Maths M.C.Q (1 Marks)

$m(\operatorname{arc} A B)=m(\operatorname{arc~BC})=120^{\circ}$

Now,

$m(\operatorname{arc} A B)+m(\operatorname{arc} B C)+m(\operatorname{arc} C A)=360^{\circ}$

$\Rightarrow 120^{\circ}+120^{\circ}+m(\operatorname{arc} C A)=360^{\circ}$

$\Rightarrow 240^{\circ}+m(\operatorname{arc} C A)=360^{\circ}$

$\Rightarrow m(\operatorname{arc} C A)=360^{\circ}-240^{\circ}=120^{\circ}$

$\therefore m(\operatorname{arc} A B)=m(\operatorname{arc} B C)=m(\operatorname{arc} C A)$

$\Rightarrow \operatorname{arc} A B \cong \operatorname{arc} B C \cong \operatorname{arc} C A$

(Two arcs are congruent if their measures are equal)

$\Rightarrow$ chord $A B \cong$ chord $B C \cong$ chord $C A$ (Chords corresponding to congruent arcs of a circle are congruent)

$\therefore \triangle \mathrm{ABC}$ is an equilateral triangle. ..(All sides of equilateral triangle are equal)

Hence, the correct answer is option Equilateral triangle.

$30^{\circ}$

We know that, $6=\frac{1}{2}(12)$ and

$6 \sqrt{3}=\frac{\sqrt{3}}{2}(12)$

$\therefore \mathrm{BC}=\frac{1}{2} \mathrm{AC}$ and $\mathrm{AB}=\frac{\sqrt{3}}{2} \mathrm{AC}$

$\therefore \angle A=30^{\circ} \quad \ldots\left(\right.$ Converse of $30^{\circ}-60^{\circ}-90^{\circ}$ theorem)

$A-Q-B$

$\triangle \mathrm{ABC} \sim \triangle \mathrm{AQR}$ and $\frac{\mathrm{AB}}{\mathrm{AQ}}=\frac{7}{5}$

$\Rightarrow \triangle A B C$ is greater and $\triangle A Q R$ is smaller.

$\angle \mathrm{A}$ is a common angle.

$\therefore$ We get, A-Q-B

1

Radii of the circle having diameters $8 \mathrm{~cm}$ and $6 \mathrm{~cm}$ are $4 \mathrm{~cm}$ and $3 \mathrm{~cm}$ respectively.

$\therefore$ The distance between their centres $=4-3=1 \mathrm{~cm}$.

50

Let $\triangle A B C D$ be the rhombus, diagonal $A C=60$ and $B D=80$ we know that the diagonals of a rhombus are perpendicular bisectors of each other.

$\therefore$ Diagonals $\mathrm{AC}$ and $\mathrm{BD}$ bisect each other at point $\mathrm{M}$.

$\therefore \ln \triangle \mathrm{AMD}, \angle \mathrm{M}=90^{\circ}, \mathrm{AM}=30, \mathrm{DM}=40$

$\therefore \mathrm{AM}^2+\mathrm{DM}^2=\mathrm{AD}^2 \quad \ldots[\text { Pythagoras theorem }]$

$\therefore(30)^2+(40)^2=A D^2$

$\therefore 900+1600=\mathrm{AD}^2$

$\therefore A D^2=2500$

$\therefore A D=50 \text { units }$

Let the two circles having centres $P$ and $R$ touch each other externally at point $Q$.

Here, $P Q=5.5 \mathrm{~cm}, Q R=4.2 \mathrm{~cm}$

The two circles touch each other externally...(P-Q-R)

If the circles touch each other externally, distance between their centres is equal to the sum of their radii.

$\therefore$ By theorem of touching circles,

$\therefore P R=P Q+Q R$

$\therefore P R=5.5+4.2$

$\therefore P R=9.7 \mathrm{~cm}$

$\therefore$ The distance between the centres of the circles is $9.7 \mathrm{~cm}$.

$\frac{4}{3}$

From the given figure, we get that

$A D=3$ units, $D B=1$ units, and $A B=4$ units ...(i)

As $\triangle \mathrm{ABC} \sim \triangle \mathrm{ADE}$, we get

$\frac{A B}{A D}=\frac{B C}{D E}=\frac{A C}{A E}$ $\quad$ $\ldots$ [Ratio of Corresponding sides of similar triangles$

...[Ratio of Corresponding sides of similar triangles]

$\therefore \frac{4}{3}=\frac{\mathrm{BC}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{AE}}$ $\quad$ $\ldots[$ From (i)]

1

Let $\square A B C D$ be the given square.

In $\triangle \mathrm{ABC}$

$\angle B=90^{\circ}$

$\therefore A C^2=A B^2+B C^2 \quad \ldots[\text { Pythagoras theorem] }$

$\therefore(\sqrt{ } 2)^2=A B^2+A B^2 \quad \ldots[\text { Sides of the square] }$

$\therefore 2=2 A B^2$

$\therefore A B^2=1$

$\therefore A B=1 \text { unit }$

16

Let $\square A B C D$ be the given rectangle.

$A C=20 \text { units and } B C=12 \text { units }$

In $\triangle \mathrm{ABC}_x$

$\angle B=90^{\circ}$

$\therefore A C^2=A B^2+B C^2 \quad \ldots[\text { Pythagoras theorem }]$

$\therefore 20^2=A B^2+12^2$

$\therefore 400=A B^2+144$

$\therefore A B^2=256$

$\therefore A B=16 \text { units }$

$A B C D$ is a cyclic quadrilateral.

$2 \angle A=3 \angle C$

Now,

$\angle \mathrm{A}+\angle \mathrm{C}=180^{\circ}$ $\quad$.......(Opposite angles of a cyclic quadrilateral are supplementary$

$\left.\Rightarrow \frac{3}{2} \angle C+\angle C=180^{\circ} \quad \text { [From }(1)\right]$

$\Rightarrow \frac{5}{2} \angle C=180^{\circ}$

$\Rightarrow \angle C=\frac{2 \times 180^{\circ}}{5}=72^{\circ}$

Thus, the measure of $\angle \mathrm{C}$ is $72^{\circ}$.

Hence, the correct answer is $72^{\circ}$.

13

In $\triangle P Q R_1$

$\angle Q=90^{\circ}$

$\therefore \mathrm{PR}^2=\mathrm{PQ}^2+\mathrm{QR}^2 \quad \ldots[$ Pythagoras theorem $]$

$\therefore P R^2=169 \quad \ldots[$ Given $]$

$\therefore \mathrm{PR}=\sqrt{169}$

Taking square root both the sides,

$\therefore \mathrm{PR}=13$

$230^{\circ}$

Explanation:

The measure of an inscribed angle is half of the measure of the arc intercepted by it. By inscribed angle theorem,

$\therefore m \angle A C B=\frac{1}{2} m(\operatorname{arc} A B)$

$\therefore \mathrm{m}(\operatorname{arc} A B)=2 \mathrm{~m} \angle A C B$

$=2 \times 65^{\circ}$

$=130^{\circ}$

$\therefore \mathrm{m}(\operatorname{arc} A C B)=360^{\circ}-\mathrm{m}(\operatorname{arc} A B)$

$=360^{\circ}-130^{\circ}$

$\left.=230^{\circ} \ldots \ldots \text{[ Measure of a circle is } 360^{\circ}\right ]$

Hence, the correct answer is $230^{\circ}$.

5 : 2

In $\triangle X Y Z$ and $\triangle P Q R$

$\triangle X Y Z \sim \triangle P Q R$

$A(\triangle X Y Z)=25 \mathrm{~cm}^2, A(\triangle P Q R)=4 \mathrm{~cm}^2$

by the theorem of areas of similar triangles,

$\frac{A(\Delta X Y Z)}{A(\triangle P Q R)}=\frac{X Y^2}{P Q^2}$

$\therefore \frac{25}{4}=\frac{X Y^2}{P Q^2}$

$\therefore \frac{X Y}{P Q}=\frac{5}{2}$

$\therefore X Y: P Q=5: 2 .$

ABCD is a parallelogram. A circle with centre $O$ touches the parallelogram at $E, F, G$ and $H$.

$A B C D$ is a parallelogram.

$\therefore A B=C D\quad ...(1)$ (Opposite sides of parallelogram are equal$

$\mathrm{AD}=\mathrm{BC}\quad ...(2)$ (Opposite sides of parallelogram are equal)

Tangent segments drawn from an external point to a circle are congruent.

$\mathrm{AE}=\mathrm{AH} \quad ...(3)$

$\mathrm{DG}=\mathrm{DH} \quad ...(4)$

$\mathrm{BE}=\mathrm{BF} \quad ...(5)$

$\mathrm{CG}=\mathrm{CF} \quad ...(6)$

Adding (3), (4), (5) and (6), we get

$A E+B E+C G+D G=A H+D H+B F+C F$

$\Rightarrow A B+C D=A D+B C\quad ...(7)$

From (1), (2) and (7), we ahve

$2 A B=2 B C$

$\Rightarrow A B=B C \quad ...(8)$

From (1), (2) and (8), we have

$A B=B C=C D=A D$

$\therefore$ Parallelogram $A B C D$ is a rhombus. $\quad$ (A rhombus is a parallelogram with all sides equal)

Hence, the correct answer is rhombus.