SSC CHSL MATHS QUIZ

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Question 1:

In a triangle $P Q R$, side $Q P$ is produced to a point $S$. If $\angle R P S=108^{\circ}$ and $\angle Q=20^{\circ}$, then $3 \angle R+2 \angle Q$ is equal to:

Question 2:

Side $\mathrm{BC}$ of a triangle $\mathrm{ABC}$ is produced to a point $\mathrm{D}$. If $\angle \mathrm{A}=40^{\circ}$ and $\mathrm{AC}$ $=\mathrm{BC}$, then $2 \angle \mathrm{ACD}+3 \angle \mathrm{B}$ is equal to:

Question 3:

In a quadrilateral $\mathrm{ABCD}$, the bisectors of $\angle C$ and $\angle D$ meet at point $\mathrm{E}$. If $\angle C E D=67^{\circ}$ and $\angle A=57^{\circ}$, then the measure of $\angle B$ is

Question 4:

In the figure, $L$ is the centre of the circle, and ML is the perpendicular to LN. If the area of the $\Delta \mathrm{MLN}$ is 36 , then the area of the circle is:

Question 5:

If the centre of the incircle of the triangle ABC is O and ∠BOC = 110°, then what is the value of ∠BAC?

Question 6:

In $\triangle A B C, \angle C=60^{\circ}, \angle A=75^{\circ}, A D \perp B C$, where $D$ is a point on $B C, B E \perp A C$ at $E$, where $E$ is a point on $A C . A D$ and $B E$ intersect each other at $\mathrm{H}$. What is the measure of $\angle \mathrm{CHD}$ ?

Question 7:

In $\triangle A B C$, the sides $\mathrm{AB}$ and $\mathrm{AC}$ are produced to $\mathrm{P}$ and $\mathrm{Q}$, respectively. The bisectors of $\angle \mathrm{PBC}$ and $\angle \mathrm{QCB}$ intersect at a point $\mathrm{O}$. If $\angle \mathrm{A}=56^{\circ}$, then $\angle \mathrm{BOC}$ equals:

Question 8:

In $\triangle A B C, \angle B=90^{\circ}, A B=28 \mathrm{~cm}$ and $B C=45 \mathrm{~cm}$. $D$ is a point on $B C$ such that $A D$ bisects $\angle A$. The length (in cm) of $B D$ is:

Question 9:

The perimeters of two similar triangles $\Delta \mathrm{ABC}$ and $\Delta \mathrm{PQR}$ are $36 \mathrm{~cm}$ and $24 \mathrm{~cm}$ respectively. If $\mathrm{PQ}=10 \mathrm{~cm}$, then $\mathrm{AB}$ is:

Question 10:

A, B, C, D are four points on a circle. AC and BD intersect at a point $\mathrm{E}$ such that $\angle \mathrm{BEC}=$ $120^{\circ}$ and $\angle \mathrm{ECD}=40^{\circ} \cdot \angle \mathrm{BAC}$ is: