The perimeters of a square are 40 cm and 32 cm.Find the perimeter of a third square whose are is equal to difference of area of two squares.

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Solution

Side of first square $= \frac{40}{4} = 10 c m$

Side of second square $= \frac{32}{4} = 8 c m$

Area of third square $= [{(10)}^{2} - {(8)}^{2}] = (100 - 64) = 36 {c m}^{2}$

Side of third square $= \sqrt{36} c m = 6 c m$

Required perimeter $= (6 \times 4) c m = 24 c m$ .

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The perimeters of a square are 40 cm and 32 cm.Find the perimeter of a third square whose are is equal to difference of area of two squares.

Side of first square =404=10cmSide of second square =324=8cmArea of third square =[(10)2−(8)2]=(100−64)=36cm2Side of third square =√36cm=6cmRequired perimeter =(6×4)cm=24cm.

Side of first square $= \frac{40}{4} = 10 c m$

Side of second square $= \frac{32}{4} = 8 c m$

Area of third square $= [{(10)}^{2} - {(8)}^{2}] = (100 - 64) = 36 {c m}^{2}$

Side of third square $= \sqrt{36} c m = 6 c m$

Required perimeter $= (6 \times 4) c m = 24 c m$ .