Lines and Angles Test

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Lines and Angles Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

Find the complements of $$\displaystyle { 68 }^{ o }$$ and $$\displaystyle { 33 }^{ o }$$ respectively.

A
$$22^o, 57^o$$

B
$$57^o, 22^o$$

C
$$33^o, 67^o$$

D
$$67^o, 33^o$$

Solution

We know, two angles are complementary when they add up to $$90^o.$$

Given, measure of one complementary angle is $$68^o.$$

$$\Rightarrow$$ Measure of other complementary angle $$=90^o-68^o$$ $$=22^o.$$

$$\therefore$$ Measure of a complementary angle of $$ 68^{o}=22^o$$.

Also given, measure of one complementary angle is $$33^o.$$
$$\Rightarrow$$ Measure of other complementary angle $$=90^o-33^o$$ $$=57^o.$$

$$\therefore$$ Measure of a complementary angle of $$ 33^{o}=57^o$$.

Hence, option $$A$$ is correct.
Question 2
1 / -0

Which of the following indicated figures are examples of adjacent angles.

A
(i) and (iv)

B
(i) and (ii)

C
(i) and (iii)

D
(ii) and (iv)

Solution

(i) and (iv) have common side, common vertex and common arm. So, these are adjacent.
Question 3
1 / -0

$$\angle AEC$$ and $$\angle CED$$ are supplementary angles. If $$\angle CED$$ is equal to $$62^{\circ}$$ ,what is the measure of $$\angle AEC$$?

A
$$28^{\circ}$$

B
$$150^{\circ}$$

C
$$118^{\circ}$$

D
$$124^{\circ}$$

Solution

Two angles are supplementary if they add up to $$180^{\circ}$$ .
As given, $$\angle AEC$$ and $$\angle CED$$ are supplementary angles,

So,
$$\angle AEC+\angle CED=180^{\circ}$$
$$\Rightarrow \angle AEC+62^{\circ}=180^{\circ}$$
$$\Rightarrow\angle AEC=180^{\circ}-62^{\circ}$$
$$\Rightarrow\angle AEC=118^{\circ}$$

Hence, the required answer is $$118^{\circ}$$.
Question 4
1 / -0

All linear pairs are

A
supplementary

B
vertically opposite

C
right angles

D
none of these

Solution

In the given diagram, $$ABC$$ is a straight line.

$$\angle ABD$$ and $$\angle CBD$$ are linear pairs.

$$\angle ABD=120^\circ$$ and $$\angle CBD=60^\circ$$

A linear pair forms a straight angle which contains $$180^\circ$$, hence the sum of $$2$$ angles measures $$180^\circ,$$ which means they are supplementary.

$$\therefore$$ $$\angle ABD+\angle CBD=120^\circ+60^\circ$$

$$=180^\circ$$

So we can say that, $$\angle ABD$$ and $$\angle CBD$$ are supplementary.

$$\therefore$$ All linear pairs are supplementary.
Question 5
1 / -0

In the given figure, $$\displaystyle \angle 1={ 70 }^{ o }$$. Find $$180^o-\displaystyle 2\angle x$$.

A
$$\displaystyle { 50 }^{ o }$$

B
$$\displaystyle { 40 }^{ o }$$

C
$$\displaystyle { 45 }^{ o }$$

D
$$\displaystyle { 60 }^{ o }$$

Solution

$$\angle x = \angle 1$$ .....................[ Alternate angles]
$$\therefore \angle x = 70^o$$
$$\therefore 180^o - 2\angle x = 40^0$$
Question 6
1 / -0

Which of the following are complementary angles?

A
$$71^o$$ and $$19^o$$

B
$$18^o$$ and $$180$$

C
$$90^o$$ and $$90^o$$

D
$$90^o$$ and $$45^o$$

E
$$15^o$$ and $$30^o$$

Solution
Question 7
1 / -0

Find the value of $$x$$ and $$y$$.

A
$$60^{\circ}, 45^{\circ}$$

B
$$45^{\circ}, 60^{\circ}$$

C
$$45^{\circ}, 45^{\circ}$$

D
$$60^{\circ}, 60^{\circ}$$

Solution

From the figure, $$45^\circ$$ is the complement of $$y$$

That is, $$y + 45^\circ = 90^{\circ}$$

$$\Rightarrow$$ $$y = 90^{\circ}-45^\circ$$

$$ \Rightarrow y = 45^{\circ}$$

Also, $$30^\circ$$ is the complement of $$x$$

$$\Rightarrow x + 30^\circ = 90^\circ$$

$$\Rightarrow x = 90^\circ-30^\circ$$

$$\Rightarrow x = 60^{\circ}$$
Question 8
1 / -0

In the figure above, if $$l$$ || $$m$$ and $$r=91^o$$, then $$t+u=$$

A
$$178^o$$

B
$$179^o$$

C
$$180^o$$

D
$$181^o$$

E
$$182^o$$

Solution

Given, $$l$$ $$||$$ $$m$$
Sum of the angles in a straight line $$=$$ $$180^o$$
Here, $$r$$ $$+$$ $$s$$ $$=$$ $$180^o$$
As $$r$$ $$=$$ $$91^o$$,
$$\Rightarrow 91^o$$ $$+$$ $$s$$ $$=$$ $$180^o$$
$$\Rightarrow s$$ $$=$$ $$180^o$$ $$-$$ $$91^o$$
$$\Rightarrow s$$ $$=$$ $$89^o$$
When two lines are parallel, corresponding angles are equal.
From the given figure,
$$s$$ and $$u$$ are corresponding angles.
$$s$$ $$=$$ $$u$$
$$s$$ $$=$$ $$u$$ $$=$$ $$89^o$$
Vertically opposite angles are equal.
Here, $$u$$ and $$t$$ are vertically opposite angles
$$u$$ $$=$$ $$t$$
$$u$$ $$=$$ $$t$$ $$=$$ $$89^o$$
$$t$$ $$+$$ $$u$$ $$=$$ $$89^o$$ $$+$$ $$89^o$$
$$t+u=$$ $$178^o$$

Therefore, the value of $$(t$$ $$+$$ $$u)$$ is $$178^o$$.
Question 9
1 / -0

In figure, If $$l_{1} \parallel l_{2}$$, calculate the value of $$x $$.

A
$$15^o$$

B
$$25^o$$

C
$$30^o$$

D
$$35^o$$

E
$$40$$

Solution

Since the two lines are parallel, the suplementary angle of angle $$(4x-30)^o$$ is equal to sum of two $$x$$ angles.(Corresponding angles of parallel lines)
$$180^o-(4x-30)= x+x$$
$$4x-30^o + 2x =180^o$$
$$\Rightarrow 6x = 210^o$$
$$\Rightarrow x = 35^o$$
Question 10
1 / -0

In quadrilateral $$PQRS$$ above, for what value of $$x$$, $$PS\parallel QR$$?

A
$$158$$

B
$$132$$

C
$$120$$

D
$$110$$

E
$$70$$

Solution

It is given that $$PS || QR$$.
Also, $$\angle SPQ=70^{0}$$

$$PQ$$ acts as a transversal since $$PS||QR$$.

Hence, $$\angle SPQ +\angle RQP=180^{0}$$

$$\Rightarrow 70^{0}+\angle Q=180^{0}$$

$$\Rightarrow \angle Q=110^{0}$$

Hence, the value of $$x$$ is $$110$$.