The perimeters of two squares are 68 cm and 60 cm. Find the perimeter of the third Square whose area is equal to the difference of the areas of these two squares.
Correct Answer: Description for Correct answer:
\( \Large a_{1} \)=\( \Large \frac{68}{4} \)=17cm
and \( \Large a_{2} \)=\( \Large \frac{60}{4} \)=15cm
According to the question,
Area of the third square
= [\( \Large (17)^{2}-(15)^{2} \)]
= (17 + 15) (17 - 15)
= \( \Large 32\times 2 \)
= 64 sq cm
Let \( \Large a_{3} \) = Side of the third square
According to the question.
\( \Large (a_{3})^{2} \) = 64 sq cm
\( \Large a_{3} \) = \( \Large \sqrt{64} \) = 8 cm
Perimeter of the third square
= \( \Large 4\times a_{3}=4\times 8 \)=32cm
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