Telangana Board Solutions for Chapter: Similar Triangles, Exercise 10: Optional Exercise

In the given figure,

$\frac{\mathrm{QT}}{\mathrm{PR}} =\frac{\mathrm{QR}}{\mathrm{QS}} \mathrm{and} \angle 1=\angle 2$

Prove that $△\mathrm{PQS}~△\mathrm{TQR}$

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? Express your answer in metre(m).

$△\mathrm{ABC} \mathrm{and} △\mathrm{AMP}$ are two right triangles right angled at $\mathrm{B}$ and $\mathrm{M}$ respectively.

Prove that $△\mathrm{ABC}~△\mathrm{AMP}$

$△\mathrm{ABC} \mathrm{and} △\mathrm{AMP}$ are two right triangles right angled at $\mathrm{B}$ and $\mathrm{M}$ respectively.

Prove that $\frac{\mathrm{CA}}{\mathrm{PA}} =\frac{\mathrm{BC}}{\mathrm{MP}}$

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 $1200 \mathrm{kmph}$. After $1\frac{1}{2} \mathrm{hour}$ if the two planes are $k \mathrm{km}$ apart. Find $k$ correct upto two decimal places.

In a right triangle $\mathrm{ABC}$ right angled at $\mathrm{C}$, $\mathrm{P} \mathrm{and} \mathrm{Q}$ are points on sides $\mathrm{AC} \mathrm{and} \mathrm{CB}$ respectively which divides these sides in the ratio of $2:1$

Prove that   $9{\mathrm{AQ}}^2=9{\mathrm{AC}}^2+4{\mathrm{BC}}^2$

In a right triangle $\mathrm{ABC}$ right-angled at $\mathrm{C}$, $\mathrm{P} \mathrm{and} \mathrm{Q}$ are points on sides $\mathrm{AC} \mathrm{and} \mathrm{CB}$ respectively which divides these sides in the ratio of $2:1$

Prove that   $9{\mathrm{BP}}^2=9{\mathrm{BC}}^2+4{\mathrm{AC}}^2$

In a right triangle $\mathrm{ABC}$ right-angled at $\mathrm{C}$, $\mathrm{P} \mathrm{and} \mathrm{Q}$ are points on sides $\mathrm{AC} \mathrm{and} \mathrm{CB}$ respectively which divides these sides in the ratio of $2:1$

Prove that   $9\left({\mathrm{AQ}}^2+{\mathrm{BP}}^2\right)=13{\mathrm{AB}}^2$