Understanding Quadrilaterals Test

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Understanding Quadrilaterals Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

If one angle of the parallelogram is $$16^{\circ}$$ less than three times the smallest angle, then the largest angle of the parallelogram is

A
$$131^{\circ}$$

B
$$136^{\circ}$$

C
$$112^{\circ}$$

D
$$108^{\circ}$$

Solution

Let $$x$$ and $$y$$ be the largest and smallest angles of parallelogram.
By hypothesis $$x = 3y - 16$$
Also in a parallelogram, we have
$$x + y = 180^{\circ}$$
$$\Rightarrow (3y - 16) + y = 180^{\circ}$$
$$\Rightarrow 4y = 196$$
$$\Rightarrow y = 49$$
Thus $$ x = 49\times 3 - 16$$
$$= 147 - 16 \\= 131^{\circ}$$
Question 2
1 / -0

If the sum of all the angles of a polygon except one angle is $$2220^\circ$$, then the number of sides of the polygon are:

A
12

B
13

C
14

D
15

Solution

Sum of angles of polygon with $$n$$ sides $$= (n - 2) \times 180^{\circ}\dots(1)$$

The sum of the angles of a polygon should be a multiple of $$180^\circ$$.

But, $$2220^{\circ} = 180^{\circ}\times 12 + 60^{\circ}$$

$$2220^{\circ}$$ + $$120^{\circ}$$ = $$2340^{\circ}$$

$$( x - 2 ) = \dfrac{2340}{180}\dots$$ (From $$(1)$$)

$$x - 2 = 13$$

$$\Rightarrow x = 15$$

Therefore, the number of sides of the polygon is $$15$$.
Question 3
1 / -0

If $$ABCD$$ is a parallelogram whose diagonals intersect at $$O$$ and $$BCD$$ is an equilateral triangle having each side of length $$6$$ cm, then the length of diagonal $$AC$$ is :

A
$$3\sqrt{3}$$ cm

B
$$6\sqrt{3}$$ cm

C
$$3\sqrt{6}$$ cm

D
$$12$$ cm

Solution

$$ABCD$$ is a parallelogram
$$\therefore BC = AD = 6$$ cm
And $$AB = DC = 6$$ cm
Hence, $$ABCD$$ becomes a rhombus.
$$\therefore$$ Diagonals $$AC$$ and $$BD$$ bisect each other at right angle.
$$\therefore OD = \dfrac{1}{2} BD = 3$$ cm
From $$\Delta OCD$$,
$$OC^2 = CD^2 - OD^2$$ .... Pythagoras theorem
$$= 36 - 9$$
$$ = 27$$
$$\therefore OC = 3 \sqrt{3}$$
and $$AC = 2\cdot OC = 6 \sqrt{3}$$

Question 4

1 / -0

Match List I with List II and select the correct answer using the codes given below the lists:

List I (Regular plane figure)	List II (Measure of interior angles)
I . Triangle	(A) $$30^{\circ}$$
II. Square	(B) $$60^{\circ}$$
III. Pentagon	(C) $$108^{\circ}$$
IV. Hexagon	(D) $$90^{\circ}$$
	(E) $$120^{\circ}$$

I-D, II-A, III-B, IV-E

I-B, II-D, III-C, IV-E

I-A, II-D, III-C, IV-B

I-B, II-C, III-A, IV-D

Solution

Each Angle (of a Regular Polygon) $$= \dfrac {(n-2) × 180°} { n}$$

Triangle = $$(3-2)×180°/3=60°$$

Square= $$(4-2)×180°/4=90°$$

Pentagon = $$(5-2)×180°/5=108°$$

Hexagon =$$(6-2)×180°/6=120°$$

$$\therefore$$ I-B, II-D, III-C, IV-E

Question 5
1 / -0

Calculate the sum of angles of a polygon having $$10$$ sides.

A
$$960^{\circ}$$

B
$$1440^{\circ}$$

C
$$900^{\circ}$$

D
$$720^{\circ}$$

Solution

Sum of interior angles of a polygon = $$180^o (n -2) $$
When $$n = 10$$,
Sum of interior angles of a polygon = $$ 180^o (n -2) = 180^o (10 -2 )= 1440^o $$
Question 6
1 / -0

If all the angles of a hexagon are equal. Find the measure of each of its angle.

A
$$100^{\circ}$$

B
$$240^{\circ}$$

C
$$140^{\circ}$$

D
$$120^{\circ}$$

Solution

All the angles of a hexagon are equal.
Let each angle is = $$x^o$$
Number of sides of hexagon $$(n) = 6$$
Sum of all interior angles of a polygon = $$ 180^o (n -2) $$
$$ => 6x = 180^o (n-2) $$
$$ => 6x = 180^o ( 6 -2) $$
$$ => x = 120^o $$
Each angle of hexagon is = $$ 120^o $$
Question 7
1 / -0

Find the number of sides in a polygon if the sum of its interior angles is $$900^{\circ}$$.

A
$$7$$

B
$$9$$

C
$$5$$

D
$$6$$

Solution

If the sum of its interior angles is $$ 900^o$$
Sum of interior angles of a polygon = $$ 180^o (n -2 ) $$
=> $$180^o (n -2 ) = 900^o $$
=> $$ ( n - 2 ) = 5 $$
=> $$n = 7$$
Number of sides in polygon $$= 7$$
Question 8
1 / -0

The sum of the interior angles of a polygon is four times the sum of its exterior angles. Find the number of sides in the polygon.

A
$$10$$

B
$$12$$

C
$$8$$

D
$$7$$

Solution

The sum of the interior angles of a polygon is four times the sum of its exterior angles.
The sum of the exterior angles of a polygon is always equal to $$360^o$$.
The sum of the interior angles of polygon = $$180 (n-2)$$
=> $$180 (n-2) = 4 \times 360$$
=> $$n -2 = 8$$
=> $$n =10$$
Number of sides in the polygon = $$10$$
Question 9
1 / -0

$$PQRS$$ is a parallelogram whose diagonals intersect at M.
If $$\angle PMS = 54^{\circ}$$, $$\angle QSR = 25^{\circ}$$ and $$\angle SQR = 30^{\circ}$$: find $$\angle PSR $$

A
$$45^{\circ}$$

B
$$55^{\circ}$$

C
$$48^{\circ}$$

D
$$56^{\circ}$$

Solution

The opposite sides of a parallelogram are
parallel. So, $$PQ \parallel SR$$ and $$QS$$ is transversal.
$$\therefore$$ $$\angle$$$$SQR$$ $$=\angle$$$$PSQ$$
Since, they are alternate interior angles.

$$\angle PSR=\angle PSQ+\angle QSR$$
$$=\angle SQR+\angle QSR$$
$$=30^\circ+25^\circ$$
$$=55^\circ$$
Question 10
1 / -0

In parallelogram ABCD, $$\angle A = 3 \angle B$$. In the same parallelogram, if AB $$= 5x-7$$ and $$CD = 3x + 1$$; find the length of CD.

A
$$CD = 16$$ units

B
$$CD = 11$$ units

C
$$CD = 13$$ units

D
$$CD = 15$$ units

Solution

AB $$= $$ CD ( Opposite sides of parallelogram)
$$\implies 5x -7 = 3x + 1$$
$$\implies x = 4 $$
$$\therefore$$ AB $$=$$ CD$$ = 13 $$ units.

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Re-Attempt Weekly Quiz Competition

Understanding Quadrilaterals Test - 14

Result

Understanding Quadrilaterals Test - 14

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SHARING IS CARING

Question 1

$$131^{\circ}$$

$$136^{\circ}$$

$$112^{\circ}$$

$$108^{\circ}$$

Solution

Question 2

12

13

14

15

Solution

Question 3

$$3\sqrt{3}$$ cm

$$6\sqrt{3}$$ cm

$$3\sqrt{6}$$ cm

$$12$$ cm

Solution

Question 4

I-D, II-A, III-B, IV-E

I-B, II-D, III-C, IV-E

I-A, II-D, III-C, IV-B

I-B, II-C, III-A, IV-D

Solution

Question 5

$$960^{\circ}$$

$$1440^{\circ}$$

$$900^{\circ}$$

$$720^{\circ}$$

Solution

Question 6

$$100^{\circ}$$

$$240^{\circ}$$

$$140^{\circ}$$

$$120^{\circ}$$

Solution

Question 7

$$7$$

$$9$$

$$5$$

$$6$$

Solution

Question 8

$$10$$

$$12$$

$$8$$

$$7$$

Solution

Question 9

$$45^{\circ}$$

$$55^{\circ}$$

$$48^{\circ}$$

$$56^{\circ}$$

Solution

Question 10

$$CD = 16$$ units

$$CD = 11$$ units

$$CD = 13$$ units

$$CD = 15$$ units

Solution

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