CBSE Questions for Class 9 Maths Quadrilaterals Quiz 3 - MCQExams.com

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.

Given, one angle is $$108^o$$ and remaining three angles are equal.

Let the remaining angles be $$x^o,y^o,z^o$$.

Since they are equal, $$x^o+y^o+z^o=$$ $$x^o+x^o+x^o=3x^o$$.

Then, $$108^o+x^o+y^o+z^o=360^o$$

$$\Rightarrow$$ $$108^o+3x^o=360^o$$

$$\Rightarrow$$ $$360^o - 108^o =3x^{\circ}$$

$$\Rightarrow$$ $$252^o =3x^{\circ}$$

$$\Rightarrow$$ $$x^o=\dfrac{252^o}{3}  =84^{\circ}$$.

Therefore, each of the three remaining angle is $$x^o=84^o$$.

Hence, option $$C$$ is correct.

The given angles are, $$75^o, 90^o, 75^o$$.

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.

The given angles are $$110^o, 80^o, 70^o, 95^o$$.

Then,

$$110^o+80^o+70^o+95^o$$.

$$=190^o + 165^o $$

$$=355^o \ne 360^o$$.

Therefore, the given angles are not the angles of a quadrilateral.

Hence, option $$B$$ is correct.

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.

The given angles are, $$70^o, 95^o, 105^o$$.

Let the fourth angle be $$x^o$$.

Then, $$70^o+95^o+105^o+x^o=360^o$$.

$$\Rightarrow$$ $$360^o - (70^o + 95^o + 105^o ) =x^{\circ}$$

$$\Rightarrow$$ $$x^o=360^o - 270^o  =90^{\circ}$$.

Therefore, the fourth angle is $$x^o=90^o$$.

Hence, option $$A$$ is correct.

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.

The given angles are, $$60^o, 110^o, 86^o$$.

Let the fourth angle be $$x^o$$.

Then, $$60^o+110^o+86^o+x^o=360^o$$.

$$\Rightarrow$$ $$360^o - (60^o + 110^o + 86^o ) =x^{\circ}$$

$$\Rightarrow$$ $$x^o=360^o - 256^o  =104^{\circ}$$.

Therefore, the fourth angle is $$x^o=104^o$$.

Hence, option $$A$$ is correct.

We know, by angle sum property, the sum of all angles of a quadrilateral is $$360^o$$.

The given angles are, $$70^o, 120^o, 65^o$$.

Let the fourth angle be $$x^o$$.

Then, $$70^o+120^o+65^o+x^o=360^o$$    ...{Angle sum property of a quadrilateral}

$$\Rightarrow$$ $$360^o - (70^o + 120^o + 65^o ) =x^{\circ}$$

$$\Rightarrow$$ $$x^o=360^o - 255^o  =105^{\circ}$$.

Therefore, the fourth angle is $$x^o=105^o$$.

Hence, option $$B$$ is correct.

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.

The given angles are $$\angle A=80^o, \angle B=70^o, \angle C=130^o, \angle D$$.

Then, $$\angle A+\angle B+\angle C+\angle D=360^o$$

$$\implies$$ $$80^o+70^o+130^o+\angle D=360^o$$.

$$\Rightarrow$$ $$\angle D=360^o - (80^o + 70^o + 130^o )$$

$$\Rightarrow$$ $$\angle D=360^o - 280^o  =80^{\circ}$$.

Therefore, the unknown angle is $$\angle D=80^o$$.

Hence, option $$A$$ is correct.

Since $$M$$ is the midpoint of $$AB$$ and $$N$$ is the midpoint of $$AC,$$ we have

$$ AB = AM+ MB $$ and $$ AB = AN + NC $$

So,

$$\begin{aligned}{}AB&=AC\quad\quad\quad\quad\quad\quad[\text{Given}]\\AM + BM& = AN + CM\\AM+BM&=AN+BM\quad\quad\quad[\text{Given }BM=CN]\\AM &= AN\end{aligned}$$

Hence proved, $$AM=AN.$$