MCQ Questions for Class 7 Maths The Triangle And Its Properties Quiz 8

Which of the following set of measurements will form a triangle

0%
$11cm,4cm,6cm$
0%
$13cm,14cm,25cm$
0%
$8cm,4cm,3cm$
0%
$5cm.16cm.5cm$

Explanation

Triangle Inequality Theorem states that the sum of two side lengths of a triangle is always greater than the third side.

If this is true for all three combinations of added side lengths.

i.e.

$a+b> c,$

$b+c> a$ and

$c+a> b$ then the lengths form a triangle

(A)

$11+4=15> 6$

And

$4+6 =10\ngtr 11$

So, it does not form a triangle

(B)

$13+14=27> 25$

And

$14+25=39> 13$

And

$25+13=38> 14$

So, it forms a triangle

(C)

$8+4=12> 3$

And

$8+3=11> 4$

And

$4+3=7\ngtr 8$

So, it does not form a triangle

(D)

$5+16=21> 5$

And

$5+5=10\ngtr 16$

So, it does not form a triangle.

Hence option B is the correct answer

Triangles with sides

$3$ cm,

$4$ cm and

$5$ cm is possible.

0%
True
0%
False

Explanation

Yes, as

$3^2+4^2=5^2$
1309449_703062_ans_488cf80961b744e9b14afe2e019e3a52.jpg

Find the values of x and y in the following figures

0%
$x=80^0 ; y=120^0$
0%
$x=50^0 ; y=30^0$
0%
$x=50^0 ; y=130^0$
0%
$x=20^0 ; y=10^0$

Explanation

$\angle CAB = 80^o$ [apposite angle]

$AB = AC$

$\therefore x = \angle ACB$

Also, by angle sum property, we have

$x + x+80 = 180$

$2x = 100$

$x = 50$

$y = 80+x = 130$ exterior angle property

$x = 50$

$y = 130$

1456060_704675_ans_3d8fb8e5753b4e7285047640234de9fa.png

Find the measure of the angle

$x$ in the given figure.

0%
${ 50 }^\circ$
0%
${ 70 }^\circ$
0%
${ 60 }^\circ$
0%
${ 30 }^\circ$

Explanation

$\textbf{Step 1: Find the relation between interior and exterior angles of triangle.}$

$x=\angle{EFD}+\angle{FED}$

$\Rightarrow \angle{x}={28^\circ}+{42^\circ}$

$\quad \textbf{[Exterior angle is equal to sum of interior angles of a triangle.]}$

$\Rightarrow \angle{x}={70^\circ}$

$\textbf{Hence, the measure of the angle x is}$

$\bf{70^\circ.}$

Four pair of showing measurements of sides

$\overline{AB}, \overline{BC}$ and

$\overline{CA}$ of

$\Delta ABC$ are given below.
Show which of the following pair/s is/are shows right angle triangle.
Pair P: AB = 25 BC = 7 AC = 24
Pair Q: AB = 8 BC = 6 AC = 10
Pair R: AB = 3 BC = 4 AC = 6
Pair S: AB = 8 BC = 6 AC = 5

0%
Pairs Q and R show right angle triangle
0%
Pairs P and Q show right angle triangle
0%
Pairs P and S show right angle triangle
0%
Pairs P, Q and S show right angle triangle

Explanation

For pair P:

$AC^2 + BC^2 = 24^2 + 7^2 = 576 + 49 = 625=25^2$
and

$AB^2 = 25^2$

$\therefore AB^2 = AC^2 + BC^2$
For pair Q:

$AB^2 + BC^2 = 8^2 + 6^2 = 64 + 36 = 100$

$AC^2 = 10^2 = 100$

$\therefore AB^2 + BC^2 = AC^2$
For pair R:

$AB^2 + BC^2 = 3^2 + 4^2 = 9 + 16 = 25$

$AC^2 = 6^2 = 36$

$\therefore AB^2 + BC^2 \neq AC^2$
For pair S:

$AC^2 + BC^2 = 5^2 + 6^2 = 25 + 36 = 61$

$AB^2 = 8^2 = 64$

$\therefore AB^2 \neq AC^2 + BC^2$

$\therefore$ Pairs P and Q show right angled triangle.

$34$ m long ladder reached in the window which is

$16$ m above from the ground on placing it against a wall. Find the distance of the foot of the ladder from the wall.

0%
$40$ m
0%
$30$ m
0%
$50$ m
0%
$10$ m

Explanation

Let

$AB=$ length of ladder,

$AC=$ height of window
In right angled

$\triangle ABC$ , we have

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

$\Rightarrow { (34) }^{ 2 }={ (16) }^{ 2 }+{ BC }^{ 2 }\quad$

$\Rightarrow { BC }^{ 2 }={ (34) }^{ 2 }-{ (16) }^{ 2 }=900$

$\Rightarrow BC=\sqrt { 900 } =30 m$

From the given figure, find the values of

$x$ and

$y$ respectively.

0%
${ 47 }^{ \circ },{ 66 }^{ \circ }$
0%
${ 66 }^{ \circ },{ 48 }^{ \circ}$
0%
${ 68 }^{ \circ },{ 47 }^{ \circ }$
0%
${ 47 }^{ \circ },{ 68 }^{ \circ }$

Explanation

$\triangle TCE$ , we have

$x=\angle TCE+\angle TEC$ (Exterior angle property)

$x={ 35 }^{ \circ }+{ 31 }^{ \circ }$

$x={ 66 }^{ \circ }\quad \quad$

$\triangle SBD$ , we have

$\ \angle AST=\angle SBD+\angle SDB\quad$

$\angle AST={ 30 }^{ \circ }+{ 36 }^{ \circ }={ 66 }^{ \circ }$

$\triangle ATS$ , we have

$y+x+\angle AST={ 180 }^{ \circ }\quad$ (Angle sum property)

$y+{ 66 }^{ \circ }+{ 66 }^{ \circ }={ 180 }^{ \circ }\quad$

$y={ 180 }^{ \circ }-\left( { 66 }^{ \circ }+{ 66 }^{ \circ } \right)$

$\Rightarrow y={ 48 }^{ \circ }$

A tree is broken at a height of

$5$ m from the ground and its top touches the ground at a distance of

$12$ m from the base of the tree. Find the original height of the tree.

0%
$10m$
0%
$15m$
0%
$13m$
0%
$18m$

Explanation

Let

$BD$ be the original height of tree.

$\therefore AD = AC$

$\triangle ABC$

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

$=5^2+12^2$

$=169$

$= 13\times 13$

$\Rightarrow AC = 13m$

$\Rightarrow AD = 13m$

$\therefore$ Original height of tree

$=(13+5)m$

$= 18m$

A tree is broken at a height of

$8$ m from the ground and its top touches the ground at a distance of

$15$ m from the base of the tree. Find the original height of the tree.

0%
$25m$
0%
$5m$
0%
$23m$
0%
$16m$

Explanation

Let

$BD$ be the original height of tree.

$\therefore AD = AC$

$\triangle ABC$

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

$=15^2+8^2$

$=225+64$

$= 289$

$\Rightarrow AC = 17m$

$\Rightarrow AD = 17m$

$\therefore$ Original height of tree

$=(17+8)m = 25m$

1111819_1059327_ans_80113fa420c248a68022b6c5ca01c7cd.jpg

Find the value of x, y and z in the adjoining figure.

0%
$x=160$ , $y=60$ , $z=80$
0%
$x=60$ , $y=80$ , $z=40$
0%
$x=80$ , $y=60$ , $z=140$
0%
$x=140$ , $y=80$ , $z=60$

Explanation

$In\triangle BCE$

$\implies\quad \angle BEC+\angle BCE+\angle CBE={ 180 }^{ \circ }$

$\implies\quad { 90 }^{ \circ }+{ 30 }^{ \circ }+\angle CBE={ 180 }^{ \circ }$

$\implies\quad \angle CBE={ 60 }^{ \circ }$

$y={ 60 }^{ \circ }$

$In\triangle APC:$

$\implies\quad { 50 }^{ \circ }+{ 30 }^{ \circ }+\angle APC={ 180 }^{ \circ }$

$\implies\quad \angle APC={ 100 }^{ \circ }$

$\therefore x=180-\angle APC$

$(\because sum\quad of\quad angles\quad on\quad a\quad straight\quad line={ 180 }^{ \circ })$

$\implies\quad x={ 180 }^{ \circ }-{ 100 }^{ \circ }={ 80 }^{ \circ }$

$In\triangle BOP:$

$z=x+y\quad (exterior\quad angle\quad of\quad a\quad triangle=sum\quad of\quad two\quad opposite\quad interior\quad angles)$

$\implies\quad z=80+60={ 140 }^{ \circ }$

$\therefore x={ 80 }^{ \circ }\quad ,\quad y={ 60 }^{ \circ }\quad ,z={ 140 }^{ \circ }$

1005693_1068116_ans_2962fefc78cf4968a64e0b6d4848aa0e.png

$\triangle XYZ$ ______________is the base.

0%
$XY$
0%
$YZ$
0%
$XZ$
0%
None of the above

$\triangle ABC,\angle A={ x }^{ \circ },\angle B={ \left( 2x-15 \right) }^{ \circ}$ and

$\angle C ={ \left( 3x+21 \right) }^{ \circ }$ . Find the value of

$x$ and the measure of each angle of the triangle.

0%
$x={ 27 }^{ \circ },\angle A={ 27 }^{ \circ },\angle B={ 41 }^{ \circ },\angle C={ 101 }^{ \circ }$
0%
$x={ 29 }^{ \circ },\angle A={ 29 }^{ \circ },\angle B={ 43 }^{ \circ },\angle C={ 108 }^{ \circ }$
0%
$x={ 27 }^{ \circ },\angle A={ 27 }^{ \circ },\angle B={ 41 }^{ \circ },\angle C={ 108 }^{ \circ }$
0%
$x={ 30 }^{ \circ },\angle A={ 30 }^{ \circ },\angle B={ 41 }^{ \circ },\angle C={ 109 }^{ \circ }$

Explanation

Given,

$\angle A=x^\circ$ ,

$\angle B=(2x-15)^\circ$ and

$\angle C=(3x+21)^\circ$ .

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

$180^\circ$ .

Then,

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

$\implies$

$x^\circ+(2x-15)^\circ+ (3x+21)^\circ=180^\circ$

$\implies$

$6x^\circ+6^\circ=180^\circ$

$\implies$

$6x^\circ=180^\circ-6^\circ$

$\implies$

$6x^\circ=174^\circ$

$\implies$

$x^\circ=29^\circ$ .

Therefore,

$\angle A=x^\circ=29^\circ$ ,

$\angle B=(2x-15)^\circ=2\times29^\circ-15^o=58^\circ-15^\circ=43^\circ$

and

$\angle C=(3x+21)^\circ=3\times29^\circ+21^\circ=87^\circ+21^\circ=108^\circ$ .

Hence, option

$B$ is correct.

From the figure, find values of

$x$ and

$y$ .

0%
$25^{o},30^{o}$
0%
$35^{o},31^{o}$
0%
$50^{o},28^{o}$
0%
$45^{o},33^{o}$

Explanation

In the given triangle,

by angle sum property,

$40^o+95^o+x^o=180^o.......(i)$

and

$x^o+y^o+102^o=180^o.......(ii)$ .

From

$(i)$ ,

$40^o+95^o+x^o=180^o$

$\implies$

$135^o+x^o=180^o$

$\implies$

$x^o=180^o-135^o$

$\implies$

$x^o=45^o.......(iii)$ .

Substitute

$(iii)$ in

$(ii)$ ,

$45^o+y^o+102^o=180^o$

$\implies$

$y^o+147^o=180^o$

$\implies$

$y^o=180^o-147^o$

$\implies$

$y^o=33^o$ .

Therefore,

$x^o=45^o$ and

$y^o=33^o$ .

Hence, option

$D$ is correct.

Find the length of the hypotenuse in a right angled triangle if the sum of the squares of the sides making right angle is 169.

0%
$15$
0%
$13$
0%
$5$
0%
$12$

Explanation

According to the Pythagoras theorem, the sum of the squares of the sides making the right angle is equal to the square of the third side (hypotenuse).

$\therefore\$ Square of the hypotenuse

$=169$

$\Rightarrow$ Hypotenuse

$=\sqrt{169}$

$=13\space\mathrm{units}$

Hence, option B is correct.

In a right-angled triangle the lengths of base and perpendicular are 6 cm and 8 cm.What is the length of the hypotenuse?

0%
9 cm
0%
10 cm
0%
11 cm
0%
12 cm

Explanation

The sides of right angled triangle are

$6,8 cm$

The hypotenuse is given as

$a^2+b^2=c^2\\6^2+8^2=c^2\\36+64=c^2\\c^2= 100\\c=\sqrt {100}\\c=10cm$

If the angles of triangle are in the ratio

$1:4:7$ , then the value of the largest angle is :

0%
$135^{\circ}$
0%
$84^{\circ}$
0%
$105^{\circ}$
0%
none of these

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

0%
$84$
0%
$74$
0%
$94$
0%
$57$

Explanation

Exterior angle = sum of 2 opposite Interior Angles

$123^0=39^0+x^0$

$x^0=123^0-39^0$

$\boxed{x^0=84^0}$

A triangle having sides of different lengths is called

0%
an isosceles triangles
0%
an equilateral triangle
0%
a scalene triangle
0%
a right triangle

The difference between the length of any two sides of a triangle is smaller than the length of third side.

0%
True
0%
False

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?

0%
(i) only
0%
(ii) only
0%
(iii) only
0%
(i) and (ii)

Explanation

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.
1670077_1789991_ans_02bbe6cbb7fd464e881ca510c3f0b1f8.png

It is possible to have a right-angled equilateral triangle.

0%
True
0%
False

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?

0%
(i) only
0%
(ii) only
0%
(iii) only
0%
(i) and (ii)

$\Delta PQR$ ,

0%
$PQ - QR > PR$
0%
$PQ + QR < PR$
0%
$PQ - QR < PR$
0%
$PQ + PR > QR$

Explanation

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.

If an isosceles triangle, each of the base angles is

$40^o$ , then the triangle is

0%
Right-angled triangle
0%
Acute angled triangle
0%
Obtuse angled triangle
0%
Isosceles right-angled triangle

Explanation

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.
1668596_1791469_ans_613dd22c032f4bf080b6d86aa940e91d.png

Which of Which of the following statements is not correct?

0%
The sum of any two sides of a triangle is greater than the third side
0%
A triangle can have all its angles acute
0%
A right-angled triangle cannot be equilateral
0%
Difference of any to sides of a triangle is greater than the third side

In Fig.

$BC = CA$ and

$\angle A = 40$ . Then,

$\angle ACD$ is equal to

0%
$40^o$
0%
$80^o$
0%
$120^o$
0%
$60^o$

Explanation

$BC = CA$

$\Longrightarrow \angle A = \angle B = 40^o$
Now,

$\angle ACD = \angle A + \angle B$ ...Exterior angle property

$= 40^o + 40^o = 80^o$

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

0%
$55^{o}$
0%
$65^{o}$
0%
$50^{o}$
0%
$60^{o}$

Explanation

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

$65^{o}$ .

$\Delta PQR$ , if

$\angle P=60^{o}$ , and

$\angle Q=40^{o}$ , then the exterior angle formed by producing

$QR$ is equal to

0%
$60^{o}$
0%
$120^{o}$
0%
$100^{o}$
0%
$80^{o}$

Explanation

Let

$QR$ is extended to point

$S$ .
(Refer image)
In triangle

$PQR$

$\angle PRS=\angle P+\angle Q$

$\Rightarrow \angle PRS=60^{o}+40^{o}$

$\Rightarrow \angle PRS=100^{o}$
1668633_1791694_ans_e20651fa4dc444079b202bd2b6f610cf.png

The measures of

$\angle x$ and

$\angle y$ in figure are respectively

0%
$30^{o},60^{o}$
0%
$40^{o}, 40^{o}$
0%
$70^{o}, 70^{o}$
0%
$70^{o}, 60^{o}$

Explanation

$120^0$ is exterior of

$\angle R$

$\therefore \angle P+\angle Q=120^0$

$\Rightarrow 120^{o}=x+50^{o}$

$\Rightarrow x=120^{o}-50^{o}$

$\Rightarrow x=70^{o}$

Now in

$\Delta PQR$

$x+y+50^{o}=180^{o}$

$\Rightarrow 70^{o}+y+50^{o}=180^{o}$

$\Rightarrow y=180^{o}-70^{o}-50^{o}$

$\Rightarrow y=60^{o}$

Therefore,

$x=60^{o}$ and

$y=70^{o}$ .

Hence, option

$D$ is correct.

1668055_1791717_ans_8fe397eddd17473cb7704bd401c421cc.png

If we join a vertex to a point on opposite side, which divides that side in the ratio

$1:1$ , then what is the special name of that line segment?

0%
Median
0%
Angle bisector
0%
Altitude
0%
Hypotenuse

Explanation

Median from an vertex divides the opposite sides into ratio of

$1:1$

So, special name of that segment is

$\text{Median}$

Hence, option

$A$ is correct.

CBSE Questions for Class 7 Maths The Triangle And Its Properties Quiz 8 - 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 8 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

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