MCQ Questions for Class 7 Maths The Triangle And Its Properties Quiz 6

The length of side of an equilateral triangular garden whose perimeter is

$66 \ m$ is:

0%
$66 \ m$
0%
$11 \ m$
0%
$3 \ m$
0%
$22 \ m$

Explanation

Let the side of the equilateral triangular garden be

$a$ ,
Then,

$3a = 66$

$\Rightarrow a\, =\, \displaystyle \frac {66}{3}\, =\, 22\, \ m$
Hence length of side of garden

$= 22 \ m.$

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is:

0%
Obtuse angle triangle
0%
Acute angle triangle
0%
Right angle triangle
0%
None of these

Explanation

$\textbf{Step-1: Apply properties of sum of angles of triangle to find it's one angle}$

$\text{Let the angles of a triabgle be}$

$\alpha ,\beta ,\gamma$

$\text{Given}$

$\alpha +\beta =\gamma$

$\text{We now that in a sum of triangles sum of angles is}$

${ 180 }^{ \circ }$

$\text{So,}$

$\alpha +\beta +\gamma ={ 180 }^{ \circ }$

$\Rightarrow 2\gamma ={ 180 }^{ \circ }$

$\Rightarrow \gamma ={ 90 }^{ \circ }$

$\text{So, it is a right angled triangle.}$

$\textbf{Hence option C is correct}$

In figure, sides QP and RQ of

$\Delta PQR$ are produced to points S and T respectively. If

$\angle SPR=135^o$ and

$\angle PQT=110^o$ , find

$\angle PRQ$ .

0%
$85^o$
0%
$65^o$
0%
$45^o$
0%
$35^o$

Explanation

Here,

$\angle QPR+135^o=180^o$ ....[Linear pair]

$\Rightarrow \angle QPR=45^o$ .

Now,

$\angle QRP+\angle QPR=\angle PQT$ ....[Exterior angle property]

$\Rightarrow \angle QRP+45^o=110^o$

$\Rightarrow \angle QRP=65^o$ .

Therefore, option

$B$ is correct.

If the angles of a triangle are in the ratio

$2:3:4$ , find the three angles.

0%
$80^o, 120^o, 160^o$
0%
$20^o, 30^o, 40^o$
0%
$40^o, 60^o, 80^o$
0%
None of these

Explanation

The angles of a triangle are in the ratio

$2:3:4$ .
Let

$x:y:z=2:3:4$ .
Then,

$x=2t, y=3t$ and

$z=4t$ .

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

$180^0$ .

$\implies$

$2t + 3t + 4t = 180^o$

$\implies$

$9t = 180^o$

$\implies$

$t = 20^o$ .

Therefore,

$x= 2\times 20^o= 40^o; y= 3\times 20^o= 60^o; z = 4\times 20^o= 80^o$ .

Hence, option

$C$ is correct.

In fig, then

$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=$

0%
$180^o$
0%
$270^o$
0%
$360^o$
0%
$540^o$

Explanation

$\triangle AEC$ ,

$\angle A + \angle E + \angle C = 180^o$ .....[Angle sum property].

Also, in

$\triangle BDF$ ,

$\angle B + \angle D + \angle F = 180^o$ .....[Angle sum property].

Adding both the equations, we get,

$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F= 180^o + 180^o = 360^o$ .

Therefore, option

$C$ is correct.

In triangle

$ABC, D$ is any point on side

$BC$ , then:

$AB + BC + AC > 2AD$

0%
True
0%
False

Explanation

In triangle ABD,

$AB + BD > AD$ [because, the sum of any two sides of a triangle is always greater than the third side]
In triangle

$ADC$ ,

$AC + DC > AD$ [because, the sum of any two sides of a triangle is always greater than the third side]
Adding these we get,

$AB + BD + AC + DC > AD + AD$

$\Rightarrow AB + (BD + DC) + AC > 2AD$

$\Rightarrow AB + BC + AC > 2AD$
Hence proved

Which of the following is true?
(i) A triangle can have two right angles.
(ii) A triangle can have all angles less than

$60^o$ .
(iii) A triangle can have two acute angles.

0%
Only (ii)
0%
Only (i)
0%
Only (iii)
0%
All are true

Explanation

(i) If the triangle has two right angles, the sum of angles of the triangle will be more than

$180^{\circ}$ , which is not possible.
(ii) If the triangle has all the angles less than

$60^{\circ}$ , the sum of the triangle will be less than

$180^{\circ}$ , which is not possible.
(iii) Thus, the only possibility is the triangle has two acute angles.

$\Delta\, ABC$ , if

$AB = BC$ and

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

$\angle C\, =$

0%
$50^{\circ}$
0%
$100^{\circ}$
0%
$130^{\circ}$
0%
None

Explanation

Given, in

$\triangle ABC, AB=BC, \angle B=80^o$

$\Rightarrow\, \angle A\, =\, \angle C$
We know

$\angle A\, +\, \angle B\, +\, \angle C\, =\, 180^{\circ}$

$\Rightarrow 2\, \angle C\, +\, 80^{\circ}=\, 180^{\circ}$

$\Rightarrow \angle\,C =\, 50^{\circ}$ .

Which of the following statement(s) is/are false with respect to quadrilateral?

0%
Each diagonal of a quadrilateral divides it into two triangles.
0%
Each side of a quadrilateral is less than the sum of the remaining three sides.
0%
A quadrilateral can at-most have three obtuse angles.
0%
A quadrilateral has four diagonals.

Explanation

Option A is true. As each diagonal of a quadrilateral divides it into two triangles.

Option B is also correct as it is not possible to construct a quadrilateral with one side greater than the sum of three other sides.

Option C is also true. A quadrilateral can at most have three obtuse angles. All the four angles cannot be obtuse and sum of 4 angles of a quadrilateral is

$360^\circ$ .

Option D is incorrect. A quadrilateral does not have four diagonals.

A quadrilateral has exactly two diagonals.

So, option (D) is false

The angles in a right angled isosceles triangle are

0%
$\displaystyle 90^{\circ},30^{\circ},60^{\circ}$
0%
$\displaystyle 90^{\circ},20^{\circ},70^{\circ}$
0%
$\displaystyle 90^{\circ},40^{\circ},50^{\circ}$
0%
$\displaystyle 90^{\circ},45^{\circ},45^{\circ}$

Explanation

${\textbf{Step - 1: Defining right angled isosceles triangle}}$

${\text{In geometry, an isosceles triangle means a triangle with two angles or sides equal to each other}}{\text{.}}$

${\text{Right angled isosceles triangle means a triangle with one angle equal to 90}}^\circ {\text{ and }}$

${\text{other two equal to each other}}{\text{.}}$

${\textbf{Step - 2: Finding all the angles}}$

${\text{Since one angle is equal to 90}}^\circ {\text{,}}$

$\therefore {\text{ x + x + 90 = 180 (interior sum property of triangle)}}$

$\Rightarrow {\text{ 2x = 90}}$

$\Rightarrow {\text{ x = 45}}^\circ$

$\mathbf{{\text{Thus, the angles in a right angles isosceles triangle are 90}}^\circ {\text{, 45}}^\circ {\text{, 45}}^\circ }$

$4$ cm and

$3$ cm are the lengths of two sides of a triangle then the length of the third side may be_____.

0%
$11$ cm
0%
$3$ cm
0%
$5$ cm
0%
$12$ cm

Explanation

In a triangle, sum of two sides

$>$ third side.

So, if two sides are

$4 cm$ and

$7 cm$ , then considering the given options,

A.

$11 cm$

$4 + 7 \ngtr 11$

Hence, this is not the correct option.

B.

$6 cm$
We see that

$4 + 7 > 6$

$4 + 6 > 7$

$7 + 6 > 4$
Hence,

$6 cm$ is the correct option.

In a right angled triangle

$ABC,\,\angle B=90^{\circ}$ such that

$AB=6\;cm,\;BC=8\;cm$ . Then find

$AC$ .

0%
$13\ cm$
0%
$14\ cm$
0%
$10\ cm$
0%
$12\ cm$

Explanation

By using Pythagoras theorem in

$\triangle ABC,$

$AC^2=AB^2+BC^2$

$\Rightarrow AC^2=6^2+8^2$

$\Rightarrow AC^2=36+64$

$\Rightarrow AC^2=100$

$\Rightarrow AC=10\ cm$

$\triangle ABC,\angle B={ 90 }^{ \circ }$ find the sides of the triangle if

$AB=(x-3)~cm,BC=(x+4)~cm$ and

$AC=(x+6)~cm$ .

0%
$8 cm, 15 cm, 17 cm$
0%
$17 cm, 24 cm, 26 cm$
0%
$9 cm, 16 cm, 18 cm$
0%
$12 cm, 19 cm, 21 cm$

In the given figure, value of

$x$ is .........

0%
45 $^o$
0%
34 $^o$
0%
64 $^o$
0%
109 $^o$

Explanation

In triangle

$\Delta ABC$ ,

$\angle A+\angle ABC+\angle BCA=180^o$

$34^o+ \angle ABC +30^o=180^o$

$\angle ABC=116^o$
So,

$\angle ABC+\angle CBD=180^o$

$116^o+\angle CBD=180^o$

$\angle CBD=180^o-116^o$

$\angle CBD=64^o$
Now use remote interior angle theorem.
So,

$\angle X = \angle CBD+ \angle D=64^o+45^o=109^o$

$ABCD$ is a rectangle. If ABP and BCQ are equilateral triangle,

$\angle PBQ$ = ....

0%
$65^0$
0%
$75^0$
0%
$60^0$
0%
$90^0$

Explanation

$ABCD$ is a rectangle. So,

$\angle ABC=90^0$
ABP and BCQ are equilateral triangle, So,

$\angle CBQ=\angle ABP = 60^0$
So,

$\angle PBQ=\angle ABC-\angle ABP+\angle CBQ\\=90^0-60^0+60^0\\=90^0$

Which of the following statements is correct?

0%
A triangle can have two right angles.
0%
All the angles of a triangle are more than 60 $^o$
0%
An exterior angle of a triangle is always greater than each of the opposite interior angles.
0%
All the angles of a triangle are less than 60 $^o$

Explanation

(A), If a triangle has two right angles, the third angle will become zero angle. The statement is false.
(B) If all angles of the triangles are greater than

$60^o$ , then sum of angles in triangle will be greater than

$180^o$ . The statement is false.
(C) An exterior angle of a triangle is equal to sum of the opposite interior angles, hence it will be greater than each of the opposite interior angles. The statement is correct.
(D) If all angles of the triangles are less than

$60^o$ , then sum of angles in triangle will be less than

$180^o$ . The statement is false.
Hence option (C) is the correct answer.

If two sides of an isosceles triangle are 4 cm and 10 cm then the length of the third side is__

0%
4 cm
0%
10 cm
0%
4 cm or 10 cm
0%
None of these

Explanation

Two sides of an isosceles triangle are

$4$ cm and

$10$ cm

In an isosceles triangle two sides are equal.

Case 1:

Let the third side be

$4$ cm.

For any triangle with sides let say

$a , b ,c$ , then

$a+b > c$ and

$c+b > a$ and

$c+a > b$

In this case

$4+4 = 8 < 10$

So, the third side of

$4$ cm will not possible.

Case 2:

Let third side be

$10$ cm

$10+4=14> 10$ and

$10+10=20>10$

Therefore,

$10$ cm is the third side.

Hence, the length of the third side is

$10$ cm.

The angles in a right-angled triangle other than the right angle are-

0%
Right
0%
Obtuse
0%
Acute
0%
None of these

Explanation

$\displaystyle \because$ One of the angle is

$\displaystyle 90^{\circ}$ and the sum of other two angles is

$\displaystyle 90^{\circ}$ So they are acute
1069006_404322_ans_c79cb1acf0a44b21a85e7dec788e10be.png

The sides of a right triangle are

$(x-1)$ ,

$x$ and

$(x+1)$ . Find the sides of the triangle.

0%
$3, 4, 5$
0%
$5, 5, 6$
0%
$2, 3, 4$
0%
None of these

Explanation

The sides of a triangle given are

$x-1, x$ and

$x+1.$

$\because x+1>x>x-1,$

$\forall x>0$

$\implies$

$x+1$ is the longest side
As it is a right angled triangle, we apply Pythagoras theorem
therefore,

$\left ( x-1 \right )^{2}+x^{2}=\left ( x+1 \right )^{2}$

$\Rightarrow x^{2}-2x+1+x^{2}=x^{2}+2x+1$

$\Rightarrow x^{2}-4x=0$

$\Rightarrow x\left ( x-4 \right )=0$

$x=4$ ,

$x=0$

$x=0$ cannot be the side of a triangle.
One side is

$4.$
Other two sides

$x-1=4-1=3$

$x+1=4+1=5$

$\therefore$ Sides of a triangle are

$3,4$ and

$5.$
Answer is option

$\text{A}$

If two angles of a triangle are acute angles, then the third angle:

0%
is less than the sum of the two angles
0%
should be an acute angle
0%
is the largest angle of the triangle
0%
is an angle with any measurement less than $180^o$

Rohit used the Pythagorean theorem to see how much time he would save taking a shortcut to home from football practice. He usually walked

$6$ blocks south and

$9$ blocks east. Which picture shows his shortcut?

Explanation

According to the Pythagorean theorem

$(AB)^2+(BC)^2=(AC)^2$

$(6)^2+(9)^2=(AC)^2$

$(AC)^2=36+81$

$AC=\sqrt{117}$
Hence, option

$A$ is answer.

The number of lines of symmetry in a scalene triangle is

0%
0
0%
1
0%
2
0%
3

Which of the following is/are true?

0%
The $\triangle$ is right angled triangle.
0%
The $\triangle$ is scalene triangle.
0%
The $\triangle$ is Isosceles triangle.
0%
The $\triangle$ is obtuse triangle.

Explanation

$9^2+40^2 = 41^2$

The given triangle satisfies Pythagoras theorem. Therefore, the triangle is right angled triangle.
Since, none of its sides are equal this is also a scalene triangle.

So,

$A$ and

$B$ both are correct.

Find the value of

$AC.$

0%
$40 \ m$
0%
$42 \ m$
0%
$41 \ m$
0%
$46 \ m$

Explanation

$\Delta ABC$ using Pythagoras theorem,

$AC^2 = AB^2 + BC^2$

$AC =\sqrt{9^2+40^2}=\sqrt{81+1600m}=\sqrt{1681}=41 \ m$

Which of the following sets of side lengths form a triangle?

0%
4 m, 3 m, 11 m
0%
7 mm, 4 mm, 4 mm
0%
3 cm, 1.1 cm, 5 cm
0%
3 m, 4 m, 8 m

Explanation

A triangle can be formed only if sum of any two sides is greater than the third side.

In option

$B$ sum of any two sides taken at a time is greater than the third side.

$7+4 = 11>4$

$4+4= 8>7$

$4+7= 11>4$
In all other options, sum of first two sides is less than the third side:
A)

$4+3<11$
C)

$3+1.1<5$
D)

$3+4<8$

So option

$B$ is correct.

Measure of

$\displaystyle x^{0}$ in the figure is

0%
$\displaystyle 60^{0}$
0%
$\displaystyle 70^{0}$
0%
$\displaystyle 80^{0}$
0%
$\displaystyle 55^{0}$

Explanation

Using exterior angle property,

$\displaystyle x^{0} + 50^{0} =120^{0}$

$\displaystyle x^{0} =120 ^{0} - 50^{0}= 70^{0}$

Which one of the following is not an acute triangle?

Explanation

Acute triangle has all angles less than

$90^o$ .
So, the second figure is not an acute triangle.

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

0%
10 cm
0%
12 cm
0%
7 cm
0%
15 cm

Explanation

$\displaystyle { AB }^{ 2 }+{ BC }^{ 2 }={ AC }^{ 2 }$

$\displaystyle { 9 }^{ 2 }+{ 12 }^{ 2 }={ AC }^{ 2 }$

$\displaystyle 81+144={ AC }^{ 2 }$

$\displaystyle 225={ AC }^{ 2 }$

$\displaystyle { 15 }^{ 2 }={ AC }^{ 2 }$

$\displaystyle AC=15$

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?

0%
$10^{o}$
0%
$20^{o}$
0%
$30^{o}$
0%
$40^{o}$

Explanation

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

$= 2 \times 20^{o} = 40^{o}.$

$\Delta PQR, \angle Q = 23^{o}$ and

$\angle R = 20^{o}.$ What is the measure of

$\angle P?$

0%
$134^{o}$
0%
$135^{o}$
0%
$136^{o}$
0%
$137^{o}$

Explanation

By angle sum property, the sum of angles is

$180^o$ .

$\angle P+ \angle Q + \angle R = 180$

$\angle P + 23^{o} + 20^{o} = 180^{o}$

$\angle P = 180^{o} - 43^{o}$

$\angle P = 137^{o}.$

CBSE Questions for Class 7 Maths The Triangle And Its Properties Quiz 6 - MCQExams.com

0:0:4

Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths The Triangle And Its Properties Quiz 6 - MCQExams.com

0:0:4

Practice Class 7 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.