Area Problems
The perimeter of two squares are 68 cm. Find the perimeter of the third square whose area is equal to the different of the areas of these two squares.
a1 = 68/4 = 17 cmand a2 = 60/4 = 15 cm [ where a1 and a2 are sides]According to the question,Area of the third square = [(17)2 - (15)2 ] = (17 + 15) (17 - 15) = 32 x 2 = 64 sq cmLet a3 = Side of the third square.According to the question, (a3)2 = 64 sq cm ? a3 = ?64 = 8 cm? Perimeter of the third square = 4 x a3 = 4 x 8 = 32 cm.
We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103. Solving the two equations, we get: l = 63 and b = 40. ∴ Area = (l x b) = (63 x 40) m2 = 2520 m2
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
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