Math Question and Answers


Question and Answers Forum

All Questions Topic List
Mensuration Questions
Previous in All Question Next in All Question
Previous in Mensuration Next in Mensuration
Question Number 145652 by imjagoll last updated on 07/Jul/21

	Answered by john_santu last updated on 07/Jul/21

	$a r e a o f s q u a r e = {(4 + y)}^{2}$ $l e t l e n g t h s i d e o f a r e a I = x$ $t h e n a r e a o f I = 4 x$ $t h e s u m a r e a o f 3, 4, 5 = 12 x = x y$ $t h e n w e g e t y = 12 a n d a r e a o f s q u a r e$ $= {(4 + 12)}^{2} = 16^{2} = 256$
	Answered by mr W last updated on 07/Jul/21

	$x = s i d e l e n g t h o f s q u a r e$ $a r e a 1 = 4 \times (x - \frac{x}{5}) = \frac{x^{2}}{5}$ $\Rightarrow x = 16$ $t o t a l a r e a = 16^{2} = 256$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum