Question

Q. Find the area of a triangle, two sides of which are 8 cm and 11 cm and the perimeter is 32 cm. ​

Answer 1

Given :-

Two sides of a triangle are 8 cm and 11 cm and the perimeter is 32 cm

To Find :-

The Area of The traingle

Solution :-

See the attachment , from their we have ;

• AB = c = 8 cm
• AC = b = 11 cm
• BC = a = ?

Also , given that Perimeter = 32 cm . It implies ;

$\quad \leadsto \quad \sf a + b + c = 32$

${ : \implies \quad \sf a + 11 + 8 = 32 }$

${ : \implies \quad \sf a + 19 = 32 }$

${ : \implies \quad \sf a = 32 - 19 }$

${ : \implies \quad \sf \therefore \: \: a = 13 \: \: cm }$

Now , we knows that Semi perimeter ( s ) is half the perimeter which implies ;

$\quad \leadsto \quad \sf s = \dfrac{Perimeter}{2}$

${ : \implies \quad \sf s = \dfrac{32}{2} }$

${ : \implies \quad \sf s = \cancel{\dfrac{32}{2}} }$

${ : \implies \quad \sf \therefore \: \: s = 16 \: \: cm}$

Now , from Heron's Formula , we knows that are of a triangle is given by :-

${ \bigstar { \underline { \boxed { \red { \underbrace { \bf { Area = \sqrt{s ( s - a ) ( s - b ) ( s - c )}}}}}}}}{\bigstar}$

Now ,

${\quad \leadsto \quad \sf Area_{(∆ABC)} = \sqrt{s ( s - a ) ( s - b ) ( s - c)}}$

${ : \implies \quad \sf Area_{(∆ABC)} = \sqrt{16 ( 16-13) ( 16-11) ( 16-8)} }$

${ : \implies \quad \sf Area_{(∆ABC)} = \sqrt{16 \times 3 \times 5 \times 8 } }$

${ : \implies \quad \sf Area_{(∆ABC)} = \sqrt{ 8 \times 2 \times 3 \times 5 \times 8 } }$

${ : \implies \quad \sf Area_{(∆ABC)} = \sqrt{\underline{8} \times 2 \times 3 \times 5 \times \underline{8} } }$

${ : \implies \quad \sf Area_{(∆ABC)} = 8 \sqrt{2 \times 3 \times 5 } }$

${ : \implies \quad \bf Area_{(∆ABC)} = 8 \sqrt{30 } \: \:cm²}$

$\quad \qquad { \bigstar { \underline { \boxed { \red { { \bf { \therefore Area_{(∆ABC)}=8 \sqrt{30} \: \:cm²}}}}}}}{\bigstar}$

Answer 2

★∴Area(∆ABC)=830cm²★.