Question

Find area of the triangle if its perimeter is
32cms, it length of its two sides is 8cm and 11cm respectively​

Answer 1

Given : The Perimeter of Triangle is 32 cm and length of its two sides is 8cm and 11cm , respectively.

To Find : The Area of Triangle.

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\sf{\bf{ Solution \:of\:Question-\::}}$ Let's consider length of third side of triangle is x cm.

⠀⠀⠀⠀⠀We know that if we are given with Perimeter and length of two sides of Triangle and we have to find the Area of triangle. So First we have to first find the third side of the triangle for Area of Triangle by using the Formula of Perimeter of Triangle that is :

⠀⠀⠀⠀⠀$\implies {\underline {\sf{ Perimeter_{( Triangle)} = Side \:A + Side \: B + Side \: C \:units}}}$

⠀⠀⠀⠀⠀Here Side A is the First Side of Triangle, Side B is the second side of Triangle & Side C is the third side of the triangle. And we are given with the Perimeter & Length of two sides of Triangle that is Perimeter of Triangle = 32 cm , First Side ( Side A ) = 8 cm , Second side ( Side B) = 11 cm & Third Side ( Side C ) = x . So by using the Required formula we can Find the third side of triangle ;

⠀⠀⠀⠀⠀$\underline {\sf{\bf\star{ Now \: By \: Substituting \: known \: Values \:: }}}$

⠀⠀⠀⠀⠀$\longmapsto { \mathrm { 11 + 8 + x = 32 \: cm}}$

⠀⠀⠀⠀⠀$\longmapsto { \mathrm { 19 + x = 32 \: cm}}$

⠀⠀⠀⠀⠀$\longmapsto { \mathrm { x = 32 - 19 \: }}$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm { x = 13 \: cm}}}}$

Therefore,

⠀⠀⠀⠀⠀$\underline {\therefore\:{\pink{ \mathrm { Third \:Side\:or\:(Side\:C) \:of\:Triangle \:is\:13\: cm}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀ For Area of Triangle first we need to find the Semi-Perimeter of the given Triangle using the Perimeter of Triangle by Applying the Formula of Semi-Perimeter of Triangle:

⠀⠀⠀⠀⠀$\implies {\underline {\sf{ Semi-Perimeter_{( Triangle)} = \dfrac{Perimeter} {2} \:units}}}$

⠀⠀⠀⠀⠀Here we have given with the Perimeter of triangle and Perimeter = 32 cm , And we can find the Semi-Perimeter using the Required formula :

⠀⠀⠀⠀⠀$\underline {\sf{\bf\star{ Now \: By \: Substituting \: known \: Values \:: }}}$

⠀⠀⠀⠀⠀$\longmapsto { \mathrm { Semi-Perimeter \: = \dfrac{\cancel {32}}{\cancel {2}} \: }}$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm { Semi-Perimeter \:= 16 \: cm}}}}$

Therefore,

⠀⠀⠀⠀⠀$\underline {\therefore\:{\pink{ \mathrm { Semi-Perimeter \:of\:Triangle \:is\:13\: cm}}}}$

⠀⠀⠀⠀⠀Now As we have found the Third side or Side C and Semi-Perimeter of Triangle and now we have all three sides and Semi-Perimeter of Triangle. And we have to find find the Area of Triangle using three sides and Semi-Perimeter of Triangle by using the Heron's Formula for Area of Triangle that is :

⠀⠀⠀⠀⠀$\implies {\underline {\sf{ Area_{( Triangle)} = \sqrt { s (s - a ) (s - b) (s -c ) }\:sq.units}}}$

⠀⠀⠀⠀⠀Here s is the Semi-Perimeter of triangle , a is First side of triangle , b is the second side of triangle & c is the third side of triangle and we have found already Semi-Perimeter and Third Side of Triangle and other Values are already Given . And Semi-Perimeter = 16 cm , First side (a) = 8 cm , Second side = 11 cm (b) & Third side (c) = 13 cm . So by Substituting the all here given Values in the Required formula we can find the Area of Triangle :

⠀⠀⠀⠀⠀$\underline {\sf{\bf\star{ Now \: By \: Substituting \: known \: Values \:: }}}$

⠀⠀⠀⠀⠀$\longmapsto { \mathrm { Area_{( Triangle)} = \sqrt { 16 (16 - 8 ) (16 - 11) (16 -13 ) } }}$

⠀⠀⠀⠀⠀$\longmapsto { \mathrm { Area_{( Triangle)} = \sqrt { 16 ( 8 ) (5) (3 ) } }}$

⠀⠀⠀⠀⠀$\longmapsto { \mathrm { Area_{( Triangle)} = \sqrt { 16 \times 8 \times 5 \times 3 } }}$

⠀⠀⠀⠀⠀$\longmapsto { \mathrm { Area_{( Triangle)} = \sqrt { (8 \times 2 ) \times 8 \times 5 \times 3 } }}$

⠀⠀⠀⠀⠀$\longmapsto { \mathrm { Area_{( Triangle)} = \sqrt { (8 \times 8 ) \times 2 \times 5 \times 3 } }}$

⠀⠀⠀⠀⠀$\longmapsto { \mathrm { Area_{( Triangle)} = \sqrt { (8 )^{2} \times 2 \times 5 \times 3 } }}$

$\sqrt{a^{2} \times y} = a\sqrt{y}$

⠀⠀⠀⠀⠀$\longmapsto { \mathrm { Area_{( Triangle)} = 8\sqrt { 2 \times 5 \times 3 } }}$

⠀⠀⠀⠀⠀$\longmapsto { \mathrm { Area_{( Triangle)} = 8\sqrt { 2 \times 15 } }}$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm { Area_{(Triangle)} =8 \sqrt {30}\: cm^{2} }}}}$

Therefore,

⠀⠀⠀⠀⠀$\underline {\therefore\:{\pink{ \mathrm { Area \:of\:Triangle \:is\: 8\sqrt{30}\: cm^{2} }}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀