MCQ Questions for Class 7 Maths The Triangle And Its Properties Quiz 1

The pair of adjacent sides in the given quadrilateral is ___.

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AB, BC
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AB, CD
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BC, AD
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None of these.

Which of the given figures is a quadrilateral?

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0%
0%
0%
none of the above

AM is a median of a

$\triangle ABC$ . Which of the following is true?

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AB + BM + MC > AC
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AB + BC + AM > AC
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AM + BC + CA > 2 AB
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None of these.

Which two triangles have

$\angle B$ as common?

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$\triangle ABD \ and \ \triangle ADC$
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$\triangle ABD \ and \ \triangle ABC$
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$\triangle ABC \ and \ \triangle ADC$
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None of these.

Some of the angles in the given figure are

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$\angle DBE$
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$\angle ACE$
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$\angle BDC$
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$\angle BCA$

Identify all the triangles in the given figure.

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ABC
0%
DEB
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BCD
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ADC

Following is pair of opposite sides in the given quadrilateral.

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AB, BC
0%
BC, AD
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AB, CD
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BC, CD

The distance from A to town B is five miles. C is six miles from B. Which of the following could be the distance from A to C?
I.

$11$
II.

$1$
III.

$7$

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I only
0%
II only
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I and II only
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II and III only
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I, II and III

Write the correct answer of the following:

The median of a triangle divides it into two

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triangles of equal area
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congruent triangles
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right triangles
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isosceles triangles

A triangle with all 3 equal sides is called

0%
isosceles
0%
equilateral
0%
scalene
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none of these

Explanation

An equilateral triangle is one in which all three sides and angles are equal.

So,

$B$ is correct.

What is the name of the closed figure with four sides?

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Hexagon
0%
Triangle
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Pentagon
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Quadrilateral

Explanation

A quadrilateral is a closed figure having four sides.

Therefore,

$D$ is the correct answer.

The construction of a triangle ABC, given that BC = 3 cm is possible when difference of AB and AC is equal to :

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3.2 cm
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3.1 cm
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3 cm
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2.8 cm

Explanation

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.

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:

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$2$
0%
$1$
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$\dfrac12$
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$\dfrac14$

Explanation

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,

$x=1$ .

$\Delta ABC$ , if AB

$=$ BC then :

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$\angle B > \angle C$
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$\angle A = \angle C$
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$\angle A = \angle B$
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$\angle A < \angle C$

Explanation

$\triangle ABC$ ,

$AB=AC$

$\therefore$ the triangle is isoceles.

In an isoceles triangle the angles opposite to equal sides are equal.

$\therefore \angle A=\angle C$

Find all the angles of an equilateral triangle.

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$60^{0},60^{0},60^{0}$
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$90^{0},30^{0},60^{0}$
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$90^{0},45^{0},45^{0}$
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$120^{0},30^{0},30^{0}$

Explanation

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,

$\angle A = \angle B = \angle C = 60^{\circ}$

In ABC, If BC=AB and

$\angle B = 80^{\circ}$ , then

$\angle A$ is equal to:

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$80^{\circ}$
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$40^{\circ}$
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$50^{\circ}$
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$100^{\circ}$

Explanation

$BC=AB$

$\Rightarrow \angle A=\angle C$ (As angles opposite to equal side are equal)

let thte angles be

$x$ each.

Now by angle sum property

$\angle A+\angle B+\angle C={ 180 }^{ \circ }\\ \Rightarrow x+{ 80 }^{ \circ }+x={ 180 }^{ \circ }\\ \Rightarrow 2x={ 180 }^{ \circ }-{ 80 }^{ \circ }\\ \Rightarrow 2x={ 100 }^{ \circ }\\ \Rightarrow x={ 50 }^{ \circ }\\ \therefore \angle A={ 50 }^{ \circ }$

Find

$x$ in the given figure.

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$80^{\circ}$
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$40^{\circ}$
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$160^{\circ}$
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$20^{\circ}$

Explanation

Exterior angle of triangle is the sum of two interior opposite angles of triangle.

$\Rightarrow x+{ 60 }^{ \circ }={ 100 }^{ \circ }\\ \Rightarrow x={ 100 }^{ \circ }-{ 60 }^{ \circ }\\ \Rightarrow x={ 40 }^{ \circ }$

The exterior angle of a triangle is equal to the sum of two

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Exterior angles
0%
Interior angles
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Interior opposite angles
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Alternate angles

Write the measure of each angle of an isosceles right-angled triangle.

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$45^o, 90^o \,\, and \,\, 45^o$
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$30^o, 90^o \,\, and \,\, 60^o$
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$60^o, 60^o \,\, and \,\, 60^o$
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$30^o, 120^o \,\, and \,\, 30^o$

Explanation

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:

$45^{\circ}, 45^{\circ}, 90^{\circ}$

Define Pythagoras theorem.

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In a right angled triangle , square of a hypotenuse is not equal to the sum of the squares of two sides.
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In a right-angled triangle, the square of a hypotenuse is equal to the sum of the squares of the other two sides.
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In a right angled triangle , hypotenuse is equal to the sum of two sides.
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In a triangle , square of a side is equal to the square of another side.

Explanation

In a right angled triangle ,square of hypotenuse (longest side) is equal to the sum of squares of other two sides.

$\triangle ABC$

${ \left( AC \right) }^{ 2 }{ =\left( AB \right) }^{ 2 }+{ \left( BC \right) }^{ 2 }$

In figure, if lines PQ and RS intersect at point T, such that

$\angle PRT=40^o, \angle RPT=95^o$ and

$\angle TSQ=75^o$ , find

$\angle SQT$ .

0%
$20^o$
0%
$60^o$
0%
$30^o$
0%
$80^o$

Explanation

$\triangle PRT$

$\angle PTS=\angle PRT+\angle RTP$ (As exterior angle is equal to sum of interior opposite angles )

$\Rightarrow \angle PTS={ 95 }^{ \circ }+{ 40 }^{ \circ }={ 135 }^{ \circ }$

Now in

$\triangle SQT$

$\angle STQ=\angle SQT+\angle TSQ$ (As exterior angle is equal to sum of interior opposite angles )

$\Rightarrow { 135 }^{ \circ }=\angle SQT+{ 75 }^{ \circ }\\ \Rightarrow \angle SQT={ 135 }^{ \circ }-{ 75 }^{ \circ }={ 60 }^{ \circ }$

The

${\triangle }$ formed by BC =7.2 cm , AC =6 cm and

${\angle C}$ =

${120^0}$ is:

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An acute angle ${\triangle }$
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An obtuse angled ${\triangle }$
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A right angled ${\triangle }$
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None of these

Explanation

Given

$\angle C=120^{\circ}$

Here one of the angles of the triangle is greater than

$90^{\circ}$ . So the

$\triangle$ is obtuse angled triangle.

Therefore option

$B$ is correct.

If two angles in a triangle are

$40^o$ and

$60^o$ , then the third angle is:

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$90^o$
0%
$80^o$
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$70^o$
0%
$60^o$

Explanation

Let the third angle be

$x$

We know, by angle sum property, the sum of all angles of a triangles is

$180^0$ .

${ 40 }^{ \circ }+{ 60 }^{ \circ }+x={ 180 }^{ \circ }\\ { 100 }^{ \circ }+x={ 180 }^{ \circ }\\ \Rightarrow x={ 80 }^{ \circ }$

One of the angles of a triangle is

$65^o$ . Find the remaining two angles, if their difference is

$25^o$ .

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$70^o, 45^o$
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$55^o, 40^o$
0%
$50^o, 85^o$
0%
$75^o, 40^o$

Explanation

Let one angle be

$\angle x$

$\therefore$ other angle is

$\angle x+25^o$
Now,
We know, by angle sum property, the sum of all angles of a triangles is

$180^0$ .

$\implies x+x+25^o+65^o=180^o$

$\implies 2x=180^o-90^o$

$\implies x=45^o$
Thus, the angles are

$45^o$ and

$70^o$

The given road sign is an equilateral triangle. What is the measure of each angle?

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$45^\circ$
0%
$90^\circ$
0%
$60^\circ$
0%
$36^\circ$

Explanation

In an equilateral triangle all the angles are equal . Let each angle be

$x$

Now by angle sum property

$x+x+x={ 180 }^{ \circ }\\ \Rightarrow 3x={ 180 }^{ \circ }\\ \Rightarrow x={ 60 }^{ \circ }$

So option

$C$ is correct.

For any triangle

$ABC$ , the true statement is

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${ AC }^{ 2 }={ AB }^{ 2 }+{ BC }^{ 2 }$
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$AC=AB+BC$
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$AC>AB+BC$
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$AC<\,AB+BC$

Explanation

For any

$\triangle ABC$ , the sum of two sides must be greater than the third side.

Hence,

$AB + BC > AC$ .

Number of interior angles formed in a triangle are:

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$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Number of interior angles formed in a triangle are

$3.$

Here,

$m\angle A, m\angle B$ and

$m\angle C$ stand for measure of angle

$A,B$ and

$C.$

The sum of all exterior angles of a triangle is

0%
$360^o$
0%
$180^o$
0%
$540^o$
0%
none of these

Explanation

$\textbf{Step 1: Naming the angles of the triangle}$

$\text{Let the angles of the triangle be }x,\;y,\;\text{and }z$

$x+y+z=180^{\circ}$

$x+y=\text{Exterior angle A}$

$y+z=\text{Exterior angle B}$

$x+z=\text{Exterior angle C}$

$\textbf{Step 2: Finding sum of exterior angles}$

$\text{Sum of exterior angles}=x+y+y+z+x+z$

$=2x+2y+2z$

$=2(x+y+z)=2\times180=360^{\circ}$

$\textbf{Final Answer: The sum of all exterior angles of a triangle is }\mathbf{360^{\circ}}.$

The triangle formed by

$BC = 5 \ cm, AC = 3 \ cm, AB = 5.8 \ cm$ is:

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a right angled $\Delta$
0%
an isosceles $\Delta$
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an equilateral $\Delta$
0%
a scalene $\Delta$

Explanation

Given three lengths of sides are different.

Also, the sides do not follow Pythagora's theorem.

So, it is a scalene triangle.

Option

$D$ is correct.

In the given figure, XYZ is a/an_________triangle.

0%
isosceles
0%
equilateral
0%
scalene
0%
none of these

Explanation

In the given

$\triangle XYZ,$

$XY = XZ = 8$ cm
Since two sides are equal of the given triangle.

$\therefore \triangle XYZ$ is an isosceles triangle.
Option A is correct.

CBSE Questions for Class 7 Maths The Triangle And Its Properties Quiz 1 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths The Triangle And Its Properties Quiz 1 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

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