Question

Suman made an arrangement with white and black coloured paper sheets as shown in figure. The total areas of the white and black paper sheets used in making the arrangement is $15c{m^2}$.
A. True
B. False

Answer 1

Hint: According to the question, we have to determine the semi-perimeter of the blacked triangular sheet which dimension has 2cm, 2cm, and 1.5cm and after finding the semi-perimeter of that blacked sheet we have to find the area of that black-triangular sheet.
So, we have to understand the formula for finding the semi-perimeter and area of a triangle.
Semi-perimeter of triangle$(s) = \dfrac{{a + b + c}}{2}..............................(A)$
Area of triangle$ = \sqrt {s(s - a)(s - b)(s - c)} .................................(B)$
Similarly, we have to determine the semi-perimeter of the bigger triangular sheet which dimension has 4cm, 4cm, and 3cm and after finding the semi-perimeter of that bigger sheet we have to find the area of that bigger-triangular sheet.
Therefore, we have to find the area of white paper sheet used in 4 white triangle design paper by the formula given below.

Formula used:
Area of white paper sheet = Area of 4 bigger triangular sheets – Area of 4 black paper sheets used$................(C)$

Complete step-by-step answer:
Step 1: first of all we have to determine the semi-perimeter of the blacked triangular sheet which dimension has 2cm, 2cm, and 1.5cm with the help of the formula (A) used in the solution hint.
Semi-perimeter of blacked-triangle sheet$(s) = \dfrac{{2 + 2 + 1.5}}{2}$
$
   \Rightarrow s = \dfrac{{5.5}}{2} \\
   \Rightarrow s = 2.75cm \\
 $
Now, we have to find the area of the blacked triangular sheet with the help of the formula (B) used in the solution hint.
Area of the blacked triangular sheet $ = \sqrt {2.75(2.75 - 2)(2.75 - 2)(2.75 - 1.5)} $
$
   \Rightarrow \sqrt {2.75 \times 0.75 \times 0.75 \times 1.25} \\
   \Rightarrow \sqrt {1.93359375} \\
   \Rightarrow 1.39c{m^2} \\
 $
Step 2: Now, we have to find the total area of all black-triangular sheet which is calculated below.
The total area of all black-triangular sheets$ = 4 \times $Area of 1 black triangular sheet
$
   \Rightarrow 4 \times 1.39 \\
   \Rightarrow 5.56c{m^2} \\
 $
Step 3: First of all we have to determine the semi-perimeter of the bigger triangular sheet which dimension has 4cm, 4cm, and 3cm with the help of the formula (A) used in the solution hint.
Semi-perimeter of bigger-triangle sheet$(s) = \dfrac{{4 + 4 + 3}}{2}$
$
   \Rightarrow \ dfrac {11}{2} \\
   \Rightarrow s=5.5cm \\
 $
Now, we have to find the area of the bigger triangular sheet with the help of the formula (B) used in the solution hint.
Area of the bigger triangular sheet $ = \sqrt {5.5(5.5 - 4)(5.5 - 4)(5.5 - 3)} $
$
   \Rightarrow \sqrt {5.5 \times 1.5 \times 1.5 \times 2.5} \\
   \Rightarrow \sqrt {30.9375} \\
   \Rightarrow 5.56c{m^2} \\
 $
Step 4: Now, we have to find the total area of all bigger-triangular sheets which is calculated below.
The total area of all bigger-triangular sheets$ = 4 \times $Area of 1 bigger triangular sheet
$
   \Rightarrow 4 \times 5.56 \\
   \Rightarrow 22.24c{m^2} \\
 $
Step 5: Now, we have to find the area of the white triangular sheet with the help of the formula (C) that is mentioned in the solution hint.
Area of white triangular sheet $ = 22.24 - 5.56$
$ \Rightarrow 16.68c{m^2}$

Hence, with the help of the formulas (A), (B), and (C) we have determined the required area which is $ \Rightarrow 16.68c{m^2}$. Therefore option (A) is correct.

Note:
It is necessary to determine the semi-perimeter of the triangle and then find the area of the triangle with the help of the formulas (A), (B) which is mentioned in the solution hint. It is necessary to find the area of the white triangular sheet by finding the difference between the area of all 4 bigger triangular sheets and the area of all 4 white triangular sheets.