The Triangle and Its Properties Test

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The Triangle and Its Properties Test - 25

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Question 1
1 / -0

Mark the correct alternative of the following.
In figure, the value of x is?

A
$$84$$

B
$$74$$

C
$$94$$

D
$$57$$

Solution

Exterior angle = sum of 2 opposite Interior Angles
$$123^0=39^0+x^0$$
$$x^0=123^0-39^0$$
$$\boxed{x^0=84^0}$$
Question 2
1 / -0

A triangle having sides of different lengths is called

A
an isosceles triangles

B
an equilateral triangle

C
a scalene triangle

D
a right triangle

Solution

A triangle having sides of different lengths is called a scalene triangle
Question 3
1 / -0

In an isosceles triangle, one angle is $$70^{o}$$. The other two angles are of
(i) $$55^{o}$$ and $$55^{o}$$
(ii) $$70^{o}$$ and $$40^{o}$$
(iii) any measure
In the given option(s) which of the above statements(s) are true?

A
(i) only

B
(ii) only

C
(iii) only

D
(i) and (ii)

Solution

Case II: Here, triangle $$ABC$$ is an isosceles triangle.
$$AB=AC$$ and vertex angle $$=70^{o}$$
$$\angle 1=\angle 2 \, \, \,...(1)\because AB=AC$$
Now,
$$\angle 1+\angle 2+\angle A=180^{o}$$
$$\Rightarrow 2(\angle 1)=180^{o}-70^{o}$$
$$\Rightarrow \angle 1=\dfrac{110^{o}}{2}=55^{o}$$
Therefore, $$\angle 1= \angle 2=55^{o}$$
In above Figure, Triangle ABC is an isosceles triangle
$$AB=AC$$, Base angle $$=\angle 2=70^{o}$$
$$\angle 1+\angle 2+\angle C=180^{o}$$
$$\Rightarrow \angle 1=180^{o}-70^{o}-70^{o}$$
$$\Rightarrow \angle 1=40^{o}$$
Therefore, $$\angle 1=40^{o}, \angle 2=70^{o}$$
Both statements are true.
Question 4
1 / -0

In a triangle, one angle is of $$90^{o}$$. Then
(i) The other two angles are of $$45^{o}$$ each
(ii) In remaining two angles, one angle is $$90^{o}$$ and other is $$45^{o}$$
(iii) Remaining two angles are complementary

In the given option(s) which is always true?

A
(i) only

B
(ii) only

C
(iii) only

D
(i) and (ii)

Solution

If one angle is right angle then the other both angles are a complementary angle in a triangle.
Question 5
1 / -0

In $$\Delta PQR$$,

A
$$PQ - QR > PR$$

B
$$PQ + QR < PR$$

C
$$PQ - QR < PR$$

D
$$PQ + PR > QR$$

Solution

In a triangle PQR,
$$\Rightarrow \,PQ + QR > PR$$
$$\Rightarrow \,QR + PR > PQ$$
$$\Rightarrow \,PR + PQ > QR$$
Since the third side of the triangle is less then the sum of any two sides of its.
Now,
$$\Rightarrow \,PQ - QR < PR$$
$$\Rightarrow \,QR - PR < PQ$$
$$\Rightarrow \,PR - PQ < QR$$
Since the length of third side of the triangle is always greater than the difference between two sides.
Question 6
1 / -0

If the exterior angle of a triangle is $$130^{o}$$ and its interior opposite angles are equal, then measure of each interior opposite angle is

A
$$55^{o}$$

B
$$65^{o}$$

C
$$50^{o}$$

D
$$60^{o}$$

Solution

Let $$y$$ and $$y$$ be the interior opposite angles.
$$\therefore 130^{o}=x+x$$ Exterior angle property
$$\Rightarrow 2x=130^{o}$$
$$\Rightarrow x=65^{o}$$
Therefore, each interior angle if of $$65^{o}$$.
Question 7
1 / -0

Which of Which of the following statements is not correct?

A
The sum of any two sides of a triangle is greater than the third side

B
A triangle can have all its angles acute

C
A right-angled triangle cannot be equilateral

D
Difference of any to sides of a triangle is greater than the third side

Solution

Since the length of the third side of the triangle is always greater than the difference between two sides.
Question 8
1 / -0

In Fig. $$BC = CA$$ and $$\angle A = 40$$. Then, $$\angle ACD$$ is equal to

A
$$40^o$$

B
$$80^o$$

C
$$120^o$$

D
$$60^o$$

Solution

$$BC = CA$$
$$\Longrightarrow \angle A = \angle B = 40^o$$
Now,
$$\angle ACD = \angle A + \angle B$$ ...Exterior angle property
$$= 40^o + 40^o = 80^o$$
Question 9
1 / -0

If an isosceles triangle, each of the base angles is $$40^o$$, then the triangle is

A
Right-angled triangle

B
Acute angled triangle

C
Obtuse angled triangle

D
Isosceles right-angled triangle

Solution

Triangle $$ABC$$ is an isosceles triangle in which base angle is $$40^o$$ and $$AB = AC$$
$$\angle A + \angle B + \angle C = 180^o$$
$$\Longrightarrow \angle A = 180^o - 40^o - 40^o$$ ...Angle sum property
$$\Longrightarrow \angle A = 100^o$$
Therefore, triangle $$ABC$$ is an obtuse-angled traingle.
Question 10
1 / -0

From Fig., the value of $$x$$ is

A
$$75^o$$

B
$$90^o$$

C
$$120^o$$

D
$$60^o$$

Solution

In triangle, $$ABC$$,
$$\angle ABC + \angle CAB = \angle ACD$$ ...Exterior angle property
$$\Longrightarrow ACD = 25^o + 35^o$$
$$\Longrightarrow \angle ACD = 60^o$$ ...(1)
Now,
$$\Longrightarrow x = \angle D + \angle ACD$$
$$\Longrightarrow x = 60^o + 60^o $$ using ...(1)
$$\therefore x = 120^o$$