The Triangle and Its Properties Test

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The Triangle and Its Properties Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

Identify the name of the triangle.

A
acute isosceles triangle

B
acute scalene triangle

C
acute obtuse triangle

D
acute right-angled triangle

Solution

We can see that, in the given triangle, the two angles are equal and all angles are less than $$90^{o}$$.

Hence, the given triangle is an acute isosceles triangle.
Question 2
1 / -0

Which one of the following is not an acute triangle?

A

B

C

D

Solution

Acute triangle has all angles less than $$90^o$$.
So, the second figure is not an acute triangle.
Question 3
1 / -0

Using the Pythagoras theorem find the length of the missing side.

A
10 cm

B
12 cm

C
7 cm

D
15 cm

Solution

$$\displaystyle { AB }^{ 2 }+{ BC }^{ 2 }={ AC }^{ 2 }$$
$$\displaystyle { 9 }^{ 2 }+{ 12 }^{ 2 }={ AC }^{ 2 }$$
$$\displaystyle 81+144={ AC }^{ 2 }$$
$$\displaystyle 225={ AC }^{ 2 }$$
$$\displaystyle { 15 }^{ 2 }={ AC }^{ 2 }$$
$$\displaystyle AC=15$$
Question 4
1 / -0

The angles of a triangle are in the ratio of $$2:3:4.$$ What is the measure of the smallest interior angle of the triangle?

A
$$10^{o}$$

B
$$20^{o}$$

C
$$30^{o}$$

D
$$40^{o}$$

Solution

By angle sum property, the sum of angles is $$180^o$$.
Let $$2x, 3x$$ and $$4x$$ be the three angles.
$$2x + 3x + 4x = 180^{o}$$
$$9x = 180^{o}$$
$$x = 20^{o}$$
Smallest interior angle $$= 2 \times 20^{o} = 40^{o}.$$
Question 5
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If m$$\angle$$ $$XYZ = 86^o$$ and m$$\angle$$ $$XZY = 23^o$$. What is m$$\angle$$ $$YXZ$$ in the triangle?

A
$$70^o$$

B
$$71^o$$

C
$$72^o$$

D
$$73^o$$

Solution

By angle sum property, the sum of angles is $$180^o$$.
$$86^o + 23^o +$$ m$$\angle$$ $$YXZ = 180^o$$
m$$\angle$$ $$YXZ = 180^o - 109^o$$
m$$\angle$$ $$YXZ = 71 ^o$$.
Question 6
1 / -0

What is the value of $$\angle$$ $$XRQ$$?

A
$$100^o$$

B
$$110^o$$

C
$$120^o$$

D
$$130^o$$

Solution

An exterior angle of a triangle is equal to the sum of the opposite interior angles.
So by this property:
$$\angle P+\angle Q = \angle XRQ$$
$$x - 30 + x - 20 = x + 30$$
$$2x - 50 = x + 30$$
$$2x - x = 30 + 50$$
$$x = 80$$
Therefore, $$\angle$$ $$XRQ = x + 30\Rightarrow 80 + 30 = 110^o$$
Question 7
1 / -0

In $$\Delta PQR, \angle Q = 23^{o}$$ and $$\angle R = 20^{o}.$$ What is the measure of $$\angle P?$$

A
$$134^{o}$$

B
$$135^{o}$$

C
$$136^{o}$$

D
$$137^{o}$$

Solution

By angle sum property, the sum of angles is $$180^o$$.
$$\angle P+ \angle Q + \angle R = 180$$
$$\angle P + 23^{o} + 20^{o} = 180^{o}$$
$$\angle P = 180^{o} - 43^{o}$$
$$\angle P = 137^{o}.$$
Question 8
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If m$$\angle$$ $$PRQ = 45^o$$ and m$$\angle$$ $$QPR = 68^o$$. What is m$$\angle$$ $$PQR$$ in the triangle?

A
$$67^o$$

B
$$68^o$$

C
$$69^o$$

D
$$70^o$$

Solution

by angle sum property, the sum of angles is $$180^o$$.

$$m\angle PRQ + m\angle QPR + m\angle PQR$$ $$= 180^o$$
$$45^o + 68^o +$$ m $$\angle$$ $$PQR = 180^o$$
m$$\angle$$ $$PQR = 180^o - 113^o$$
m$$\angle$$ $$PQR = 67^o$$ .
Question 9
1 / -0

In $$\Delta ABC, \angle A = 56^{o}$$ and $$\angle B = 60^{o}.$$ What is the measure of $$\angle C?$$

A
$$64^{o}$$

B
$$65^{o}$$

C
$$66^{o}$$

D
$$67^{o}$$

Solution

By angle sum property, the sum of angles is $$180^o$$.
$$\angle A+ \angle B + \angle C = 180^{o}$$
$$56^{o} + 60^{o} + \angle C = 180^{o}$$
$$\angle C = 180^{o} - 116^{o}$$
$$\angle C = 64^{o}$$
Question 10
1 / -0

Find $$\angle$$CAD.

A
$$102^o$$

B
$$110^o$$

C
$$180^o$$

D
$$122^o$$

Solution

An exterior angle of a triangle is equal to the sum of the opposite interior angles.
So by this property:
$$71^o$$ and $$51^o$$.
So, $$\angle$$ $$CAD = 71 + 51$$
$$\angle$$ $$CAD = 122^o$$