M.C.Q

Question 511 Mark

ABCD is a rhombus such that $\angle\text{ACB}=50^{\circ}.$ Then, $\angle\text{ADB}=?$

40°
25°
65°
130°

Answer

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,
$\angle\text{AOD}+\angle\text{DAO}+\angle\text{ADO}=180^{\circ}$
$\Rightarrow90^{\circ}+\angle\text{50}^{\circ}+\angle\text{ADO}=180^{\circ}$
$\Rightarrow\angle\text{ADO}=40^{\circ}$
$\Rightarrow\angle\text{ADB}=40^{\circ}$

View full question & answer→

Question 521 Mark

The angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. The smallest of these angles is:

45°
60°
36°
48°

Answer

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°.

View full question & answer→

MCQ 531 Mark

Three angles of a quadrilateral are $80^\circ , 95^\circ$ and $112^\circ.$ Its fourth angle is:

A
$78^\circ$
✓
$73^\circ$
C
$85^\circ$
D
$100^\circ$

Answer

Correct option: B.

$73^\circ$

Let the measure of the fourth angle be $x^\circ.$
We know that, the sum of the angles of a quadrilateral is $360^\circ.$
So, $80^\circ + 95^\circ + 112^\circ + x = 360^\circ$
$\Rightarrow 287^\circ + x = 360^\circ$
$\Rightarrow x = 73^\circ$
$\therefore$ Its fourth angle is $73^\circ.$

View full question & answer→

Question 541 Mark

The figure formed by joining the mid-points of the adjacent sides of a rectangle is a:

Rhombus.
Square.
Rectangle.
Parallelogram.

Answer

Rhombus.

Solution:
The figure formed by joining the mid points of the adjacent sides of a rectangle is a rhombus.

View full question & answer→

Question 551 Mark

In a trapezium ABCD, if E and F be the mid-point of the diagonals AC and BD respectively. Then, EF = ?

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

Answer

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

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

View full question & answer→

Question 561 Mark

In the given figure, ABCD is a parallelogram, M is the mid-point of BD and BD bisects $\angle\text{B}$ as well as $\angle\text{D}.$ Then, $\angle\text{AMB}= ?$

45°
60°
90°
30°

Answer

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.
$\therefore\text{AM}\perp\text{BD}$
$\therefore\angle\text{AMB}=90^{\circ}$

View full question & answer→

MATHS M.C.Q