The Triangle and Its Properties Test

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The Triangle and Its Properties Test - 5

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Question 1
1 / -0

In a right triangle, the square of the hypotenuse is $$x$$ times the sum of the squares of the other two sides. The value of $$x$$ is:

A
$$2$$

B
$$1$$

C
$$\dfrac12$$

D
$$\dfrac14$$

Solution

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Its a standard pythagoras theorem for right angle triangles.
$$\mbox {Hyp}^2 = \mbox {Perpendicular}^2 + \mbox {Base}^2$$
Accordig to question we have
$$\mbox {Hyp}^2 = x(\mbox {Perpendicular}^2 + \mbox {Base}^2)$$
So on comparing the above two equations,
We get, $$x=1$$.
Question 2
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The exterior angle of a triangle is equal to the sum of two

A
Exterior angles

B
Interior angles

C
Interior opposite angles

D
Alternate angles

Solution

The exterior angle of a triangle is equal to the sum of two interior opposite angles.
Question 3
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In the given figure, the value of $$(\angle x+\angle y)$$ is :

A
$$110^0$$

B
$$100^0$$

C
$$120^0$$

D
$$60^0$$

Solution

$$\textbf{Step -1: Stating exterior angle theorem.}$$
$$\text{The sum of two interior adjacent angles is equal to its opposite exterior angle.}$$
$$\therefore \angle x + \angle y = 100^\circ$$
$$\textbf{Hence, option B is correct.}$$
Question 4
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In $$\Delta ABC$$, if AB $$=$$ BC then :

A
$$\angle B > \angle C$$

B
$$\angle A = \angle C$$

C
$$\angle A = \angle B$$

D
$$\angle A < \angle C$$

Solution

In $$\triangle ABC$$ , $$AB=AC$$
$$\therefore$$ the triangle is isoceles.
In an isoceles triangle the angles opposite to equal sides are equal.
$$\therefore \angle A=\angle C$$
Question 5
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Find all the angles of an equilateral triangle.

A
$$60^{0},60^{0},60^{0}$$

B
$$90^{0},30^{0},60^{0}$$

C
$$90^{0},45^{0},45^{0}$$

D
$$120^{0},30^{0},30^{0}$$

Solution

Given, $$\triangle ABC$$ is an equilateral triangle. In an equilateral triangle all the angles are equal.
$$\angle A = \angle B = \angle C = x$$
Sum of angles = 180
$$\angle A + \angle B + \angle C = 180$$
$$x + x + x = 180$$
$$x = 60$$
Thus, $$\angle A = \angle B = \angle C = 60^{\circ}$$
Question 6
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The construction of a triangle ABC, given that BC = 3 cm is possible when difference of AB and AC is equal to :

A
3.2 cm

B
3.1 cm

C
3 cm

D
2.8 cm

Solution

Let the length of $$AB$$ be $$x$$ and $$AC$$ be $$y$$
A triangle can be formed if the sum of any two sides is greater then the third
$$\Rightarrow BC+AC>AB\\ \Rightarrow 3+AC>AB\\ \Rightarrow 3>AB-AC\\ \Rightarrow AB-AC<3$$
So only option $$D$$is correct.
Question 7
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In given figure, value of $$(\angle x+ \angle y)$$ is

A
$$80^{\circ}$$

B
$$30^{\circ}$$

C
$$100^{\circ}$$

D
$$60^{\circ}$$

Solution

Exterior angle of triangle is equal to the sum of two interior opposite angles
$$\therefore x+y={ 100 }^{ \circ }$$
Option $$C$$ is correct.
Question 8
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In ABC, If BC=AB and $$\angle B = 80^{\circ}$$, then $$\angle A$$ is equal to:

A
$$80^{\circ}$$

B
$$40^{\circ}$$

C
$$50^{\circ}$$

D
$$100^{\circ}$$

Solution

$$BC=AB$$
$$\Rightarrow \angle A=\angle C$$ (As angles opposite to equal side are equal)
let thte angles be $$x$$ each.
Now by angle sum property
$$\angle A+\angle B+\angle C={ 180 }^{ \circ }\\ \Rightarrow x+{ 80 }^{ \circ }+x={ 180 }^{ \circ }\\ \Rightarrow 2x={ 180 }^{ \circ }-{ 80 }^{ \circ }\\ \Rightarrow 2x={ 100 }^{ \circ }\\ \Rightarrow x={ 50 }^{ \circ }\\ \therefore \angle A={ 50 }^{ \circ }$$
Question 9
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Find $$x$$ in the given figure.

A
$$80^{\circ}$$

B
$$40^{\circ}$$

C
$$160^{\circ}$$

D
$$20^{\circ}$$

Solution

Exterior angle of triangle is the sum of two interior opposite angles of triangle.
$$\Rightarrow x+{ 60 }^{ \circ }={ 100 }^{ \circ }\\ \Rightarrow x={ 100 }^{ \circ }-{ 60 }^{ \circ }\\ \Rightarrow x={ 40 }^{ \circ }$$
Question 10
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Write the measure of each angle of an isosceles right-angled triangle.

A
$$45^o, 90^o \,\, and \,\, 45^o$$

B
$$30^o, 90^o \,\, and \,\, 60^o$$

C
$$60^o, 60^o \,\, and \,\, 60^o$$

D
$$30^o, 120^o \,\, and \,\, 30^o$$

Solution

Since, the angle is right angled isosceles triangle. One of the angles will be $$90^{\circ}$$. let the other two angles be $$x$$
Thus, sum of angles = 180
$$90 + x + x = 180$$
$$2x = 90$$
x = $$45^{\circ}$$
The measure of each angle is: $$45^{\circ}, 45^{\circ}, 90^{\circ}$$