M.C.Q

MCQ 11 Mark

Write the correct answer in the following: The length of each side of an equilateral triangle having an area of $9\sqrt{3}\text{cm}^2$ is:

A
$8\ cm$
B
$36\ cm$
C
$4\ cm$
✓
$6\ cm$

Answer

Correct option: D.

$6\ cm$

Area of equilateral $\triangle\text{ i.e., } 9\sqrt{3}=\frac{\sqrt{3}}{4}(\text{side})^2$
$\Rightarrow(\text{side})^2=\frac{9\sqrt{3}\times4}{\sqrt{3}}=36$
$\therefore\ \text{side}=\pm\sqrt{36}=6\text{cm}$

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MCQ 21 Mark

Write the correct answer in the following: The area of an equilateral triangle with side $2\sqrt{3}\text{cm}$ is:

✓
$5.196 \mathrm{~cm}^2$
B
$0.866 \mathrm{~cm}^2$
C
$3.496 \mathrm{~cm}^2$
D
$1.732 \mathrm{~cm}^2$

Answer

Correct option: A.

$5.196 \mathrm{~cm}^2$

Area of equilateral $\triangle=\frac{\sqrt{3}}{4}(\text{side})^2$
$=\frac{\sqrt{3}}{4}\big(2\sqrt{3}\big)^2=3\sqrt{3}=3\times1.732$
$=5.196\text{cm}^2$

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MCQ 31 Mark

Write the correct answer in the following: If the area of an equilateral triangle is $16\sqrt{3}\text{cm}^2,$ then the perimeter of the triangle is:

A
$48\ cm$
✓
$24\ cm$
C
$12\ cm$
D
$36\ cm$

Answer

Correct option: B.

$24\ cm$

Given, area of an equilateral triangle $=16\sqrt{3}\text{cm}^2$
Area of an equilateral triangle $=\frac{\sqrt{3}}{4}(\text{side})^2$
$\frac{\sqrt{3}}{4}(\text{side})^2={16}\sqrt{3}$
$\Rightarrow(\text{side})^2=64$
$\Rightarrow\text{sides}=8\text{cm}$
[taking positive square root because side is always positive]
Perimeter of an equilateral triangle $= 3 \times $ Side$= 3 \times 8 = 24\ cm$
Hence, the perimeter of an equilateral triangle is $24\ cm$.

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MCQ 41 Mark

Write the correct answer in the following: The perimeter of an equilateral triangle is $60\ m$. The area is:

A
$10\sqrt{3}\text{m}^2$
B
$15\sqrt{3}\text{m}^2$
C
$20\sqrt{3}\text{m}^2$
✓
$100\sqrt{3}\text{m}^2$

Answer

Correct option: D.

$100\sqrt{3}\text{m}^2$

Perimeter of triangle $= 3a$
Now, $3\text{a}=60$
$\Rightarrow\text{ a}=60\div3=20\text{m}$
Area of equilateral $\triangle=\frac{\sqrt{3}}{4}(\text{side})^2$
$=\frac{\sqrt{3}}{4}\times(20)^2=100\sqrt{3}\text{m}^2$

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MCQ 51 Mark

Write the correct answer in the following: The sides of a triangle are $35\ cm, 54\ cm$ and $61\ cm$, respectively. The length of its longest altitude:

A
$16\sqrt{5}\text{cm}$
B
$10\sqrt{5}\text{cm}$
✓
$24\sqrt{5}\text{cm}$
D
$28\text{cm}$

Answer

Correct option: C.

$24\sqrt{5}\text{cm}$

Sides of the triangle are $35\ cm, 54\ cm$ and $61\ cm$
$\text{s}=\frac{35+54+61}{2}=75\text{cm}$
Area of $\triangle=\sqrt{75(75-35)(75-54)(75-61)}$
$=\sqrt{75\times40\times21\times14}$
$=\sqrt{5\times5\times3\times2\times2\times2\times5\times3\times7\times7\times2}$
$=5\times3\times2\times2\times7\sqrt{5}$
$=420\sqrt{5}\text{cm}^2$
Now, longest altitude will be the perpendicular on the smallest side of the triangle from the opposite vertex.
$\therefore$ Length of longest altitude $=\frac{2(\text{Area of }\triangle)}{35}$
$=\frac{2\times420\sqrt{5}}{35}=24\sqrt{5}\text{cm}$

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MCQ 61 Mark

Write the correct answer in the following:
The sides of a triangle are $56\ cm, 60\ cm$ and $52\ cm$ long. Then the area of the triangle is:

A
$1322 \mathrm{~cm}^2$
B
$1311 \mathrm{~cm}^2$
✓
$1344 \mathrm{~cm}^2$
D
$1392 \mathrm{~cm}^2$

Answer

Correct option: C.

$1344 \mathrm{~cm}^2$

Since, the three sides of a triangle are $a = 56\ cm, b = 60\ cm$ and $c = 52\ cm$
Then, semi-perimeter of a triangle,
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{56+60+52}{2}=\frac{168}{2}=84\text{cm}$
Area of triangle $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$ [by Heron's formula]
$=\sqrt{84(84-56)(84-60)(84-52)}$
$=\sqrt{4\times7\times3\times4\times7\times4\times2\times3\times4\times4\times2}$
$=\sqrt{(4)^6\times(7)^2\times(3)^2}$
$=(4)^3\times7\times3=1344\text{cm}^2$
Hence, the area of triangle is $1344 \mathrm{~cm}^2$

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MCQ 71 Mark

Write the correct answer in the following:
The edges of a triangular board are $6 \mathrm{~cm}, 8 \mathrm{~cm}$ and $10\ cm$. The cost of painting it at the rate of $9$ paise per $\mathrm{cm}^2$ is:

A
$Rs.\ 2.00$
✓
$Rs.\ 2.16$
C
$Rs.\ 2.48$
D
$Rs.\ 3.00$

Answer

Correct option: B.

$Rs.\ 2.16$

Since, the edges of a triangular are $a = 6\ cm, b = 8\ cm$ and $c = 10\ cm$
Now, semi-perimeter of a triangular board.
$\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}$
$=\frac{6+8+10}{2}=\frac{24}{2}=12\text{cm}$
Now, area of a triangular board $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{12(12-6)(12-8)(12-10)}$
$=\sqrt{12\times6\times4\times2}$
$=\sqrt{(12)^2\times(2)^2}$
$=12\times2=24\text{cm}^2$
Since, the cost of painting for area $1 \mathrm{~cm}^2=$ $Rs.\ 0.09$
$\therefore$ Cost of paint for area $24 \mathrm{~cm}^2=0.09 \times 24=$ $Rs.\ 2.16$
Hence, the cost of a triangular board is $Rs.\ 2.16$

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MCQ 81 Mark

Write the correct answer in the following: An isosceles right triangle has area $8\ cm^2$.
The length of its hypotenuse is:

✓
$\sqrt{32}\text{cm}$
B
$\sqrt{16}\text{cm}$
C
$\sqrt{48}\text{cm}$
D
$\sqrt{24}\text{cm}$

Answer

Correct option: A.

$\sqrt{32}\text{cm}$

$ABC$ is an isosceles right triangle. We have,

$\text{AB}=\text{BC}=\text{a cm}$
Area of $\triangle=\frac{1}{2}\text{base}\times\text{Height}$
$\Rightarrow\ 8=\frac{1}{2}\times\text{a}\times\text{a}$ $\big[\because\ \text{AB}=\text{BC}=\text{a cm}\big]$
$\Rightarrow\ \text{a}^2=16$ $\therefore\ \text{a}=+\sqrt{16}=4\text{cm}$
Using Pythagoras theorem,
Hypotenuse $\text{AC}=\sqrt{4^2+4^2}$
$=\sqrt{16+16}=\sqrt{32}\text{cm}$

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MCQ 91 Mark

Write the correct answer in the following: The area of an isosceles triangle having base $2\ cm$ and the length of one of the equal sides $4\ cm$, is:

✓
$\sqrt{15}\text{cm}^2$
B
$\sqrt{\frac{15}{2}}\text{cm}^2$
C
$2\sqrt{15}\text{cm}^2$
D
$4\sqrt{15}\text{cm}^2$

Answer

Correct option: A.

$\sqrt{15}\text{cm}^2$

Here, $\text{s}=\frac{4+4+2}{2}=5\text{cm}$
Area of $\triangle=\sqrt{5(5-2)(5-4)(5-4)}$
$=\sqrt{5\times3\times1\times1}=\sqrt{15}\text{cm}^2$

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