Question

find the area of a triangle two sides of which are 8cm and 11 cm and the perimeter is 32 cm

Answer 1

Step-by-step explanation:

One side= 8cm

Second side = 11cm

Perimeter = 32cm

perimeter of a triangle

=first side + second side + third side

=> 32 = 8 + 11+ third side

=>32=19+ third side

=> third side + 19 = 32

=> third side = 32 - 19

=> third side =13

Answer 2

Answer:

$\sf{Area \ of \ the \ triangle \ is \ 6\sqrt5 \ cm^{2}.}$

To find:

$\textsf{The area of a triangle two sides of which are} \\ \textsf{8 cm and 11 cm and the perimeter is 32 cm.}$

Solution:

$\sf{Let \ the \ third \ side \ of \ the \ triangle \ be \ x.} \\ \\ \sf{According \ to \ the \ given} \\ \\ \sf{8+11+x=32} \\ \\ \sf{\therefore{x=32-19}} \\ \\ \sf{\therefore{x=13 \ cm}} \\ \\ \sf{Now, \ sides \ of \ triangle \ are} \\ \\ \sf{8 \ cm, \ 11 \ cm \ and \ 13 \ cm} \\ \\ \sf{Here, \ a = 8 \ cm, \ b=11 \ cm \ and \ c=13 \ cm} \\ \\ \sf{By \ heron's \ formula} \\ \\ \sf{s=\dfrac{Perimeter}{2}} \\ \\ \sf{\therefore{s=\dfrac{32}{2}=16}}$

$\boxed{\sf{A(\triangle)=\sqrt{s(s-a)(s-b)(s-c)}}} \\ \\ \sf{\therefore{A(\triangle)=\sqrt{16(16-8)(16-11)(16-13)}}} \\ \\ \sf{\therefore{A(\triangle=\sqrt{6\times2\times5\times3}}} \\ \\ \sf{\therefore{A(\triangle)=6\sqrt{5} \ cm^{2}}} \\ \\ \purple{\tt{\therefore{Area \ of \ the \ triangle \ is \ 6\sqrt5 \ cm^{2}.}}}$