M.C.Q (1 Marks)

MCQ 2011 Mark

In the given figure, there are two concentric circles of radii $6 \ cm$ and $4 \ cm$ with centre $O$. If $A P$ is a tangent to the larger circle and $B P$ to the smaller circle and length of $A P$ is $8 \ cm$, then the length of $B P$ is

A
$21 \ cm$
B
$26$
✓
$2 \sqrt{21} cm$
D
None of these

Answer

Correct option: C.

$2 \sqrt{21} cm$

In right $\triangle \text{A O P, O P}^2=A P^2+O A^2$
$=8^2+6^2=100$
In right $\triangle B O P, O P^2=B P^2+O B^2$
$\Rightarrow 100=B P^2+4^2$
$\Rightarrow B P^2=100-16=84$
$\Rightarrow B P=2 \sqrt{21} \ cm$

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MCQ 2021 Mark

A chord of a circle of radius $10 \ cm$ subtends a right angle at its centre. The length of the chord $($in $cm)$ is

A
$5 \sqrt{2}$
✓
$10 \sqrt{2}$
C
$\frac{5}{\sqrt{2}}$
D
$10 \sqrt{3}$

Answer

Correct option: B.

$10 \sqrt{2}$

Let $A B$ is a chord of circle which subtends right angle at its centre.
$\therefore$ In $\triangle \text{O A B}$, by Pythagoras theorem, we have
$(A B)^2=(O A)^2+(O B)^2$
$\Rightarrow(A B)^2=(10)^2+(10)^2$
$\Rightarrow(A B)^2=200$
$\Rightarrow A B=10 \sqrt{2} \ cm$

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MCQ 2031 Mark

In the given figure, $O$ is the centre of a circle, $P Q$ is a chord and $P T$ is the tangent at $P, \angle P O Q=70^{\circ}$, then $\angle T P Q$ is equal to

A
$55^{\circ}$
B
$70^{\circ}$
C
$45^{\circ}$
✓
$35^{\circ}$

Answer

Correct option: D.

$35^{\circ}$

(d) : In $\triangle O P Q, O P=O Q \quad$ (Radii of same circle)
$
\Rightarrow \angle O Q P=\angle O P Q
$
(Angles opposite to equal sides are equal)
$
\Rightarrow \angle O Q P=\angle O P Q=55^{\circ}
$
[By using angle sum property]
Also, $\angle O P T=90^{\circ}$
$[\because$ Tangent is perpendicular to the radius through the point of contact.]
$
\Rightarrow \angle T P Q=90^{\circ}-55^{\circ}=35^{\circ}
$
[From (i) and (ii)]

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Maths M.C.Q (1 Marks)