M.C.Q

Question 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:

8cm
36cm
4cm
6cm

Answer

Solution:
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:

✓
$5.196\ cm^2$
B
$0.866\ cm^2$
C
$3.496\ cm^2$
D
$1.732\ cm^2$

Answer

Correct option: A.

$5.196\ cm^2$

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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Question 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:

48cm
24cm
12cm
36cm

Answer

24cm

Solution:
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 × Side= 3 × 8 = 24cm
Hence, the perimeter of an equilateral triangle is 24cm.

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Question 41 Mark

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

$10\sqrt{3}\text{m}^2$
$15\sqrt{3}\text{m}^2$
$20\sqrt{3}\text{m}^2$
$100\sqrt{3}\text{m}^2$

Answer

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

Solution:
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 edges of a triangular board are $6\ cm, 8\ cm$ and $10\ cm$. The cost of painting it at the rate of $9$ paise per $\ cm^2$ 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 $1\ cm^2 = Rs. 0.09$
$\therefore$ Cost of paint for area $24\ cm^2 = 0.09 \times 24 = Rs. 2.16$
Hence, the cost of a triangular board is $Rs. 2.16$

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Question 61 Mark

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

$16\sqrt{5}\text{cm}$
$10\sqrt{5}\text{cm}$
$24\sqrt{5}\text{cm}$
$28\text{cm}$

Answer

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

Solution:
Sides of the triangle are 35cm, 54cm and 61cm
$\text{s}=\frac{35+54+61}{2}=75\text{cm}$
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 71 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
$1322\ cm^2$
B
$1311\ cm^2$
✓
$1344\ cm^2$
D
$1392\ cm^2$

Answer

Correct option: C.

$1344\ cm^2$

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 $1344\ cm^2$

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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.

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

$\text{ABC}$ is an isosceles right triangle. We have,

$\text{AB}=\text{BC}=\text{a cm}$
Area of $\triangle=\frac{1}{2}$ base $\times$ 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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Question 91 Mark

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

$\sqrt{15}\text{cm}^2$
$\sqrt{\frac{15}{2}}\text{cm}^2$
$2\sqrt{15}\text{cm}^2$
$4\sqrt{15}\text{cm}^2$

Answer

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

Solution:
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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