Lines and Angles Test

Result

Lines and Angles Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

The angle formed by the pages of an open book is:

A
Acute

B
Obtuse

C
Right

D
Straight

Solution

The angle formed by the pages of an open book is obtuse.
Since the angle will always be more than $$90$$ but less than $$180$$.
Question 2
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 3
1 / -0

When the clock shows time $$5:20$$, the angle formed between the two hands is

A
obtuse

B
right

C
acute

D
none of these

Solution

Angles

between $$ {0}^{o} $$ and $$ {90}^{o} $$ are acute angles.

Since at the time $$ 5:20$$, the hands of the clock form an angle less than $$ {90}^{o} $$, the angle formed between the two hands is an acute angle.
Question 4
1 / -0

If the angles of a triangle are in the ratio $$2:3:4$$, find the three angles.

A
$$80^o, 120^o, 160^o$$

B
$$20^o, 30^o, 40^o$$

C
$$40^o, 60^o, 80^o$$

D
None of these

Solution

The angles of a triangle are in the ratio $$2:3:4 $$.
Let $$x:y:z=2:3:4$$.
Then, $$x=2t, y=3t$$ and $$z=4t$$.

We know, by angle sum property, the sum of all angles of a triangles is $$ 180^0$$.
$$\implies$$ $$ 2t + 3t + 4t = 180^o $$
$$\implies$$ $$ 9t = 180^o$$
$$\implies$$ $$ t = 20^o $$.

Therefore, $$ x= 2\times 20^o= 40^o; y= 3\times 20^o= 60^o; z = 4\times 20^o= 80^o $$.

Hence, option $$C$$ is correct.
Question 5
1 / -0

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is:

A
Obtuse angle triangle

B
Acute angle triangle

C
Right angle triangle

D
None of these

Solution

$$\textbf{Step-1: Apply properties of sum of angles of triangle to find it's one angle}$$
$$\text{Let the angles of a triabgle be}$$ $$\alpha ,\beta ,\gamma $$
$$\text{Given}$$ $$\alpha +\beta =\gamma $$
$$\text{We now that in a sum of triangles sum of angles is}$$ $${ 180 }^{ \circ }$$
$$\text{So,}$$$$ \alpha +\beta +\gamma ={ 180 }^{ \circ }$$
$$\Rightarrow 2\gamma ={ 180 }^{ \circ }$$
$$\Rightarrow \gamma ={ 90 }^{ \circ }$$
$$\text{So, it is a right angled triangle.}$$
$$\textbf{Hence option C is correct}$$
Question 6
1 / -0

In fig, then $$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=$$

A
$$180^o$$

B
$$270^o$$

C
$$360^o$$

D
$$540^o$$

Solution

In $$\triangle AEC$$,
$$ \angle A + \angle E + \angle C = 180^o $$ .....[Angle sum property].

Also, in $$\triangle BDF$$,
$$ \angle B + \angle D + \angle F = 180^o $$ .....[Angle sum property].

Adding both the equations, we get,
$$ \angle A+\angle B+\angle C+\angle D+\angle E+\angle F= 180^o + 180^o = 360^o $$.

Therefore, option $$C$$ is correct.
Question 7
1 / -0

Complementary angle of $$72\displaystyle \frac{1}{2}^{\circ}$$ is:

A
$$17^{\circ}$$

B
$$18\displaystyle \frac{1}{2}^{\circ}$$

C
$$21\displaystyle \frac{1}{2}^{\circ}$$

D
$$17\displaystyle \frac{1}{2}^{\circ}$$

Solution

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

Given, measure of one complementary angle is $$72\dfrac{1}{2}^o=\dfrac{145}{2}^o.$$

$$\Rightarrow$$ Measure of other complementary angle $$=90^o-\dfrac{145}{2}^o$$ $$=\dfrac{35}{2}^o=17\dfrac{1}{2}^o.$$

$$\therefore$$ Measure of a complementary angle of $$72\dfrac{1}{2}^o=$$ $$17\dfrac{1}{2}^o.$$
Therefore, option $$D$$ is correct.
Question 8
1 / -0

In the given figure, value of $$x$$ is .........

A
45$$^o$$

B
34$$^o$$

C
64$$^o$$

D
109$$^o$$

Solution

In triangle $$\Delta ABC$$,
$$\angle A+\angle ABC+\angle BCA=180^o$$
$$34^o+ \angle ABC +30^o=180^o$$
$$\angle ABC=116^o$$
So, $$\angle ABC+\angle CBD=180^o$$
$$116^o+\angle CBD=180^o$$
$$\angle CBD=180^o-116^o$$
$$\angle CBD=64^o$$
Now use remote interior angle theorem.
So, $$\angle X = \angle CBD+ \angle D=64^o+45^o=109^o$$
Question 9
1 / -0

A reflex angle is .........a straight angle.

A
Smaller than

B
Larger than

C
Equal to

D
One-half of

Solution
Question 10
1 / -0

The measure of angle $$y$$ in the given figure is ......... .

A
$$\angle y = 65^{\circ}$$

B
$$\angle y = 85^{\circ}$$

C
$$\angle y = 75^{\circ}$$

D
$$\angle y = 80^{\circ}$$

Solution

$$\textbf{Step 1: Using property of angle of straight line calculate y}$$
$$\text{We know that a straight line extends an angle of }{180^\circ}$$
$$\implies \angle \text{ABC}+\angle \text{CBD}=180^\circ$$
$$\implies \angle \text{CBD
}=180^\circ - \angle \text{ABC
}$$
$$\implies y^\circ=180^\circ - 115^\circ$$
$$\implies y^\circ=65^\circ$$

$$\mathbf{\text{Thus, the measure of angle y}=65^\circ}$$