MCQ

MCQ 11 Mark

In the given figure, $\text{AB}$ is a diameter of the circle. If $\text{AC=BC}$, then $\angle \text{CAB}$ is equal to

A
$30^{\circ}$
✓
$45^{\circ}$
C
$60^{\circ}$
D
$90^{\circ}$

Answer

Correct option: B.

$45^{\circ}$

In circle with centre $\text{O, A B}$ is its diameter.
$\therefore \angle C=90^{\circ} ($Angle in a semi$-$circle$)$
By $\angle$ sum property of $\triangle$
$\angle A+\angle B+\angle C=180^{\circ}$
$\angle A+\angle B+90^{\circ}=180^{\circ}$
$\therefore \angle A+\angle B=90^{\circ}$
$\because C A=C B$
$\therefore \angle A=\angle B=\frac{90^{\circ}}{2}=45^{\circ}$
$\therefore \angle \text{CAB}=45^{\circ}$

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MCQ 21 Mark

If $AB = 12 \ cm, BC = 16 \ cm$ and $AB$ is perpendicular to $BC,$ then the radius of the circle passing through the points $A, B$ and $C$ is

A
$6 \ cm$
B
$8 \ cm$
✓
$10 \ cm$
D
$20 \ cm$

Answer

Correct option: C.

$10 \ cm$

$A B=12 \ cm , BC =16 \ cm$
$A C$ is the diagonal of $\triangle \text{ABC}$
and $AC$ is the diameter of the circle $\left(\because \angle B =90^{\circ}\right)$

Now $A C=\sqrt{(A B)^2+(B C)^2}$
$=\sqrt{12^2+16^2}$
$=\sqrt{144+256}$
$=\sqrt{400}$
$=20 \ cm$
$\therefore$ Radius of the circle $=\frac{\text { Diameter }}{2}$
$=\frac{20}{2} \ cm$
$ =10 \ cm$

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MCQ 31 Mark

If $P$ is a point in the interior of a circle with centre $O$ and radius $r$, then