J P Mohindru and Bharat Mohindru Solutions for Chapter: Circles, Exercise 5: NATIONAL TALENT SEARCH EXAMINATION (NTSE)

In the figure, $\mathrm{ABC}$ is an isosceles triangle in which $\mathrm{AB}=\mathrm{AC}$ and $\angle \mathrm{ABC}=50°$. Then $\angle \mathrm{BDC}$ is equal to:

In the figure, $\mathrm{O}$ is the centre of the circle with $\mathrm{AC}$ as diameter. If $\angle \mathrm{ADB}=120°$ and the length of the radius of the circle is $6 \mathrm{cm}$, then the area of the triangle $\mathrm{ABC}$ is

In the figure, $\mathrm{ABC}$ is a right-angled triangle, right-angled at $\mathrm{A}$. If $\mathrm{AB}=3 \mathrm{cm}, \mathrm{BC}=5 \mathrm{cm}$ and $\mathrm{AC}=4 \mathrm{cm}$ then the radius of the circle is:

In the figure, $PQ$ is the tangent to the circle and the segment $PR$ intersects the circle at $M \mathrm{and} R$. If $PQ=15 \mathrm{cm} \mathrm{and} PR=25 \mathrm{cm}$, then the length of $PM$ is equal to

In the figure, $\mathrm{CD}$ is a direct common tangent to the circles intersecting each other at $\mathrm{A} \mathrm{and} \mathrm{B}$, then value of $\angle \mathrm{CAD}+\angle \mathrm{CBD}$ equals: