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In a triangle ABC where DE is a line parallel to BC . DE parallel BC and AD is 1 cm and BD is 2 cm. What is the ratio of ar triangle(ABC) to the ar triangle (ADE)?

Posted by Ritika Pallai 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

We know that, the ratio of the areas of two similar triangles is equal to the square of their corresponding sides.
Also,
{tex}\Delta ADE{/tex} {tex}\sim{/tex} {tex}\triangle{/tex}ABC [given]
Now AD = 1, DB = 2
{tex}\therefore{/tex} AB = AD + DB = 1 + 2 = 3
{tex}\therefore{/tex} {tex}\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle A D E)}{/tex} = {tex}\frac{A B^{2}}{A D^{2}}{/tex} = {tex}\left(\frac{3}{1}\right)^{2}{/tex} = {tex}\frac{9}{1}{/tex}

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