Andhra Pradesh Board Solutions for Chapter: Similar Triangles, Exercise 15: OPTIONAL EXERCISE

In the figure, $\frac{QT}{PR}=\frac{QR}{QS}$ and $\angle 1=\angle 2$, prove that $∆PQS~∆TQR$.

Ravi is $1.82 \mathrm{m}$ tall. He wants to find the height of a tree in his backyard. From the tree's base he walked $12.20 \mathrm{m}$ along the tree's shadow to a position where the end of his shadow exactly overlaps the end of the tree's shadow. He is now $6.10 \mathrm{m}$ from the end of the shadow. How tall is the tree?

The diagonal $AC$ of a parallelogram $ABCD$ intersects $DP$ at the point $Q$, where $P$ is any point on side $AB$. Prove that $CQ\times PQ=QA\times QD$.

$∆ABC$ and$∆AMP$ are two right triangles right-angled at $B$ and $M$ respectively. Prove that $∆ABC~∆AMP$.

$∆ABC$ and$∆AMP$ are two right triangles right-angled at $B$ and $M$ respectively. Prove that $\frac{CA}{PA}=\frac{BC}{MP}$.

An aeroplane leaves an airport and flies due north at a speed of $1000 \mathrm{kmph}$. At the same time another aeroplane leaves the same airport and flies due west at a speed of $1200 \mathrm{kmph}$. How far apart will the two planes be after $1\frac{1}{2}$ hours?

In a right triangle $ABC$ right-angled at $C$. $P$ and $Q$ are points on sides $AC$ and $CB$ respectively which divide these sides in the ratio of $2:1$.

Prove that $9A{Q}^2=9A{C}^2+4B{C}^2$

In a right triangle $ABC$ right-angled at $C$. $P$ and $Q$ are points on sides $AC$ and $CB$ respectively which divide these sides in the ratio of $2:1$.

Prove that $9B{P}^2=9B{C}^2+4A{C}^2$.