MCQ
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 $24\ cm^2 = 0.09 × 24 = Rs. 2.16$
Hence, the cost of a triangular board is $Rs. 2.16$
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 $24\ cm^2 = 0.09 × 24 = Rs. 2.16$
Hence, the cost of a triangular board is $Rs. 2.16$
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