(solved)Question 10 SSC-CGL 2020 March 3 Shift 2

Given that ΔABC ~ ΔDEF, and AM and DN are the altitudes of these triangles. The ratio of areas of ΔABC and ΔDEF is 9 : 25. Then (DN + AM)/(DN - AM) =?

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Parveen,

Question 10
SSC-CGL 2020 Mar 3 Shift 2

Given that ΔABC ~ ΔDEF, and AM and DN are the altitudes of these triangles. The ratio of areas of ΔABC and ΔDEF is 9 : 25. Then (DN + AM)/(DN - AM) =?

Solution in Detail

REMEMBER: The ratio of the areas of two similar triangles is equal to the square of the ratio of any pair of their corresponding sides and altitudes

$\displaystyle \therefore \bigg(\frac{DN}{AM}\bigg)^2 = \frac{25}{9}$

$\displaystyle \implies \frac{DN}{AM} = \frac 53$

By componendo dividendo

$\displaystyle \frac xy = \frac ab \implies \frac{x + y}{x - y} = \frac{a + b}{a - b}$

$\displaystyle \therefore \frac{DN + AM}{DN - AM} = \frac {5 + 3}{5 - 3}= 4\:\underline {Ans}$

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