Question

The area of triangle with given two sides 18cm and 10cm respectively and perimeter equal to 42 CM is


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Answer 1

Answer:

$\sf{The \ area \ of \ the \ triangle \ is \ 24\sqrt{35} \ cm^{2}}$

Given:

• Two sides of triangle are 18 cm and 10 cm

• Perimeter of the triangle is 42 cm

To find:

• The area of the triangle.

Solution:

$\sf{Let \ the \ third \ side \ of \ the \ triangle \ be \ x.}$

$\sf{Sum \ of \ all \ sides=Perimeter}$

$\sf{\therefore{18+10+x=42}}$

$\sf{\therefore{x=42-28}}$

$\sf{\therefore{x=14}}$

$\sf{Hence, \ three \ sides \ of \ triangle \ are \ 18 \ cm,}$

$\sf{10 \ cm \ and \ 14 \ cm}$

$\sf{Let, \ a=18 \ cm, \ b=10 \ cm \ and \ c=14 \ cm}$

$\sf{By \ heron's \ formula}$

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

$\sf{\therefore{s=\dfrac{42}{2}}}$

$\sf{\therefore{s=24}}$

$\boxed{\sf{A(\triangle)=\sqrt{s(s-a)(s-b)(s-c)}}}$

$\sf{\therefore{A(\triangle)=\sqrt{24(24-18)(24-10)(24-14)}}}$

$\sf{\therefore{A(\triangle)=\sqrt{(24\times6\times14\times10}}}$

$\sf{\therefore{A(\triangle)=\sqrt{6\times4\times6\times7\times4\times5}}}$

$\sf{\therefore{A(\triangle)=24\sqrt{35} \ cm^{2}}}$

$\sf\purple{\tt{\therefore{The \ area \ of \ the \ triangle \ is \ 24\sqrt{35} \ cm^{2}}}}$

Answer 2

Answer:

The area of the triangle is 24root35 cm²s

Hope it helps you.