Question

Suman made a picture with some white paper and a black paper as shown in figure. White paper is available at her home and free of cost. The cost of black paper used is at the rate of $10$ paise per $c{m}^2$, then the total cost of the black paper used. (Take $\sqrt{3}=1.732$ and $\sqrt{11}=3.31$) is $Rs.14.92$.

• A. True
• B. False

Answer 1

The correct option is A True

Consider the triangular sheet whose each side is $10cm$ long.
We know that the area of equilateral triangle $=\frac{\sqrt{3}}{4}{side}^2$
$=\frac{\sqrt{3}}{4}{\left(10\right)}^2=25\sqrt{3}{cm}^2$
So, $A_1$ = $25\sqrt{3}{cm}^2$
Now, consider the two identical triangular strip having sides $10cm,9cm$ and $3cm$, respectively.
Here, we have $a=10cm,b=9cm$ and $c=3cm$.
Semi-perimeter, $s=\frac{a+b+c}{2}=\frac{10cm+9cm+3cm}{2}=11cm$
Area of the triangle $=\sqrt{11\times (11-10)\times (11-9)\times (11-3)}c{m}^2$
$=\sqrt{11\times 1\times 2\times 8}$

$=4\sqrt{11}{cm}^2$
So, area of two identical triangular strips, $A_2=2\times 4\sqrt{11}{cm}^2=8\sqrt{11}{cm}^2$

Again, consider the two identical triangular strip having sides $20cm,20cm$ and $4cm$, respectively.
Here, we have $a=20cm,b=20cm$ and $c=4cm$
Semi-perimeter, $s=\frac{a+b+c}{2}=\frac{20cm+20cm+4cm}{2}=22cm$
Area of the triangle $=\sqrt{22\times (22-20)\times (22-20)\times (22-4)}c{m}^2$
$=\sqrt{22\times 2\times 2\times 18}$

$=12\sqrt{11}{cm}^2$
So, area of two identical triangular strip, $A_3=2\times 12\sqrt{11}{cm}^2=24\sqrt{11}{cm}^2$
Therefore, area of total black coloured sheet used $=A_1+A_2+A_3$

$=25\sqrt{3}{cm}^2+8\sqrt{11}{cm}^2+24\sqrt{11}{cm}^2$
$=25\sqrt{3}{cm}^2+32\sqrt{11}{cm}^2$
$=25\times 1.732+32\times 3.31=149.22{cm}^2$

Given, cost of black paper used per ${cm}^2=10$ paise or $Rs.0.1$

So, cost of using $149.22{cm}^2$ of black paper $=Rs.0.1\times 149.22=Rs.14.92$