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  • 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 (B) Rs 2.16 (C) Rs 2.48 (D) Rs 3.00

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

(B) Rs 2.16

(C) Rs 2.48

(D) Rs 3.00

Answers (1)

[B]

Solution.        

To find the cost of painting, we have to find the area of the triangular board

Let the sides be denoted as, a = 6 cm, b = 8 cm, c = 10 cm

semi-perimeter,S= \frac{a+b+c}{2}

S= \frac{6+8+10}{2}= \frac{24}{2}= 12cm

We know that,

Using Heron’s formula, area of triangle = \sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}

= \sqrt{12\left ( 12-6 \right )\left ( 12-8 \right )\left ( 12-10\right )}

= \sqrt{12\times 6\times 4\times 2}

= \sqrt{6\times 2\times 6\times 2\times 2\times 2}

6\times 2\times 2= 24cm^{2}

Now\; cost\; of\; painting\; 1cm^{2}= 9\; paise= \frac{9}{100}Rs= 0.09Rs.

\left [ 1Rs= 100paise,1\; paise= \frac{1}{100}Rs \right ]

\therefore\; cost\; of\; painting\; 24cm^{2}= 24\times 0.09= 24\times \frac{9}{100}= Rs2.16

Hence,cost of painting Rs. 2.16.

Hence option (B) is correct.

Posted by

infoexpert21

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