MCQ Questions for Class 7 Maths The Triangle And Its Properties Quiz 5

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

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Obtuse angled triangle
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Acute angled triangle
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Right angled triangle
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Equilateral triangle

In the figure, if

$\angle A + \angle B + \angle C + \angle D + \angle E + \angle F = k\times$ right angle, then

$k$ is :

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$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

$\Rightarrow$ 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 )]

$\Rightarrow$

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

$\Rightarrow$

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

$\therefore$

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

$\therefore$

$k=4$

An exterior angle of a triangle is

$105^0$ and its two interior opposite angles are equal. Each of these equal angles is

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$37\dfrac{1^{0}}{2}$
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$52\dfrac{1^{0}}{2}$
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$72\dfrac{1^{0}}{2}$
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$75^0$

Explanation

Given: Exterior angle =

$105^{\circ}$
Let the two Interior angles be

$x$ .
Sum of Interior opposite angles = Exterior angle

$x + x = 105$

$2x = 105$

$x = 52 \dfrac{1}{2} ^{\circ}$

If one of the angles of a triangle is

$130^0$ , then the angle between the bisectors of the other two angles can be

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$50^0$
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$65^0$
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$145^0$
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$155^0$

Explanation

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

$=155^o$ .

$\Delta ABC$ , if

$AB=AC$ and

$\angle B=50^{0}$ , then

$\angle A$ is equal to:

0%
$40^{0}$
0%
$50^{0}$
0%
$80^{0}$
0%
$130^{0}$

Explanation

Given,

$AB=AC$ and

$\angle B=50^o$

We have to find

$\angle A$

$\angle B=50^o$ , we have

$\angle C=50^o$

Therefore,

$\angle A+\angle B+\angle C=180^o$

$\Rightarrow 50+50+\angle A=180^o$

$\Rightarrow 100+ \angle A=180^o$

$\Rightarrow \angle A=80^o$

$\Delta ABC, BC=AB$ and

$\angle B=80^{o}.$ Then

$\angle A$ is equal to

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$80^{o}$
0%
$40^{o}$
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$50^{o}$
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$100^{o}$

Explanation

Given,

$BC=AB$ and

$\angle B=80^o$

$\triangle ABC$ , we have

$\angle A+ \angle B+ \angle C=180$ .....(1) (Sum of all the angles in triangle is

$180^o$ )

$BC=AB$

$\Rightarrow \angle A= \angle C$ .......... [Angles opposite to equal sides are equal]

and

$\angle B= 80^o$ .....(2)
From (1) and (2), we get

$2\angle A= 100$

$\therefore \angle A= 50^{o}$

$\Delta PQR, \angle R=\angle P$ and

$QR$

$=4$ cm and

$PR$

$= 5$ cm. then what is the length of

$PQ$ ?

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$4$ cm
0%
$5$ cm
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$2$ cm
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$2.5$ cm

Explanation

Given,

$\angle R= \angle P$ and

$QR=4$ cm,

$PR=5$ cm

In an isosceles triangle two sides are congruent and their base angles are also congruent.

Here,

$\angle R=\angle P$

$\therefore QR=PQ$ .

It is given that

$QR=4$ cm, therefore

$PQ=4$ cm.

Two sides of a triangle are of lengths

$4$ cm and

$1.5$ cm. The length of the third side of the triangle cannot be

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$3.6$ cm
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$4.1$ cm
0%
$3.8$ cm
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$5.8$ cm

Explanation

Sum of given sides of the triangle is

$5.5$ cm.

The length of the third side cannot be greater than the sum of the two given sides.
Here, one option gives length as

$5.8$ cm is not possible.

The construction of a triangle

$ABC$ , given that

$BC =$

$6$ cm,

$B =$

$45 ^{\circ}$ is not possible when difference of

$AB$ and

$AC$ is equal to:

0%
$6.9$ cm
0%
$5.2$ cm
0%
$5.0$ cm
0%
$4.0$ cm

Explanation

According to the theorem of inequalities, the sum of any two sides of the triangle is greater than the third side.

Therefore,

$AC+BC>AB$

$\Rightarrow BC>AB-AC$

Therefore, only the first option that is

$6.9$ cm does not satisfy the above equation. Rest all the options satisfy the equation.

If lengths of sides of a triangle are 6, 10, 8, then ABC is .........triangle.

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isosceles
0%
equilateral
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right angled
0%
none of these

Which of the following is NOT the length of a

median in triangle ABC with vertices A(1,3), B(1,1) and C(5,1)?

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$2\sqrt{6}$
0%
$\sqrt{26}$
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$2\sqrt{5}$
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$\sqrt{2}$

In given figure, value of

$(\angle x+ \angle y)$ is

0%
$80^{\circ}$
0%
$30^{\circ}$
0%
$100^{\circ}$
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$60^{\circ}$

Explanation

Exterior angle of triangle is equal to the sum of two interior opposite angles

$\therefore x+y={ 100 }^{ \circ }$

Option

$C$ is correct.

The sides of a triangle are in the ratio 4 : 6 :Then

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The triangle is obtuse -angled
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The triangle is acute-angled
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The triangle is right-angled
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The triangle `Is impossible

Explanation

In the given ratio of sides,

$7^2 < 4^2 + 6^2$ , thus the triangle is an acute triangle.

$\Delta$ ABC,

$\angle B = 90 ^\circ$ , AB = 6 cm and BC = 8 cm. The length of AC is

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3 cm
0%
10 cm
0%
4 cm
0%
7 cm

Explanation

Given that in the

$\triangle ABC, \angle B = 90 ^\circ$

$\therefore \triangle ABC$ is a right angle triangle.

So we can apply pythagoras theorem here.

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

$\therefore$ Side, AC =

$\sqrt{(6^2 + 8^2)}$ = 10 cm

If length of the largest side of a triangle is 12 cm then other two sides of triangle can be

0%
4.8 cm, 8.2 cm
0%
3.2 cm, 7.8 cm
0%
6.4 cm, 2.8 cm
0%
7.6 cm, 3.4 cm

Explanation

Sum of any two sides of a triangle is greater than the third side.

Here the sum must be greater than

$12\ \ cm$

In option

$A$

$4.8\ \ cm+8.2\ \ cm=13\ \ cm$

$\Rightarrow 13\ \ cm>12 \ \ cm$

In rest of the options sum is less than

$12\ \ cm$

So option

$A$ is correct.

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If both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
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If both Assertion and Reason are correct, but Reason is not the correct explanation for Assertion.
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If Assertion is correct, but Reason is incorrect.
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If Assertion is incorrect, but Reason is correct.

Explanation

In right angled

$\triangle ABC$ ,

$AB^2 = AC^2 + BC^2$ ...(By Pythagorus Theorem)

$= AC^2+AC^2 [\because BC =AC]$

$= 2AC^2$

$\therefore AB^2 = 2AC^2$

$\therefore$ Assertion (A) is false.
Again, since

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

$=AC^2+BC^2$ ....

$(\because AC=BC\:given)$

$\therefore \angle C=90^o$ ....(By converse of Pythagoras Theorem )

$\therefore$ Reason is true.

If m

$\angle$

$XYZ = 86^o$ and m

$\angle$

$XZY = 23^o$ . What is m

$\angle$

$YXZ$ in the triangle?

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$70^o$
0%
$71^o$
0%
$72^o$
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$73^o$

Explanation

By angle sum property, the sum of angles is

$180^o$ .

$86^o + 23^o +$ m

$\angle$

$YXZ = 180^o$

$\angle$

$YXZ = 180^o - 109^o$
m

$\angle$

$YXZ = 71 ^o$ .

The angles of a triangle are in the ratio 2: 1:Is the triangle right-angled triangle?

0%
True
0%
False

Explanation

The angles of the triangle are in the ratio,

$2:1: 3$ .
Let the angles be

$2x, x$ and

$3x$ .

By angle sum property, sum of the angles

$= 180^o$

$2x + x+ 3x = 180^o$

$\implies$

$6x = 180^o$

$\implies$

$x = 30^o$ .

Hence, the angles will be

$x=30^o, \ 2x=60^o$ and

$3x=90^o$ .

Since, one of the angles is

$90^o$ , the triangle is a right angled triangle.

Hence, the statement is true and option

$A$ is correct.

In the given figure, the value of

$(\angle x+\angle y)$ is :

0%
$110^0$
0%
$100^0$
0%
$120^0$
0%
$60^0$

Explanation

$\textbf{Step -1: Stating exterior angle theorem.}$

$\text{The sum of two interior adjacent angles is equal to its opposite exterior angle.}$

$\therefore \angle x + \angle y = 100^\circ$

$\textbf{Hence, option B is correct.}$

In the given figure,

$AB \;|\;|\; DC$ ,

$BO=6\; cm$ and

$DQ= 8\; cm$ then find the value of

$\displaystyle BP \times DO.$

0%
$\displaystyle 48 cm^{2}$
0%
$\displaystyle 24 cm^{2}$
0%
$\displaystyle 56 cm^{2}$
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None of the above

From the adjoining figure, calculate the values of

$a$ .

0%
$46^o$
0%
$38^o$
0%
$56^o$
0%
none of the above

Explanation

We know, by angle sum property, the sum of angles of triangle

$= 180^o$
Then,

$a + 110^o + 32^o = 180^o$

$\implies$

$a + 142^o = 180^o$

$a =180^o-142^o=38^{\circ}$ .

Thus measure of angle

$a$ is

$38^o$ .

Hence, option

$B$ is correct.

State true or false:

$ED= EC$ , then

$AB\, +\, AD> BC$ .(Refer given fig.)

0%
True
0%
False

Explanation

In $\triangle ABD$ , we have sum of two sides $>$ third side
$=> AB + AD > BD$ OR

$=> AB + AD > BE + ED -- (1)$

Also, in $\triangle BCE$ , we have sum of two sides $>$ third side

$=> EB + EC > BC$ ---- (2)

Adding equations $1, 2$ we get

$=> AB + AD + EB +EC > BE + ED +BC$

$=> AB + AD + EC > ED + BC$

But $EC = ED$

Hence, $AB + AD > BC$

In the above figure, point

$D$ divides

$AB$ in the ratio

$3: 5$ . If

$BC= 4.8 \;cm$ , find the length of

$DE$ .

0%
$1.8 cm$
0%
$2.6 cm$
0%
$1.2 cm$
0%
$4 cm$

In the following figure, point

$D$ divides

$AB$ in the ratio

$3: 5$ . Find

$\displaystyle \dfrac{AE}{EC}$ .

0%
$\dfrac{3}{5}$
0%
$\dfrac{3}{7}$
0%
$\dfrac{2}{5}$
0%
$\dfrac{7}{5}$

In the given figure,

$ABC$ is a triangle, side

$CB$ is produced to

$E$ and

$\angle A$ :

$\angle B$ :

$\angle C$

$= 2: 1: 3$ . Find

$\angle DBE$ , if

$DB$ is perpendicular to

$AB$ .

0%
$66^o$
0%
$46^o$
0%
$60^o$
0%
none of the above

Explanation

In triangle

$ABC$ ,

$\angle A : \angle B : \angle C = 2: 1: 3$
Let the angles be

$2x, x, 3x$
By angle sum property, the sum of the angles

$2x + x+ 3x = 180^o$

$\implies$

$6x = 180^o$

$\implies$

$x = 30^o$ .

Hence,

$\angle B = 30^o$ .

Now,

$\angle DBE + \angle DBA + \angle ABC = 180^o$ ...[Straight angle property]

$\angle DBE + 90^o + 30^o = 180^o$

$\angle DBE = 180^o - 120^o$

$\angle DBE = 60^{\circ}$ .

Therefore, option

$C$ is correct.

In the given figure,

$DE\;|\;|\;BC$ ,

$AE= 15\;cm$ ,

$EC= 9\;cm$ ,

$NC= 6\; cm$ and

$BN=24\; cm$ . Find the length of

$ME$ is

0%
$3.75\,cm$
0%
$2.5\,cm$
0%
$4\,cm$
0%
$5.5\,cm$

Which of the following best describes the given triangle.

0%
Equilateral obtuse triangle
0%
Isosceles right triangle
0%
Isosceles obtuse triangle
0%
Isosceles acute triangle.

Explanation

Let the third angle be

$x$

It is known that sum of interior angles of a triangle is

$180^o$

Thus,

$x+{ 36 }^{ \circ }+{ 36 }^{ \circ }={ 180 }^{ \circ }$

$\Rightarrow x={ 108 }^{ \circ }$

Now two of angles of the triangle are equal and one angle is greater than

${ 90 }^{ \circ }$

So the given triangle is an isosceles obtuse triangle.

An exterior angle of a triangle is less than either of its interior opposite angles.

0%
True
0%
False
0%
Ambiguous
0%
Data Insufficient

$\Delta ABC$ , if

$\angle A = 50^0$ and

$\angle B = 60^0$ , then the shortest and largest sides of the triangle, respectively are:

0%
$AC$ and $AB$
0%
$BC$ and $AB$
0%
$AC$ and $BC$
0%
$BC$ and $AC$

Explanation

We have,

$\angle A = 50$ and

$\angle B = 60$

$\therefore \angle A + \angle B + \angle C = 180$

$[$ Angle sum property of a

$\Delta$

$]$

$\Rightarrow 50 + 60 + \angle C = 180$

$\Rightarrow 110 + \angle C = 180$

$\Rightarrow \angle C = 180 - 110$

$\Rightarrow \angle C = 70$
Since

$\angle A$ and

$\angle C$ are the smallest and largest angles respectively, therefore corresponding sides

$BC$ and

$AB$ are the smallest and largest sides of the triangle respectively.

Hence, option

$B$ is the correct answer.

In a right angled triangle

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

$AC=13\;cm,\;BC=5\;cm$ . Then, find

$AB$ .

0%
$12 cm$
0%
$17 cm$
0%
$15 cm$
0%
$14 cm$

Explanation

In a right angled triangle

$ABC$ , if

$\angle B={ 90 }^{ 0 }$

using Pythagoras theorem, we have

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

$AC=13cm,BC=5cm$

$\Rightarrow {13}^{2} = {AB}^{2} + {5}^{2}$

$\Rightarrow 169 = {AB}^{2} + 25$

$\Rightarrow 169 - 25 = {AB}^{2}$

$\Rightarrow AB = \sqrt {144} = 12 cm$

CBSE Questions for Class 7 Maths The Triangle And Its Properties Quiz 5 - 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 5 - MCQExams.com

0:0:2

Practice Class 7 Maths Quiz Questions and Answers

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