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Question Number 152204 by cherokeesay last updated on 26/Aug/21

Commented by Paradoxical last updated on 26/Aug/21

Commented by cherokeesay last updated on 26/Aug/21

$N i c e!$ $t h a n k y o u!$
	Answered by Rasheed.Sindhi last updated on 26/Aug/21

	$\underset{―}{\overset{―}{}}$ $L e n g t h = x, W i d t h = y$ $S m a l l t r i a n v l e s a r e s i m i l a r :$ $(∵ ∢ s a r e c o n g r u e n t)$ $∴ \frac{x}{3} = \frac{4}{y} \Rightarrow x y = 12 \Rightarrow A = 12 sq units .$ $\underset{―}{\overset{―}{}}$
	Commented by cherokeesay last updated on 26/Aug/21

	$v e r y c l e v e r!$ $t h a n k y o u!$
	Answered by Paradoxical last updated on 26/Aug/21

	Commented by Rasheed.Sindhi last updated on 27/Aug/21

	$p a r a d o x i c a l s i r, W h y h a v e y o u p u t$ $t w o i d e n t i c a l v e r s i o n s o f y o u r$ $a n s w e r ? :)$
	Answered by Rasheed.Sindhi last updated on 27/Aug/21

	$⟨ ⟨ T h r o u g h H y p o t e n u s e s ⟩ ⟩$ $Sum of Hypotenuses of small triangles$ $= Hypotenuse of big triangle$ $L e n g t h = x, W i d t h = y$ $\sqrt{4^{2} + x^{2}} + \sqrt{y^{2} + 3^{2}} = \sqrt{{(4 + y)}^{2} + {(x + 3)}^{2}}$ $16 + x^{2} + y^{2} + 9 + 2 \sqrt{4^{2} + x^{2}} \sqrt{y^{2} + 3^{2}}$ $= 16 + 8 y + y^{2} + x^{2} + 6 x + 9$ $2 \sqrt{4^{2} + x^{2}} \sqrt{y^{2} + 3^{2}} = 8 y + 6 x$ $\sqrt{4^{2} + x^{2}} \sqrt{y^{2} + 3^{2}} = 4 y + 3 x$ $(16 + x^{2}) (y^{2} + 9) = 16 y^{2} + 24 x y + 9 x^{2}$ $x^{2} y^{2} + 9 x^{2} + 16 y^{2} + 144 = 16 y^{2} + 24 x y + 9 x^{2}$ ${(x y)}^{2} - 24 (x y) + 144 = 0$ ${(x y - 12)}^{2} = 0$ $x y = 12$
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