MCQ Questions for Class 7 Maths The Triangle And Its Properties Quiz 9

How many altitudes does a triangle have?

0%
$1$
0%
$3$
0%
$6$
0%
$9$

Explanation

$\textbf{Step-1: Represent the altitudes of a triangle using a diagram. }$

$\text{A triangle has three altitudes. From each vertex perpendicular to the opposite side can be drawn.}$

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

2165280_1791703_ans_65886f4209e04d658530c81e42b00c4e.png

Which of the following can be the length of the third side of triangle whose two sides measure

$18 cm$ and

$14 cm$ ?

0%
$4 \,cm$
0%
$3 \,cm$
0%
$5 \,cm$
0%
$32 \,cm$

Explanation

Let

$y \,cm$ be the third side of the triangle.

Hence,

$\Rightarrow 18+14>y$

$\Rightarrow 32>y$

$18-14 <y$

$y>4 \ and\ 32>y$

$4<y<32$

Therefore, the third side of a triangle can be

$5 cm$ .

From Fig., the value of

$x$ is

0%
$75^o$
0%
$90^o$
0%
$120^o$
0%
$60^o$

Explanation

In triangle,

$ABC$ ,

$\angle ABC + \angle CAB = \angle ACD$ ...Exterior angle property

$\Longrightarrow ACD = 25^o + 35^o$

$\Longrightarrow \angle ACD = 60^o$ ...(1)
Now,

$\Longrightarrow x = \angle D + \angle ACD$

$\Longrightarrow x = 60^o + 60^o$ using ...(1)

$\therefore x = 120^o$

In a right-angled triangle

$ABC$ , if angle

$B = 90^o$ , then which of the following is true?

0%
$AB^2 = BC^2 + AC^2$
0%
$AC^2 = AB^2 + BC^2$
0%
$AB = BC + AC$
0%
$AC = AB + BC$

Explanation

Hypotenuse is the longest side of triangle which is opposite to right angle.
By pythagoras theorm,

$\text{(hypotenuse)}^2=\text{(side 1)}^2+\text{(side 2)}^2$

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

So, option

$B$ is correct.

1668116_1791850_ans_5d2af03034cf4eaa9b57fb55580556c4.png

Write the correct answer from the given four options.
The hypotenuse of a right triangle with its legs of lengths

$3x \times 4x$ is

0%
$5x$
0%
$7x$
0%
$16x$
0%
$25x$

Explanation

$(Hypotenuse)^2 = (3x)^2 + (4x)^2 = 9x^2 + 16x^2 = 25x^2$

$Hypotenuse = \sqrt{25x^2} = 5x$

Which of the following figures will have its altitude outside the triangle?

Explanation

In obtuse angled triangle, two of its altitudes lie outside the triangle.

Also, from figure it is clear that

$D$ has its altitude outside triangle, which is an obtuse angled triangle.

Hence, option D is correct.

1668138_1791872_ans_a03da46c8c644ae78439c8a59f87ffd7.png

State whether the following statements are true (T) or false (F):
An altitude of a triangle always lies outside the triangle.

0%
True
0%
False

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

0%
Isosceles triangle
0%
Obtuse angled triangle
0%
Equilateral triangle
0%
Right angled triangle

Altitudes of

$\triangle ABC$ are

0%
AB
0%
AC
0%
BC
0%
None of these.

Does altitude of a triangle always lie inside the triangle?

0%
Yes
0%
No
0%
It may lie inside or outside the triangle.
0%
None of these.

In a

$\Delta \mathrm{PQR}$ , if

$\angle \mathrm{P}+\angle \mathrm{Q}=\angle \mathrm{R}$ then type of triangle is:

0%
scalene triangle
0%
equilateral triangle
0%
isosceles triangle
0%
right-angled triangle

Explanation

Given,

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

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

$\Rightarrow 180^\circ=2\angle R$

$\therefore \angle R=90^\circ$

$\triangle PQR$ is right-angled triangle

$\triangle PQR$ ,

$PS$ is an altitude and

$PT$ is median.

Which of the following statement is correct?

0%
$PT = PS$
0%
$RT = QT$
0%
$RS = TQ$
0%
None of these

$\triangle ABC$ ,

$D$ and

$E$ are points on

$BC$ such that

$BD = DC$ and

$AD$ is perpendicular to

$BC$ .

Which of the following is/are correct statement/s?

0%
$AD$ is an altitude of $\triangle ABC$
0%
$AE$ is an altitude of $\triangle ABC$
0%
$AD$ is an median of $\triangle ABC$
0%
$AE$ is an median of $\triangle ABC$

Let ABC is a triangle. Then, which of the following is possible?

0%
AC + BC > AB
0%
AB + BC > AC
0%
AB + AC > BC
0%
None of these.

In given

$\triangle ABC$ , AE, BF and CD are

0%
Bisectors
0%
Altitudes
0%
Sides
0%
Medians

Let AM is the median of

$\triangle ABC$ . Then,

0%
BM = MC
0%
AB = AC
0%
AB = AM
0%
AC = AM

Is it possible to have a triangle with the following sides?
3 cm, 6 cm, 8 cm

0%
Yes
0%
No
0%
May be
0%
Information insufficient.

The lengths of two sides of a triangle are 6 cm and 8 cm. Which of the following can be length the third side of the triangle?

0%
1 cm
0%
5 cm
0%
15 cm
0%
None of the above

Let two sides of a triangle are 2 cm and 5 cm. Which of the following can be the length of its third side?

0%
6 cm
0%
7 cm
0%
2 cm
0%
8 cm

Given

$\triangle ABC$ in the figure. Then which is NOT true for a, b, c?
(Assume a > c > b)

0%
a + b > c
0%
c + b > a
0%
a + c > b
0%
None of these

Given

$\triangle ABC$ in the figure. Then which is true for a, b, c?

0%
a + b > c
0%
b + c > a
0%
a + c > b
0%
None of these

Choose all correct options to fill in the blank.
In the given

$\triangle ABC$ ,

$\angle APC$ is a right angle. Then, AP is a/an _________ of

$\triangle ABC$ .

0%
Median
0%
Altitude
0%
Side
0%
None of these.

An isosceles triangle has a

$10$ inch base and two

$13$ inch sides. What other value can the base have and still yield a triangle with the same area?

0%
$18"$
0%
$19"$
0%
$24"$
0%
$27"$

Explanation

Let

$\Delta ABC$ with

$AB = AC = 13$ inch and base

$BC = 10$ inch.

Therefore,

$\cfrac { 1 }{ 2 } BC = 5$ inch

Let AD be its height.

Applying Pythagoras theorem, its height is

$\sqrt { { 13 }^{ 2 }-{ 5 }^{ 2 } } = 12$ inch

Therefore, area of the triangle

$= \cfrac { 1 }{ 2 } \times BC \times AD = \cfrac { 1 }{ 2 } \times 10\times 12$

$= 60$ sq.inch.

Now, we can divide the triangle along AD and rearrange the parts so that

$AD + AD$ becomes base,

$\cfrac { 1 }{ 2 }$ BC becomes height and the equal sides are

$AB =13$ inch each.

The base of the new

$\Delta$ is

$AD + AD = 12 + 12 = 24$ inch.

Height

$= \cfrac { 1 }{ 2 } BC = \cfrac { 1 }{ 2 } \times 10 inch = 5$ inch.

And its two equal sides are

$13$ inch, each.

Its area

$=\cfrac { 1 }{ 2 } \times 24 \times 5 = 60$ sq.inch, which is the area of the previous

$\triangle ABC.$

In the given figure, if AD is a median of

$\triangle ABC$ , the

0%
AB + AC < 2 AD
0%
AB + AC > 2 AD
0%
AB + AC < AE
0%
AC + CE < AE

In a triangle

$ABC$ ,

$\angle ABC={ 90 }^{ o },\angle ACB={ 30 }^{ o }$ ,

$AB=5cm$ . What is the length of

$AC$ ?

0%
$10cm$
0%
$5cm$
0%
$5\sqrt { 2 } cm$
0%
$5\sqrt { 3 } cm$

The hypotenuse of a right triangle is

$3\sqrt{10}$ units. If the smaller side is tripled and the lower side is doubled, new hypotenuse becomes

$9\sqrt{5}$ units. What are the lengths of the smaller and longer sides of the right triangle respectively?

0%
$5$ units, $9$ units
0%
$5$ units, $6$ units
0%
$3$ units, $9$ units
0%
$3$ units, $6$ units

$ABC$ is a right-angled triangle, right angled at

$C$ and

$p$ is the length of perpendicular from

$C$ on

$AB$ . If

$a,b,c$ are the sides of the triangle, then which one of the following is correct?

0%
$({ a }^{ 2 }+{ b }^{ 2 }){ p }^{ 2 }={ a }^{ 2 }{ b }^{ 2 }$
0%
${ a }^{ 2 }+{ b }^{ 2 }={ a }^{ 2 }{ b }^{ 2 }{ p }^{ 2 }$
0%
${ p }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }$
0%
${ p }^{ 2 }={ a }^{ 2 }-{ b }^{ 2 }$

Which of the following is a right angled triangle?

0%
$\triangle APD$
0%
$\triangle DOC$
0%
$\triangle BDC$
0%
$\triangle AOD$

State and prove isosceles angle properly
If

$AB = AC$ , then what is value of

$\angle a$ ?

0%
$40^o$
0%
$20^o$
0%
$120^o$
0%
$100^o$

Which of the following describes the given triangle?

0%
Isosceles, obtuse
0%
Equilateral, acute
0%
Isosceles, right
0%
Equilateral, obtuse

CBSE Questions for Class 7 Maths The Triangle And Its Properties Quiz 9 - 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 9 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

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