Maths M.C.Q

1. 40°

Solution:

ABCD is a rhombus.

$\Rightarrow\text{AD || BC}$ and $\text{AC}$ is the transversal.

$\Rightarrow\angle\text{DAC}=\angle\text{ACB}$ (alternate angles)

$\Rightarrow\angle\text{DAC}=50^{\circ}$

In $\triangle\text{AOD},$ by angle sum property,

$\angle\text{AOD}+\angle\text{DAO}+\angle\text{ADO}=180^{\circ}$

$\Rightarrow90^{\circ}+\angle\text{50}^{\circ}+\angle\text{ADO}=180^{\circ}$

$\Rightarrow\angle\text{ADO}=40^{\circ}$

$\Rightarrow\angle\text{ADB}=40^{\circ}$

2. 60°

Solution:

Let tha common multipal be x.

$\therefore$ The angle measure 3x, 4x, 5x and 6x.

Since the sum of the angles of a quadrilateral being 360°, we have

3x + 4x + 5x + 6x = 360°

⇒18x = 360°

⇒ x = 20°

$\therefore$ The angles of the quadrilateral are

3x = 3(20) = 60°,

4x = 4(20) = 80°,

5x = 5(20)= 100°,

6x = 6(20) = 120°,

$\therefore$ The smallest angle is 60°.

2. 73°

Solution:

Let the measure of the fourth angle be x^o.

We know that, the sum of the angles of a quadrilateral is 360^o.

So, 80^o + 95^o + 112^o + x = 360^o

⇒ 287^o + x = 360^o ​

⇒ x = 73^o

$\therefore$ Its fourth angle is 73^o.

4. $\frac{1}{2}(\text{AB}-\text{CD})$

Solution:

Construction: join CF and extend it to meet AB at G.

In $\triangle\text{CDF}$ and $\triangle\text{GFB},$

$\angle\text{CDF}=\angle\text{GFB}$ ...(Vertically opposite angles)

$\text{AB || CD},$

So, $\angle\text{DCF}=\angle\text{GFB,}$ ...(alternate angles)

$\text{DF = FB}$ ...(F is the mid-point of BD)

$\Rightarrow\triangle\text{CDF}\cong\triangle\text{GFB}$ ...(ASA congruence criterion)

So, $\text{CD = GB}$ and $\text{CF = GF}$ ...(C.P.C.T.)

Since E and F are the mid-points of CA and CG respectively,

we have $\text{EF}=\frac{1}{2}\text{AG}$

$=\frac{1}{2}(\text{AB}-\text{GB})$

$=\frac{1}{2}(\text{AB}-\text{CD})$

3. 90°,

Solution:

$\angle\text{ABC}=\angle\text{ADC}$ ...(Opposite angles of a parallelogram are equal)

$\Rightarrow\frac{1}{2}\angle\text{ABC}=\frac{1}{2}\angle\text{ADC}$

$\Rightarrow\angle\text{ABD}=\angle\text{ADB}$

So, $\text{AD = AB}$ ...(Sides opposite equal angles are equal.)

$\therefore\triangle\text{ABD}$ is isosceles

Also, M is the mid-point of BD.

$\therefore\text{AM}\perp\text{BD}$

$\therefore\angle\text{AMB}=90^{\circ}$