The Triangle and Its Properties Test

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The Triangle and Its Properties Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

A right angled triangle has $$24,7cm $$ as its sides . What will be its hypotenuse?

A
$$25$$ cm

B
$$37$$ cm

C
$$29$$ cm

D
$$53$$ cm

E
None of these

Solution

The sides of right angled triangle are $$24,7cm$$

The hypotenuse is given as

$$a^2+b^2=c^2\\7^2+24^2=c^2\\49+576=c^2\\c^2=625\\c=\sqrt {625}\\c=25cm$$
Question 2
1 / -0

If two angles of a triangle are $$60^o$$ each, then the triangle is

A
Isosceles but not equilateral

B
Scalene

C
Equilateral

D
Righe-angled

Solution

Let triangle $$ABC$$, in which the base angle is $$60^o$$
$$\angle A + \angle B + \angle C = 180^o$$
$$\Longrightarrow \angle A = 180^o - 60^o - 60^o$$
$$\Longrightarrow \angle A = 60^o$$
Therefore, $$\Delta ABC$$ is $$60^o$$ equilateral triangle.
Question 3
1 / -0

An isosceles right triangle has area $$8\ cm^{2}$$. The length of its hypotenuse is

A
$$\sqrt{32} cm$$

B
$$\sqrt{16} cm$$

C
$$\sqrt{48} cm$$

D
$$\sqrt{24} cm$$

Solution

Explanation:
Let height of triangle = $$h$$

As the triangle is isosceles,

Let base = height =$$h$$

According to the question, Area of triangle = $$8cm^{2}$$

$$\implies \dfrac{1}{2} \times Base \times Height = 8$$

$$\implies \dfrac{1}{2} \times h \times h = 8$$

$$\implies h^{2}= 16$$

$$\implies h = 4cm$$

Base = Height = $$4cm$$

Since the triangle is right angled,

$$Hypotenuse^{2} = Base^{2} + Height^{2}$$

$$\implies Hypotenuse^{2} = 4^{2} + 4^{2}$$

$$\implies Hypotenuse^{2} = 32$$

$$\implies Hypotenuse = \sqrt{32}$$

Hence, Options $$A$$ is the correct answer.
Question 4
1 / -0

A triangle formed by the sides of lengths $$4.5\ cm,6\ cm,$$ and $$4.5\ cm$$ is

A
scalene

B
isosceles

C
equilateral

D
none of these

Solution

A triangle formed by the sides of lengths $$4.5\ cm,6\ cm,$$ and $$4.5\ cm$$ is isosceles. As, two sides are of equal length.
Question 5
1 / -0

Select the correct alternative . If $$a , b ,c$$ are sides of a triangle and $$a^{2} + b ^{2}= c^{2} $$, name the type of triangle

A
Obtuse angled triangle

B
Acute angled triangle

C
Right angled triangle

D
Equilateral triangle

Solution

In a triangle , if the square of one side is equal to the sum of the squares of the remaining two sides, then the triangle is right angled triangle.
Hence, C is the correct option
Question 6
1 / -0

An exterior angle of a triangle is $$105^0$$ and its two interior opposite angles are equal. Each of these equal angles is

A
$$37\dfrac{1^{0}}{2}$$

B
$$52\dfrac{1^{0}}{2}$$

C
$$72\dfrac{1^{0}}{2}$$

D
$$75^0$$

Solution

Given: Exterior angle = $$105^{\circ}$$
Let the two Interior angles be $$x$$.
Sum of Interior opposite angles = Exterior angle
$$x + x = 105$$
$$2x = 105$$
$$x = 52 \dfrac{1}{2} ^{\circ}$$
Question 7
1 / -0

Write the correct answer of the following:
The median of a triangle divides it into two

A
triangles of equal area

B
congruent triangles

C
right triangles

D
isosceles triangles

Solution

The median of a triangle divides it into two triangles of equal area.
Hence, (a) is the correct answer.
Question 8
1 / -0

In the figure, if $$\angle A + \angle B + \angle C + \angle D + \angle E + \angle F = k\times $$ right angle, then $$k$$ is :

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

$$\Rightarrow$$ In $$\triangle ABC$$

$$\Rightarrow$$ $$\angle A+\angle B+\angle C=180^\circ$$ [Sum of interior angles of triangle is $$180^\circ$$] --- ( 1 )

$$\Rightarrow$$ In $$\triangle DEF$$

$$\Rightarrow$$ $$\angle D+\angle E+\angle F=180^\circ$$ [Sum of interior angles of triangle is $$180^\circ$$] ---( 2 )

$$\Rightarrow$$ $$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=180^\circ+180^\circ$$ [Adding ( 1 ) and ( 2 )]

$$\Rightarrow$$ $$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=360^\circ$$

$$\Rightarrow$$ $$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=k\times (right\, angle)$$.

$$\therefore$$ $$k=\dfrac{360^o}{90^o}$$

$$\therefore$$ $$k=4$$
Question 9
1 / -0

If one of the angles of a triangle is $$130^0$$, then the angle between the bisectors of the other two angles can be

A
$$50^0$$

B
$$65^0$$

C
$$145^0$$

D
$$155^0$$

Solution

Let $$x$$ and $$y$$ be the bisected angles.
So, in the original triangle,
sum of angles $$=180^o$$
Therefore, $$2x+2y+130=180$$
$$\Rightarrow 2(x+y)=50$$
$$\Rightarrow x+y=25$$
In the smallest triangle, consisting of original side opposite $$130^o$$.
Therefore, $$25^o+$$ Angle between bisectors $$=180^o$$
$$\Rightarrow $$ Angle between bisectors of other two angles $$=155^o$$.
Question 10
1 / -0

In $$\Delta ABC$$, if $$AB=AC$$ and $$\angle B=50^{0}$$, then $$\angle A$$ is equal to:

A
$$40^{0}$$

B
$$50^{0}$$

C
$$80^{0}$$

D
$$130^{0}$$

Solution

Given, $$AB=AC$$ and $$\angle B=50^o$$
We have to find $$\angle A$$
As $$\angle B=50^o$$, we have $$\angle C=50^o$$
Therefore, $$\angle A+\angle B+\angle C=180^o$$
$$\Rightarrow 50+50+\angle A=180^o$$
$$\Rightarrow 100+ \angle A=180^o$$
$$\Rightarrow \angle A=80^o$$

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Re-Attempt Weekly Quiz Competition

The Triangle and Its Properties Test - 15

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The Triangle and Its Properties Test - 15

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SHARING IS CARING

Question 1

$$25$$ cm

$$37$$ cm

$$29$$ cm

$$53$$ cm

None of these

Solution

Question 2

Isosceles but not equilateral

Scalene

Equilateral

Righe-angled

Solution

Question 3

$$\sqrt{32} cm$$

$$\sqrt{16} cm$$

$$\sqrt{48} cm$$

$$\sqrt{24} cm$$

Solution

Question 4

scalene

isosceles

equilateral

none of these

Solution

Question 5

Obtuse angled triangle

Acute angled triangle

Right angled triangle

Equilateral triangle

Solution

Question 6

$$37\dfrac{1^{0}}{2}$$

$$52\dfrac{1^{0}}{2}$$

$$72\dfrac{1^{0}}{2}$$

$$75^0$$

Solution

Question 7

triangles of equal area

congruent triangles

right triangles

isosceles triangles

Solution

Question 8

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 9

$$50^0$$

$$65^0$$

$$145^0$$

$$155^0$$

Solution

Question 10

$$40^{0}$$

$$50^{0}$$

$$80^{0}$$

$$130^{0}$$

Solution

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