Telangana Board Solutions for Chapter: Similar Triangles, Exercise 7: Exercise

$\text{D}\mathit{\text{,}}\text{E}\mathit{\text{,}}\text{F}$ are midpoints of sides $\text{BC}\mathit{\text{, }}\text{CA}\mathit{\text{, }}\text{AB}$ of $△\mathrm{ABC}$. Find the ratio of areas of $△\mathrm{DEF}$ and $△\mathrm{ABC}$.

In $∆ABC, XY \Vert AC$ and $XY$ divides the triangle into two parts of equal area. Find the ratio of  $\frac{AX}{XB}$.

The ratio of $\frac{\text{AX}}{\text{XB}}$ is $\sqrt{k}-1:1$ . Find the value of k.

Prove that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

$△ABC~△DEF\mathit{,}BC=3 \mathrm{cm},EF=4 \mathrm{cm}$ and area of $△ABC=54 {\mathrm{cm}}^2$. Determine the area of $△DEF$ ( in ${\mathrm{cm}}^2$ ).

$ABC$ is a triangle and $PQ$ is a straight line meeting $AB$ in $P$ and $AC$ in $Q$. If $AP=1 \mathrm{cm}$, $BP=3 \mathrm{cm},\mathit{ }AQ=1.5 \mathrm{cm}$ and $CQ=4.5 \mathrm{cm},$ prove that area of $△APQ=\frac{1}{16}$( area of $△ABC$).

The areas of two similar triangles are $81 {\mathrm{cm}}^2$ and $49 {\mathrm{cm}}^2$ respectively. If the altitude of the bigger triangle is $4.5 \mathrm{cm}$. Find the corresponding altitude of the smaller triangle ( in $\mathrm{cm}$ ).