 ?y = 1So, the sides are 5 cm , 5 cm and 4 cm Now s= 1/2 (5 + 5 + 4) cm = 7 cm (s-a) = 2 cm (s-b) = 2 cm and (s-c) = 3 cm? Required Area = ? (7 x 2 x 2 x 3) cm2=2?21 cm2
EXAMIANS