In - ABC- a line is drawn parallel to BC to meet sides AB and AC in D and E respectively- If the area of the - ADE is 19 times area of the - ABC- then the value of ADAB is equal to-

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Standard X

Mathematics

Relation between Areas and Sides of Similar Triangles

In Δ ABC, a...

Question

In $Δ A B C$ , a line is drawn parallel to $B C$ to meet sides $A B$ and $A C$ in $D$ and $E$ respectively. If the area of the $Δ A D E$ is $\frac{1}{9}$ times area of the $Δ A B C$ , then the value of $\frac{A D}{A B}$ is equal to:

\frac{1}{3}

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\frac{1}{4}

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\frac{1}{5}

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\frac{1}{6}

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Solution

The correct option is D $\frac{1}{3}$
By theorem on ratio of areas of similar triangles, we get
$\frac{A (△ A D E)}{A (△ A B C)} = {(\frac{A D}{D B})}^{2}$
$∴ \frac{1}{9} = \frac{A D^{2}}{D B^{2}}$
$∴ \frac{A D}{D B} = \frac{1}{3}$ .

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Similar questions

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If a line intersects sides AB and AC of a $Δ A B C$ at D and E respectively and is parallel to BC, prove that $\frac{A D}{A B} = \frac{A E}{A C}$ . [2 MARKS]

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In ΔABC, a line is drawn parallel to BC to meet sides AB and AC in D and E respectively. If the area of the ΔADE is 19 times area of the ΔABC, then the value of ADAB is equal to:

The correct option is D 13By theorem on ratio of areas of similar triangles, we getA(△ADE)A(△ABC)=(ADDB)2∴19=AD2DB2∴ADDB=13.

In $Δ A B C$ , a line is drawn parallel to $B C$ to meet sides $A B$ and $A C$ in $D$ and $E$ respectively. If the area of the $Δ A D E$ is $\frac{1}{9}$ times area of the $Δ A B C$ , then the value of $\frac{A D}{A B}$ is equal to:

The correct option is D $\frac{1}{3}$
By theorem on ratio of areas of similar triangles, we get
$\frac{A (△ A D E)}{A (△ A B C)} = {(\frac{A D}{D B})}^{2}$
$∴ \frac{1}{9} = \frac{A D^{2}}{D B^{2}}$
$∴ \frac{A D}{D B} = \frac{1}{3}$ .