2 Marks Questions

Question 12 Marks

Three angles of a quadrilateral are $75^\circ , 90^\circ $ and $75^\circ $. Find the measure of the fourth angle.

Answer

Given: Three angles of a quadrilateral are $75^\circ , 90^\circ $ and $75^\circ .$
Let the fourth angle be $x.$
Using angle sum property of quadrilateral,
$75^\circ + 90^\circ + 75^\circ + x = 360^\circ $
$\Rightarrow 240^\circ + x = 360^\circ $
$\Rightarrow x = 360^\circ - 240^\circ $
$\Rightarrow x = 120^\circ $
So, the measure of the fourth angle is $120^\circ .$

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Question 22 Marks

The angles of a quadrilateral are in the ratio $2 : 4 : 5 : 7$. Find the angles.

Answer

Let $\angle\text{A}=2\text{x}^{\circ}.$
Then $\angle\text{B}=(4\text{x})^{\circ},\angle\text{C}=(5\text{x})^{\circ}$ and $\angle\text{D}=(7\text{x})^{\circ}$
Since the sum of the angles of a quadrilateral is $360^{\circ}$,
we have: $2\text{x}+4\text{x}+5\text{x}+7\text{x}=360^{\circ}$
$\Rightarrow18\text{x}=360^{\circ}$
$\Rightarrow\text{x}=20^{\circ}$
$\therefore\angle\text{A}=40^{\circ};\angle\text{B}=80^{\circ};\angle\text{C}=100^{\circ};\angle\text{D}=140^{\circ}$

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Question 32 Marks

In the adjoining figure, $ABCD$ is a trapezium in which $AB \| DC$. If $\angle\text{A}=55^{\circ}$ and $\angle\text{B}=70^{\circ},$ find $\angle\text{C}$ and $\angle\text{D}.$

Answer

We have $AB \| DC.$
$\angle\text{A}$ and $\angle\text{D}$ are the interior angles on the same side of transversal line $AD$, whereas $\angle\text{B}$ and $\angle\text{C}$ are the interior angles on the same side of transversal line $BC.$
Now, $\angle\text{A}+\angle\text{D}=180^{\circ}$
$\Rightarrow\angle\text{D}=180^{\circ}-\angle\text{A}$
$\therefore\angle\text{D}=180^{\circ}-55^{\circ}=125 ^{\circ}$
Again, $\angle\text{B}+\angle\text{C}=180^{\circ}$
$\Rightarrow\angle\text{C}=180^{\circ}-\angle\text{B}$
$\therefore\angle\text{C}=180^{\circ}-70^{\circ}=110^{\circ}$

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Question 42 Marks

In a parallelogram $ABCD$, points $M$ and $N$ have been taken on opposite sides $AB$ and $CD$ respectively such that $AM = CN.$ Show that $AC$ and $MN$ bisect each other.

Answer

In $\triangle\text{AMO}$ and $\triangle\text{CNO}$
$\angle\text{MAO = }\angle\text{NCO}$ ($AB \| CD$, alternate angles)
$\text{AM = CN}$ (given)
$\angle\text{AOM}=\angle\text{CON}$ (vertically opposite angles)
$\therefore\triangle\text{AMO}\cong\triangle\text{CNO}$ (by $ASA$ congruence criterion)
$\Rightarrow\text{AO = CO}$ and $\text{MO = NO} (CP.C.T.)$
$\Rightarrow\text{AC}$ and $\text{MN}$ bisect each other.

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Maths 2 Marks Questions

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