Lines and Angles Test

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Lines and Angles Test - 22

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Question 1
1 / -0

In figure, if $$AB\parallel CD$$, then find the value of $$\angle PRG$$

A
$$138^o$$

B
$$118^o$$

C
$$108^o$$

D
$$98^o$$

Solution

In the figure, $$AB\parallel CD$$

$$ \angle PRB = 2x$$ ....... (Given)

As lines $$AB$$ and $$PQ$$ intersect at $$R$$

$$ \angle PRB = \angle GRS $$ .......(Vertically opposite angles)

$$\Rightarrow \angle GRS = 2x$$

Also, $$AB \parallel CD$$

$$\angle GRS + \angle HSR = 180^o$$ ...... (Sum of Co-interior angle is $$180^o$$)

$$ \angle HSR = 3x$$ ....... (Given)

$$ \Rightarrow 2x + 3x =180^o$$

$$ \Rightarrow 5x = \dfrac { 180^o }{ 5 } $$

$$ \Rightarrow x = 36^o$$

$$\angle RSH=3x$$ ....... (Given)

Also, $$\angle RSH=\angle PRG$$ ........ (Corresponding angles are equal)

$$\Rightarrow \angle PRG=3x$$

$$\Rightarrow \angle PRG=3\times 36^o=108^o$$

Hence, $$\angle PRG=108^o$$
Question 2
1 / -0

In figure if $$BA\parallel DF, AD\parallel FG, \angle BAC=65^o$$ and $$\angle ACB=55^o$$, then find $$\angle FGH$$.

A
$$\angle FGH=125^o$$

B
$$\angle FGH=145^o$$

C
$$\angle FGH=65^o$$

D
$$\angle FGH=55^o$$

Solution

In $$\triangle ABC$$,
$$\angle ACB + \angle ACE = 180^{\circ}$$ (Linear pair)
$$55 + \angle ACE = 180^{\circ}$$
$$\angle ACE = 125^{\circ}$$
Now, given, $$AD \parallel FG$$,
$$\angle ACE = \angle FGH = 125^{\circ}$$ (Corresponding angles of parallel lines)
Question 3
1 / -0

The measure of an angle is three times the measure of its complement. The angles are:

A
$$20^\circ,\, 60^\circ$$

B
$$30^\circ,\, 60^\circ$$

C
$$45^\circ,\, 135^\circ$$

D
$$22.5^\circ,\, 67.5^\circ$$

Solution

Let the complement be $$x$$.
Then the angle $$=3x$$.
We know, the sum of the complementary angles is $${ 90 }^{ \circ }$$
$$\Rightarrow x+3x={ 90 }^{ \circ }\\ \Rightarrow 4x={ 90 }^{ \circ }\\ \Rightarrow x=22.5^{ \circ }$$.
Hence, the angles are:
$$x=1\times 22.5^{ \circ }=22.5^{ \circ }$$
and $$3x=3\times 22.5^{ \circ }={ 67.5 }^{ \circ }$$.
Hence, option $$D$$ is correct.
Question 4
1 / -0

$$AB\, ||\, CD\, ||\, EF$$ and $$GH\, ||\, DI$$. If $$\angle DIH\, =\, 95^{\circ}$$ then the measure of $$\angle GCD$$ is equal to

A
$$95^{\circ}$$

B
$$105^{\circ}$$

C
$$90^{\circ}$$

D
$$85^{\circ}$$

Solution

$$\angle CDI\, =\, 180^{\circ}\, -\, 95^{\circ}\, =\, 85^{\circ}$$ (CD || H1, co-interior angles are supplementary)
$$\therefore\, \angle GCD\, =\, \angle CDI\, =\, 85^{\circ}$$ (GH || DI, alternate angles)
Question 5
1 / -0

In figure, if $$AB\parallel CD$$, then find the value of $$\angle CHF$$

A
$$188^o$$

B
$$178^o$$

C
$$118^o$$

D
$$108^o$$

Solution

$$\quad In\quad figure\quad AB\parallel CD\\ \angle PRB\quad =\quad 2x(Given)\\ As\quad lines\quad AB\quad and\quad PQ\quad intersect\quad at\quad R\\ \angle PRB\quad =\quad \angle GRS\quad \\ \Longrightarrow \angle GRS\quad =\quad 2x\\ Also,AB\quad \parallel \quad CD\\ \quad \angle GRS\quad +\quad \angle HSR\quad =\quad (sum\quad of\quad interior\quad angle\quad is\quad supplementary\\ \angle HSR\quad =\quad 3x\quad (Given)\\ \Longrightarrow 2x\quad +\quad 3x\quad =180\\ \Longrightarrow 5x\quad =\quad \frac { 180 }{ 5 } \\ \Longrightarrow x\quad =\quad 36\\ \angle EGA\quad =\quad 2x\quad (Given)\\ \angle EGA\quad +\quad \angle EGR\quad =\quad 180\\ \Longrightarrow 2x+\quad \angle EGR\quad =180\\ \Longrightarrow \angle EGR\quad =\quad 180\quad -\quad 2x\\ Also\quad \angle GHS\quad =\quad \angle EGR(Corrosponding\quad angles\quad are\quad equal)\\ \Longrightarrow \angle GHS\quad =\quad 180\quad -\quad 2x\\ Putting\quad value\quad of\quad x=36\\ \angle GHS\quad =\quad 180\quad -\quad 2\times 36\quad =180\quad -\quad 72\quad =108\\ \Longrightarrow \angle GHS\quad =\quad 108\\ Now\quad \angle GHS\quad =\quad \angle CHF(Vertically\quad opposite\quad angles)\\ Hence\quad \angle CHF\quad =\quad 108$$
Question 6
1 / -0

In figure, if $$AB\parallel CD$$, then find the value of $$x$$

A
$$86^o$$

B
$$56^o$$

C
$$36^o$$

D
$$16^o$$

Solution

Given, $$ AB\parallel CD $$
$$ \angle PRG = 3x$$ .......... (Corresponding angles)
$$ \angle PRG + 2x = 180^o $$ (Linear pair)
$$ 3x + 2x = 180^o $$
$$\Rightarrow x = 36^o $$
Question 7
1 / -0

Lines $$m$$ and $$n$$ are cut by a transversal so that $$\angle 1$$ and $$\angle 5$$ are corresponding angles. If $$\angle 1=26x-{7}^{\circ}$$ and $$\angle 5=20x+{17}^{\circ}.$$ What value of $$x$$ makes the lines $$m$$ and $$n$$ parallel?

A
$$5$$

B
$$4$$

C
$$4\cfrac{1}{2}$$

D
$$3\cfrac{1}{4}$$

Solution

For the lines $$m$$ and $$n$$ to be parallel, corresponding angles should be equal, i.e, $$\angle 1=\angle 5$$

$$\Rightarrow$$ $$26x-{7}^{\circ}=20x+{17}^{\circ}$$

$$\Rightarrow$$ $$6x={24}^{\circ}$$

$$\Rightarrow$$ $$x={4}^{\circ}$$
Question 8
1 / -0

If ray $$PQ$$ and $$RS$$ are parallel as given in the figure, then fine $$a+b+c$$.

A
$${360}^{o}$$

B
$$270^o$$

C
$$180^o$$

D
$$315^o$$

Solution

Draw a ray $$TU$$ paralell to $$PQ$$ and $$RS$$. Now,
$$\angle QPT+\angle PTU={180}^{o}$$ ($$PQ\parallel TU$$, co-int $$\angle s$$)........(i)
and $$\angle UTR+\angle TRS={180}^{o}$$ ($$TU\parallel RS$$, co-int, $$\angle s$$)..........(ii)
Adding (i) and (ii) we get $$\angle QPT+\angle PTU+\angle UTR+\angle TRS={180}^{o}+{180}^{o}={360}^{o}$$
$$a+\angle PTR+c={360}^{o}$$ $$\Rightarrow$$ $$a+b+c={360}^{o}$$
Question 9
1 / -0

In the given figure, $$AB\parallel CD$$. Find $$x$$.

A
$${26}^{o}$$

B
$${13}^{o}$$

C
$${23}^{o}$$

D
$${46}^{o}$$

Solution

Through point $$E$$ draw a line $$EF$$ parallel to $$CD$$ and $$AB$$
Now, $$\angle CEF=4x$$ ($$CD\parallel EF$$, corr. $$\angle s$$)
$$\angle AEF=6x$$($$EF\parallel AB$$, corr. $$\angle s$$)
Given, $$\angle CEA={130}^{o}$$ $$\Rightarrow$$ $$\angle CEF+\angle AEF={130}^{o}$$
$$\Rightarrow$$ $$4x+6x={130}^{o}$$ $$\Rightarrow$$ $$10x={130}^{o}$$
$$\Rightarrow$$ $$x={13}^{o}$$
Question 10
1 / -0

Three lines intersect at a point generating six angles. If one of these angles is $${90}^{o}$$, then the number of other distinct angles is:

A
$$1$$ or $$2$$

B
$$1$$ or $$3$$

C
$$2$$ or $$3$$

D
$$2$$ or $$4$$