Maths M.C.Q

2. 50º

Solution:

We know that, digonals of a rectangle are equal in length.

$\therefore\ \text{AD}=\text{BD}$

$\Rightarrow\ \frac{1}{2}\text{AC}=\frac{1}{2}\text{BD}$ [dividing both sides by 2]

$\Rightarrow\ \text{OA}=\text{OB}$ [since, O is the mid-point of AC and BD]

$\Rightarrow\ \angle2=\angle1$ [angles opposite to equal sides are equal]

$=25^\circ$

$\therefore\ \angle3=\angle1+\angle2$

[exterior angle is equal to the sum of two opposite intersite interior angles]

$=25^\circ+25^\circ=50^\circ$

Hence, the acute angle between the diagonals is 50°

3. Trapezium.

Solution:

Given, ratio of angles of quadrilateral ABCD is 3 : 7 : 6 : 4.

Let angles of quadrilateral ABCD be 3x, 7x, 6x and 4x, respectively. We know that, sum of all angles of a quadrilateral is 360°.

3x + 7x + 6x + 4x = 360°

⇒ 20x = 360°

$\Rightarrow\frac{\text{x}=360^\circ}{20^\circ=18^\circ}$

$\therefore$ Angles of the quadrilateral are

$\angle\text{A}=3\times18=54^\circ$

$\angle\text{B}=7\times18=126^{\circ} $

$\angle\text{C}=6\times18=108^\circ$

$\angle\text{D}=4\times18=72^\circ$

From figure, $\angle\text{BCE}=180^\circ-\angle\text{BCD}$ [linear pair axiom]

$\Rightarrow\ \angle\text{BCE}=180^\circ-108^\circ=72^\circ$

Here, $\angle\text{BCE}=\angle\text{ADC}=72^\circ$

Since, the of cointerior angles,

$\therefore\ \text{BC}||AD$

Now, sum of cointerior angles,

$\angle\text{A}+\angle\text{B}=126^\circ+54^\circ=180^\circ$

$\angle\text{C}+\angle\text{D}=108^\circ+72^\circ=180^\circ$

Hence, ABCD is a trapezium.

3. $\text{DE}=\text{EF}$

Solution:

We need $\text{DE}=\text{EF}.$

Hence, (c) is the correct answer.

3. 38º

Solution:

Given, $\angle\text{AOB}=70^\circ$ and $\angle\text{DAC}=32^\circ$

$\therefore\angle\text{ACB}=32^\circ$ [AD||BC and AC is transver sal]

Now, $\angle\text{AOB}+\angle\text{BOC}=180^\circ$ [linear pair axiom]

$\Rightarrow\ \angle\text{BOC}=180^\circ-\angle\text{AOB}=180^\circ-70^\circ=110^\circ$

Now, in $\Delta\text{BOC},$ we have

$\angle\text{BOC}+\angle\text{BCO}+\angle\text{OBC}=180^\circ$[by angle sum property of a tringle]

$\Rightarrow\ 110^\circ+32^\circ\angle\text{OBC}=180^\circ$ $[\because\angle\text{BCO}=\angle\text{ACB}=32^\circ]$

$\Rightarrow\ \angle\text{OBC}=180^\circ-(110^\circ+32^\circ)=38^\circ$

$\therefore\ \angle\text{DBC}-\angle\text{OBC}=38^\circ$

4. A parallelogram.

Solution:

Since the line segment joiing the mid-poients of any two sides of a triangle is parallel to third side and is half to it, so

$\therefore\ \text{DE}=\frac{1}{2}\text{BC}$ and $\text{DE}||\text{BC}$

Similarly, $\text{DP}=\frac{1}{2}\text{AO}$ and $\text{DP||AO}$

And $\text{EQ}=\frac{1}{2}\text{AO}$ and $\text{EQ||AO}$

$\therefore\ \text{DP}=\text{EQ}[\therefore\text{Each}=\frac{1}{2}\text{AO}]$

And $\text{DP||EQ}[\therefore\text{Each}=\frac{1}{2}\text{AO}]$

Now, DEQP is quadrilateral in which one pair of its opposite is equal and parallel.

Therefore, quadrilateral DEQP is a parallelogram.

Hence, (d) is the correct answer.

3. 50º

Solution:

ABCD is a rhombus such thet $\angle\text{ACB}=40^\circ.$

We know that diagonnals of rhombus bisect each other right angles.

In right $\Delta\text{BOC},$ we have

$\angle\text{OBC}=180^\circ-(\angle\text{BOC}+\angle\text{BCO})$ (angle sum property)

$=180^\circ-(90^\circ+40^\circ)=50^\circ$

$\therefore\ \angle\text{DBC}=\angle\text{OBC}=50^\circ$

Now,

$\angle\text{ADB}=\angle\text{DBC}$ [Alt. int. $\angle\text{s}$ ]

$\therefore\ \angle\text{ADB}=50^\circ[\therefore\angle\text{DBC}=50^\circ]$

Hence, (c) is the correct answer.