3 Marks Question

Question 13 Marks

Find the cost of laying grass in a triangular field of sides $50\ m, 65\ m$ and $65\ m$ at the rate of $Rs. 7$ per $m^2.$

Answer

We have, $2\text{s}=50\text{ m}+65\text{m}+65\text{ m}=180\text{ m}$
$\text{s}=180\div2$
$=90\text{ m}$
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 $m^2 = Rs. 7$
Cost of laying grass $1500\ m^2 $
$= Rs. (1500 \times 7) $
$= Rs. 10,500$

View full question & answer→

Question 23 Marks

The area of a trapezium is $475\ cm^2$ and the height is $19\ cm.$ Find the lengths of its two parallel sides if one side is $4\ cm$ 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$) \times$ 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 $23\ cm$ and $(23 + 4)\ cm\ i.e., 23\ cm$ and $27\ cm.$

View full question & answer→

Question 33 Marks

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

Answer

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

View full question & answer→

Question 43 Marks

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $13\ m, 14\ m$ and $15\ m$. The advertisements yield an earning of $Rs. 2000$ per $m^2$ 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 $a = 13\ m, b = 14\ m$ and $c = 15\ m$
$\therefore$ 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 $1\ m^2 = Rs. 2000$
$\therefore$ Advertisement yield earning per year on $84\ m^2 $
$= 2000 \times 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 company $6$ paid tent $Rs. 84000$

View full question & answer→

MATHS 3 Marks Question

Test yourself on this topic