Understanding Quadrilaterals(8th Grade > Mathematics ) Questions and answers for exam Preparation

8th Grade > Mathematics

UNDERSTANDING QUADRILATERALS MCQs

Total Questions : 464 | Page 45 of 47 pages

Question 180
ABCD is a parallelogram. Point P and Q are taken on the sides AB and AD, respectively and the parallelogram PRQA is formed. If $∠ C = 45^{\circ}$ , find $∠ R$ .

Discuss Question

Answer: Option A. ->
:
Let ABCD be a parallelogram,
where

∠ C = 45^{\circ}

Question 180ABCD Is A Parallelogram. Point P And Q Are Taken...

Since, ABCD is a parallelogram,

∠ A = ∠ C

[opposite angles of parallelogram are equal]
Again, since PRQA is a parallelogram,

∠ A = ∠ R

[opposite angles of parallelogram are equal]

\Rightarrow ∠ R = 45^{\circ}

[

∴ ∠ A = ∠ C = 45^{\circ}

]

Question 442.

Question 182
A regular pentagon ABCDE and a square ABFG are formed on opposite sides of AB. Find $∠ B C F$ ?

Discuss Question

Answer: Option A. ->
:

Question 182A Regular Pentagon ABCDE And A Square ABFG Are F...

Given, ABCDE is a regular pentagon.
Then, measure of each interior angle of the regular pentagon

= \frac{S u m o f i n t e r i o r a n g l e s}{N u m b e r o f s i d e s} = \frac{(x - 2) \times 180^{\circ}}{5} = \frac{(5 - 2) \times 180^{\circ}}{5} = \frac{540^{\circ}}{5} = 108^{\circ} ∴ ∠ C B A = 108^{\circ}

Join CF.
Now,

∠ F B C = 360^{\circ} - (90^{\circ} + 108^{\circ}) = 360^{\circ} - 198^{\circ} = 162^{\circ}

Δ F B C

, by the angle sum property, we have

∠ F B C + ∠ B C F + ∠ B F C = 180^{\circ} \Rightarrow ∠ B C F + ∠ B F C = 180^{\circ} - 162^{\circ} \Rightarrow ∠ B C F + ∠ B F C = 18^{\circ}

Since,

Δ F B C

is an isosceles triangle and BF = BC.

∴ ∠ B C F = ∠ B F C = 9^{\circ}

Question 443.

Question 181
In parallelogram ABCD, the angle bisector of $∠ A$ bisects BC. Will angle bisector of B also bisect AD? Give reason.

Discuss Question

Answer: Option A. ->
:
Given, ABCD is a parallelogram, bisector of

∠ A

, bisects BC at F, i.e.

∠ 1 = ∠ 2

, CF = FB.
Draw

F E ∥ B A

ABFE is a parallelogram by construction [

∵ F E ∥ B A

]

\Rightarrow ∠ 1 = ∠ 6

[alternate angle]

Question 181In Parallelogram ABCD, The Angle Bisector Of ∠...

But

∠ 1 = ∠ 2

[given]

∴ ∠ 2 = ∠ 6

AB = FB [opposite sides to equal angles are equal] ..... (i)

∴ A B F E i s a r h o m b u s

.
Now, in

Δ A B O

and

Δ B O F

, AB = BF          [from Eq. (i)]
BO = BO         [common]
AO = FO         [diagonals of rhombus bisect each other]

∴ Δ A B O ≅ Δ B O F

[by SSS]

∠ 3 = ∠ 4

[by CPCT]
Now,

B F = \frac{1}{2} B C

[given]

\Rightarrow B F = \frac{1}{2} A D

[BC = AD]

\Rightarrow A E = \frac{1}{2} A D

[BF = AE]

∴

E is the mid-point of AD.

Question 444.

Question 184
In the following figure, $F D ∥ B C ∥ A E$ and $A C ∥ E D$ . Find the value of x.

Discuss Question

Answer: Option A. ->
:

Question 184In The Following Figure, FD∥BC∥AE And AC∥E...

Δ A B C

∠ A B C + ∠ B C A + ∠ C A B = 180^{\circ}

[Angle sum property of triangle]

\Rightarrow 64^{\circ} + ∠ B C A + 52^{\circ} = 180^{\circ}

\Rightarrow ∠ B C A = 180^{\circ} - 64^{\circ} - 52^{\circ} = 64^{\circ}

∠ F A E = ∠ B C A

[

∵

Alternate angles;

A E ∥ B C

, AC is the transversal]

\Rightarrow ∠ F A E = 64^{\circ}

Now,

∠ F A E + x = 180^{\circ}

[Adjacent angles in a parallelogram are supplementary]

\Rightarrow x = 180^{\circ} - 64^{\circ} ∴ x = 116^{\circ}

Question 445.

Question 183
Find the maximum number of acute angles which a convex quadrilateral, a pentagon and a hexagon can have. Observe the pattern and generalise the result for any polygon.

Discuss Question

Answer: Option A. ->
:
Let us assume that a convex polygon has four or more acute angles. If an interior angle is acute, then the corresponding exterior angle will be obtuse ( greater than

90^{\circ}

). So, the sum of the exterior angles of such a polygon will be greater than

4 \times 90^{\circ} = 360^{\circ}

.
However, this is impossible, since the sum of the exterior angles of a polygon must always be equal to

360^{\circ}

. Hence, a polygon can have, at most, 3 obtuse exterior angles.

Question 446.

Question 185
In the following figure, $A B ∥ D C$ and $A D = B C$ . Find the value of x.

Discuss Question

Answer: Option A. ->
:
Given, an isosceles trapezium, where

A B ∥ D C

, AD = BC and

∠ A = 60^{\circ}

.
Then,

∠ B = 60^{\circ}

.
Draw a line parallel to BC through D which intersects the line AB at E (say).

Question 185In The Following Figure, AB∥DC And AD=BC. Find...

Then, DEBC is a parallelogram, where
BE = CD = 20 cm and DE = BC = 10 cm
Now,

∠ D E B + ∠ C B E = 180^{\circ}

[adjacent angles are supplementary in parallelogram]

\Rightarrow ∠ D E B = 180^{\circ} - 60^{\circ} = 120^{\circ}

∴ I n Δ A D E, ∠ A D E = 60^{\circ}

[exterior angle]
Also,

∠ D E A = 60^{\circ}

[

∵

AD = DE = 10cm and

∠ D A E = 60^{\circ}

]
Then,

Δ A D E

is an equilateral triangle.

∴ A E = 10 c m

\Rightarrow A B = A E + E B = 10 + 20 = 30 c m

Hence, x = 30 cm.

Question 447.

Question 186
Construct a trapezium ABCD in which $A B ∥ D C, ∠ A = 105^{\circ}$ , AD = 3 cm, AB = 4 cm and CD = 8 cm.

Discuss Question

Answer: Option A. ->
:
We know that.

∠ A + ∠ D = 180

[

∵

sum of adjacent angle of a trapezium is

180^{\circ}

]

105^{\circ} + ∠ D = 180^{\circ}

∠ D = 75^{\circ}

Question 186Construct A Trapezium ABCD In Which AB∥DC,∠A...

Steps of Construction
Step I Draw AB = 4 cm.
Step II Draw

¯ A X

such that

∠ B A X = 105^{\circ}

.
Step III Mark a point D on AX such that AD = 3 cm.
Step IV Draw

¯ D Y

such that

∠ A D Y = 75^{\circ}

.
Step V Mark a point C such that CD = 8 cm.
Step VI Join BC.
Hence, ABCD is the required trapezium.

Question 448.

Question 188
Construct a rhombus whose side is 5 cm and one angle is of $60^{\circ}$ .

Discuss Question

Answer: Option A. ->
:

Question 188Construct A Rhombus Whose Side Is 5 Cm And One A...

∠ B = 60^{\circ}

[suppose]

∠ A + ∠ B = 180^{\circ}

[sum of cointerior angles]

∠ A + 60^{\circ} = 180^{\circ}

∠ A = 120^{\circ}

AB = BC = CD = DA = 5 cm
Steps of Construction
Step I Draw AB = 5 cm.
Step II Draw ray AY such that

∠ B A Y = 120^{\circ}

.
Step III Mark a point D such that AD = 5 cm.
Step IV Draw a ray BX such that

∠ A B X = 60^{\circ}

Step V Mark a point C such that BC = 5 cm.
Step VI Join C and D.

∴

ABCD is required rhombus.

Question 449.

Question 187
Construct a parallelogram ABCD in which AB = 4 cm, BC = 5 cm and $∠ B = 60^{\circ}$ .

Discuss Question

Answer: Option A. ->
:
We know that the opposite sides of a parallelogram are equal.
So, AB = DC = 4 cm
BC = AD = 5 cm

∠ B = 60^{\circ}

∠ A + ∠ B = 180^{\circ}

[sum of cointerior angles]

∠ A = 120^{\circ}

Question 187Construct A Parallelogram ABCD In Which AB = 4 C...

Steps of Construction
Step I Draw AB = 4 cm.
Step II Draw ray BX such that

∠ A B X = 60^{\circ}

.
Step III Mark a point C such that BC = 5 cm.
Step IV Draw a ray AY such that

∠ Y A B = 120^{\circ}

.
Step V Mark a point D such that AD = 5 cm.
Step VI Join C and D.
Hence, ABCD is required parallelogram.

Question 450.

Question 191
Construct a rhombus CLUE in which CL = 7.5 cm and LE = 6 cm.

Discuss Question

Answer: Option A. ->
:
We know that, all sides of a rhombus are equal and opposite sides are parallel to each other.
Steps of Construction
Step I Draw a line segment CL = 7.5 cm.
Step II With C as a centre, draw an arc CE = 7.5 cm.
Step III With L as a centre, draw another arc LU = 7.5 cm.

Question 191Construct A Rhombus CLUE In Which CL = 7.5 Cm An...

Step IV Now, with centre L, draw an arc LE = 6 cm, which cut-off previous arc CE.
Step V With E as a centre, draw an arc UE = 7.5 cm, which cut off the previous arc LU.
Step VI Now join UL, CE and EU.
Thus, we have required rhombus CLUE.