Question

Find the area of triangle, two of its sides are 16 and 22 cm and the perimeter is 64 cm.

Answer 1

Answer:

why we find

find yourself

Answer 2

Solution :

$\underline{\bf{Given\::}}}}$

• Two sides of a triangle are 16 cm & 22 cm .
• Perimeter of triangle, (P) = 64 cm

$\underline{\bf{Explanation\::}}}}$

Diagram :

$\setlength{\unitlength}{0.8cm}\begin{picture}(6,5)\thicklines\put(1,0.5){\line(2,1){3}}\put(4,2){\line(-2,1){2}}\put(2,3){\line(-2,-5){1}}\put(0.7,0.3){ \bf A }\put(4.05,1.9){ \bf B }\put(1.7,2.95){ \bf C }\put(3.1,2.5){ \sf r\:cm }\put(1.3,1.7){ \sf \:\:\:22\:cm }\put(2.5,1.05){ \sf16\:cm }\end{picture}$

Let the third side of triangle be r cm;

As we know that perimeter of triangle :

$\boxed{\bf{Perimeter=Side+Side+Side}}}$

$\longrightarrow\sf{64=16+22+r}\\\\\longrightarrow\sf{64=38+r}\\\\\longrightarrow\sf{r=64-38}\\\\\longrightarrow\bf{r=26\:cm}$

Now;

$\underline{\boldsymbol{By\:using\:Heron's\:formula\::}}}$

• A = 16 cm
• B = 22 cm
• C = 26 cm

$\longrightarrow\sf{Semi-Perimeter=\dfrac{Side+Side+Side}{2} }\\\\\\\longrightarrow\sf{Semi-Perimeter=\dfrac{A+B+C}{2} }\\\\\\\longrightarrow\sf{Semi-Perimeter=\dfrac{16+22+26}{2} }\\\\\\\longrightarrow\sf{Semi-Perimeter=\cancel{\dfrac{64}{2} }}\\\\\\\longrightarrow\bf{Semi-Perimeter=32\:cm}$

&

$\boxed{\bf{Area\:of\:\triangle = s\sqrt{(s-a)(s-b)(s-c)} }}}$

$\mapsto\sf{Area\:of\:\triangle = \sqrt{32(32-16)(32-22)(32-26)} }\\\\\mapsto\sf{Area\:of\:\triangle =\sqrt{32\times 16 \times 10\times 6} }\\\\\mapsto\sf{Area\:of\:\triangle =\sqrt{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 \times 2\times 2\times 5 \times 2\times 3} }\\\\\mapsto\sf{Area\:of\:\triangle =2 \times 2\times 2\times 2\times 2 \sqrt{30} }\\\\\mapsto\bf{Area\:of\:\triangle =32\sqrt{30} \:cm^{2} }$

Thus;

The area of triangle will be 32√30 cm² .