Understanding Quadrilaterals Test

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Understanding Quadrilaterals Test - 21

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Question 1
1 / -0

In the given parallelogram, find the opposite angle $$y$$?

A
$$60^o$$

B
$$50^o$$

C
$$70^o$$

D
$$120^o$$

Solution

$$y=120^o$$ (as opposite angles in a parallelogram are equal)
Therefore, D is the correct answer.
Question 2
1 / -0

A Rhombus is also a

A
Parallelogram

B
Trapezium

C
rectangle

D
None of these

Solution

A Rhombus has two pairs of parallel side, opposite angles are equal and all the sides are equal, which are the properties of a parallelogram.
Hence, a rhombus is also a parallelogram.

Whereas, a Rhombus is not a trapezium as in a trapezium four siders are not equal.

Also, a Rhombus is not a rectangle as all the four siders are not equal.

Therefore, A is the correct answer.
Question 3
1 / -0

A four sided figure with all sides equal and opposite angles also equal.

A
kite

B
Rhombus

C
Parallelogram

D
Rectangle

Solution

Among all, rhombus has all sides equal with opposite angles equal.
Therefore, B is the correct answer.
Question 4
1 / -0

If one angle of a parallelogram is $$65^o$$. What are the other angles?

A
$$50^o, 120^o, 50^o$$

B
$$115^o, 65^o, 115^o$$

C
$$65^o, 115^o, 65^o$$

D
None of the above
Question 5
1 / -0

If two adjacents angles of a parallelogram are in ratio 3 : 9, then measure of smallest angle is

A
$$105$$

B
$$60$$

C
$$45$$

D
None

Solution

Sum of adjacent angles of parallelogram $$=180^o$$
One angle of a parallelogram $$= 3x$$
other angle of a parallelogram $$= 9x$$
$$3x + 9x = 180$$
$$12 x = 180, x = \dfrac{180}{12} = 15$$
one angle $$3x = 3 \times 15 = 45$$
Question 6
1 / -0

In fig. ABCD is a parallelogram. What are the values of $$p$$ and $$q$$ ?
If $$\angle CBE$$ = $$100^{\circ}$$.

A
$$\angle p = 100^o, \angle q = 80^o$$

B
$$\angle p = 80^o, \angle q = 100^o$$

C
$$\angle p = \angle q$$

D
None

Solution

$$\angle CBE = 100, \angle p + \angle CBE = 180^o, \angle p = 180 - 100, \angle p = 80^o$$
$$\angle q = \angle CBE, [\because AD || BC]$$
$$\therefore \angle q = 100^o$$
Question 7
1 / -0

In parallelogram, both pairs of opposite sides are ....................

A
equal

B
parallel and equal

C
perpendicular

D
perpendicular and equal

Solution

In $$ABCD$$ is a parallelogram,

$$\Rightarrow$$ We know that $$AB\parallel CD$$

Here $$AC$$ is a transversal for the parallel lines $$AB$$ and $$CD$$

$$\Rightarrow$$ So, $$\angle BAC = \angle DCA$$ [Alternate interior angles are equal] $$... ( 1 )$$

$$\Rightarrow$$ Similarly, we know that $$BC\parallel AD$$

$$\Rightarrow \angle BCA = \angle DAC$$ $$... ( 2 )$$

In $$\triangle ABC$$ and $$\triangle ADC,$$

We have $$\angle BAC = \angle DCA$$ [From ( 1 )]

$$\Rightarrow$$ $$AC$$ is the common side

$$\Rightarrow \angle BCA = \angle DAC$$ [From ( 2 )]

$$\Rightarrow$$ So, $$\triangle ABC \cong \triangle CDA$$ [As per ASA Congruence rule]

$$\therefore AB = CD$$ and $$BC = AD$$ [Corresponding sides of congruent triangle are equal ]

Hence, both pairs of opposite sides in a parallelogram are parallel and equal.
Question 8
1 / -0

PQRS is a parallelogram & PA & RB are perpendicular from P & R in diagonal SQ. Which of following is the correct option?

A
$$AP = BQ$$

B
$$PA = BR$$

C
$$AS = RB$$

D
None

Solution

In $$\triangle APQ$$ & $$\triangle RSB$$,
$$\angle PAQ = \angle SBR $$ ......each $$90^o$$
$$PQ = SR$$ ......(opposite sides of parallelogram)
$$\angle AQP = \angle RSB$$ .....(Alternate angle)
$$\therefore \triangle APQ \cong \triangle RSB$$
$$\therefore PA = BR$$ .....cpct
Question 9
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PQRS is a parallelogram and side RS is smaller than side QR. Then the relation between $$\angle 1$$ and $$\angle 2$$ is ___ .

A
$$\angle 1 < \angle 2$$

B
$$\angle 1 > \angle 2$$

C
$$\angle 1 = \angle 2$$

D
None

Solution

$$\angle 1 = \angle RSQ$$ ....alternate angles
In $$\triangle RSQ$$,
$$RS<QR$$ ...given
$$\therefore \angle 2 < \angle RSQ$$ ....angle opp to shorter side is smaller
i.e. $$\therefore \angle 2 < \angle 1$$
Question 10
1 / -0

$$WXYZ$$ is a parallelogram & its diagonals intersect at point $$O$$. Find $$\angle OZY$$.

A
$$115^o$$

B
$$65^o$$

C
$$60^o$$

D
$$70^o$$

Solution

In parallelogram $$WXYZ,$$

$$\angle ZWX = 30^o + 35^o = 65^o$$

We know opposite angles of a parallelogram are equal

Hence $$\angle ZWX = \angle XYZ = 65^o$$

We know the sum of any two adjacent angles of a parallelogram is $$180^\circ$$

$$\Rightarrow\angle Z = 180^o - \angle W $$

$$= 180^o - 65^o $$

$$= 115^o$$

In $$\triangle WOX,$$

$$\Rightarrow \angle OXW + \angle WXO + \angle WOX = 180^o$$

$$\Rightarrow \angle WXO = 180^o - 30^o - 80^o$$

$$\Rightarrow \angle WXO = 70^o$$

$$\therefore \angle XZY = \angle OZY = 70^o\ \ \ \ \ \ \ (\because WX \parallel ZY)$$