3 Marks Question

Question 13 Marks

Find the cost of laying grass in a triangular field of sides 50m, 65m and 65m at the rate of Rs. 7 per m².

Answer

We have, $2\text{s}=50\text{m}+65\text{m}+65\text{m}=180\text{m}$
$\text{s}=180\div2=90\text{m}$
$\text{Area of }\triangle=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{90(90-50)(90-65)(90-65)}$
$=\sqrt{90\times40\times25\times25}=60\times25$
$=1500\text{m}^2$
Cost of laying grass at the rate of Rs. 7 per m² = Rs. 7
Cost of laying grass 1500m² = Rs. (1500 × 7) = Rs. 10,500

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Question 23 Marks

The area of a trapezium is 475cm² and the height is 19cm. Find the lengths of its two parallel sides if one side is 4cm greater than the other.

Answer

Let one of the parallel sides be x cm, then orter parallel side be = (x + 4)cm
Area of trapezium $=\frac{1}{2}\times$ (Sum of the parallel side) × height
$\Rightarrow475=\frac{1}{2}\times(\text{x}+\text{x}+4)\times19\text{cm}$
$\Rightarrow2\text{x}+4=\frac{950}{19}=50$
$\Rightarrow2\text{x}=50-4=46$
$\Rightarrow\text{x}=46\div2=23$
Hence, the length of two parallel sides are 23cm and (23 + 4)cm i.e., 23cm and 27cm.

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Question 33 Marks

A rectangular plot is given for constructing a house, having a measurement of 40m long and 15m in the front. According to the laws, a minimum of 3m, wide space should be left in the front and back each and 2m wide space on each of other sides. Find the largest area where house can be constructed.

Answer

The length of the rectangular plot = 40m
And the breath of the plot = 15m
As a minimum of 3m wide space should be left in the front and back 2m wide space each of other side, so the largest area where the house can be constructed.
Length x Breadth = [40 - 2(3)][15 - 2(2)] = 34 × 11 = 374m²

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Question 43 Marks

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 13m, 14m and 15m. The advertisements yield an earning of Rs. 2000 per m² a year. A company hired one of its walls for 6 months. How much rent did it pay?

Answer

Since, the sides of a triangular walls are a = 13m, b = 14m and c = 15m
$\therefore$ Semi-perimeter of triangular side wall, $\text{s}=\frac{\text{a}+\text{b}+\text{c}}{2}=\frac{13+14+15}{2}=\frac{42}{2}=21\text{m}$
$\therefore$ Area of triangular side wall,
$=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$ [by Heron's formula]
$=\sqrt{21(21-13)(21-14)(21-15)}$
$=\sqrt{21\times8\times7\times6}$
$=\sqrt{21\times4\times2\times7\times3\times2}$
$=\sqrt{(21)^2\times(4)^2}$
$=21\times4=84\text{m}^2$
Since, the advertisement yield eaming per year for 1m² = Rs. 2000
$\therefore$ Advertisement yield earning per year on 84m² = 2000 × 84 = Rs. 168000
As the company hired one of its walls for 6 moths, therefore company pay the rent
$=\frac{1}{2}(168000)=\text{Rs. }84000$
Hence, the company6 paid tent Rs. 84000

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