Areas of two similar triangles are 36, cm^{2} | vedclass

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(D) Given,area of smaller triangle $= 36\, cm^{2}$ and area of larger triangle $= 100\, cm^{2}$.Also,length of a side of the larger triangle $= 20\, c

Vedclass · Accepted Answer

$12$

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(D) Given,area of smaller triangle $= 36\, cm^{2}$ and area of larger triangle $= 100\, cm^{2}$.Also,length of a side of the larger triangle $= 20\, cm$.Let the length of the corresponding side of the smaller triangle be $x\, cm$.By the property of areas of similar triangles,the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.$\frac{	ext{Area of larger triangle}}{	ext{Area of smaller triangle}} = \frac{(	ext{Side of larger triangle})^{2}}{(	ext{Side of smaller triangle})^{2}}$$\Rightarrow \frac{100}{36} = \frac{(20)^{2}}{x^{2}}$$\Rightarrow x^{2} = \frac{20^{2} 	imes 36}{100} = \frac{400 	imes 36}{100} = 4 	imes 36 = 144$$x = \sqrt{144} = 12\, cm$.Hence,the length of the corresponding side of the smaller triangle is $12\, cm$.

Areas of two similar triangles are $36\, cm^{2}$ and $100\, cm^{2}$. If the length of a side of the larger triangle is $20\, cm$,find the length of the corresponding side of the smaller triangle. (in $cm$)

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