The Triangle and Its Properties Test

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The Triangle and Its Properties Test - 21

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Question 1
1 / -0

Find the value of $$X.$$

A
$$60^{o}$$

B
$$70^{o}$$

C
$$80^{o}$$

D
$$90^{o}$$

Solution

The exterior angle is equal to the sum of the two non-adjacent interior angles which are $$70^{o}$$ and $$10^{o}.$$
So, $$X = 70^{o} + 10^{o}$$
$$X = 80^{o}.$$
Question 2
1 / -0

The angles of a triangle are in the ratio of $$3:5:7.$$ What is the measure of the largest interior angle of the triangle?

A
$$18^{o}$$

B
$$48^{o}$$

C
$$84^{o}$$

D
$$96^{o}$$

Solution

By angle sum property, the sum of angles is $$180^o$$.
Let $$3x, 5x$$ and $$7x$$ be the three angles of the triangle.
So,
$$3x + 5x + 7x = 180^{o}$$
$$15x = 180^{o}$$
$$x = 12^{o}$$

Hence, Largest interior angle $$= 7 \times 12^{o} = 84^{o}$$
Question 3
1 / -0

In $$\Delta ABC,$$ $$\angle A = 43^{o}$$ and $$\angle C = 70^{o}.$$ What is the measure of $$\angle B?$$

A
$$63^{o}$$

B
$$65^{o}$$

C
$$66^{o}$$

D
$$67^{o}$$

Solution

By angle sum property, the sum of angles is $$180^o$$.
$$\angle A+ \angle B + \angle C = 180^{o}$$
$$43^{o} +\angle B + 70^{o} = 180^{o}$$
$$\angle B = 180^{o} - 113^{o}$$
$$\angle B = 67^{o}.$$
Question 4
1 / -0

Find the value of exterior angle.

A
$$100^{\circ}$$

B
$$110^{\circ}$$

C
$$120^{\circ}$$

D
$$115^{\circ}$$

Solution

An exterior angle of a triangle is equal to the sum of the opposite interior angles.
So by this property:
$$x - 20^{\circ} + 2x - 45^{\circ} = 2x - 5^{\circ}$$
$$3x - 65^{\circ} = 2x - 5^{\circ}$$
$$3x - 2x = 65^{\circ} - 5^{\circ}$$
$$x = 60^{\circ}$$
Therefore, $$\angle$$ $$F$$ exterior angle $$= 2x - 5^{\circ}$$
$$= 2\times 60^{\circ} - 5^{\circ} = 115^{\circ}$$
Question 5
1 / -0

Find the measure of an exterior angle at the base of an isosceles triangle measures $$78^o$$

A
$$100^o$$

B
$$101^o$$

C
$$102^o$$

D
$$103^o$$

Solution

Base angles of an isosceles triangle are equal
$$\therefore \angle B=\angle C=78^{\circ}$$
Now $$x+\angle C=180^{\circ}$$ (Angles made on straight line)
$$x=180^{\circ}-78^{\circ}=102^{\circ}$$
So exterior angle is $$102^{\circ}$$
Question 6
1 / -0

What is the value of $$x$$ using Pythagoras theorem?

A
10

B
12

C
5

D
4

Solution

In a right triangle, $$a^{2} + b^{2} = h^{2}$$, where $$a$$ and $$b$$ are the lengths of the legs and $$c$$ is the length of the hypotenuse. This is called Pythagoras Theorem.
$$a = 5, b = x, c =$$ hypotenuse side $$= 13$$
$$5^{2} + x^{2} = 13^{2}$$
$$169 - 25 =$$ $$x^{2}$$
$$x^{2}$$ $$= 144$$
Taking root on both the sides, we get
$$x = 12$$
Question 7
1 / -0

In $$\Delta$$ $$PQR$$, an exterior angle at $$R$$ is represented by $$5x + 10$$. If the two non-adjacent interior angles are represented by $$3x + 15$$ and $$3x - 20$$, find the value of the exterior angle, $$\angle$$ $$R$$.

A
$$90^o$$

B
$$80^o$$

C
$$84^o$$

D
$$85^o$$

Solution

The exterior angle is equal to the sum of the two non-adjacent interior angles.
$$3x + 15 + 3x - 20 = 5x + 10$$
$$6x - 5 = 5x + 10$$
$$6x - 5x = 10 + 5$$
$$x = 15$$
Therefore, $$\angle$$ $$R = 5x + 10 => 5\times15 + 10 = 85^o$$
Question 8
1 / -0

In $$\triangle ABC$$, the line segment $$AM_1$$ is the median from vertex ____ to the opposite side $$BC$$ at $$M_1$$.

A
$$A$$

B
$$B$$

C
$$C$$

D
$$D$$

Solution

A median of a triangle is the line segment that joins any vertex of the triangle with the mid-point of its opposite side.
Therefore, $$A$$ is the vertex for the median $$M_{1}$$.
Question 9
1 / -0

How many medians(drawn in fig.) are there in a triangle $$ABC$$?

A
1

B
2

C
3

D
4

Solution

A median of a triangle is the line segment that joins any vertex of the triangle with the mid-point of its opposite side.
So, here the triangle is having only one median.
Question 10
1 / -0

Find the value of $$x$$ and $$y$$. (Use Pythagoras theorem).

A
$$x = 12$$ and $$y = 3$$

B
$$x = 10$$ and $$y = 5$$

C
$$x = 12$$ and $$y = 5$$

D
$$x = 11$$ and $$y = 5$$

Solution

In a right triangle, $$a^{2} + b^{2} = c^{2}$$, where a and b are the lengths of the legs and c is the length of the hypotenuse. This is called Pythagoras Theorem.
$$a = 3, b = 4, c =$$ hypotenuse side $$= y$$
$$3^{2} + 4^{2} = c^{2}$$
$$9 + 16 =$$ $$c^{2}$$
$$c^{2}$$ $$= 25$$
Taking root on both the sides, we get
$$c = 5 = y$$
So, $$x^{2}+y^{2}= 13^{2}$$
$$x^{2}+5^{2}= 13^{2}$$
$$x^{2} = 169 - 25$$
$$x^{2} = 144$$
$$x = 12$$
So, the value of $$x = 12$$ and $$y = 5$$.