Question

find the area of a trangle whose perimeter is 24 and two sides are 6m and 8m.
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Answer 1

Answer:

40 cm^2

Step-by-step explanation:

area=1/2*b*h

perimeter= a+b+c sides

                 6+8+x=24

                         x=24-(6+8)

                         x=24-14

                       x=10

area=1/2*8*10

       =40 cm^2

Answer 2

★GIVEN★

A triangle whose perimeter is 24 and two sides are 6m and 8m.

★To Find★

The area.

★SOLUTION★

We know that,

All sides of a triangle sum up to its perimeter.

Therefore,

• Perimeter = 24 m.
• Two sides are:- 6 m and 8 m.
• Let the other side be x.

★SOLUTION★

$\large\implies{\sf{6+8+x=24}}$

$\large\implies{\sf{14+x=24}}$

$\large\implies{\sf{x=24-14}}$

$\large\therefore\boxed{\bf{x=10.}}$

The third side is 10 m.

★Now the area★

We have to find the semiperimeter.

$\large\implies{\sf{Semiperimeter=\dfrac{Perimeter}{2}}}$

$\large\implies{\sf{Semiperimeter=\dfrac{24}{2}}}$

$\large\implies{\sf{Semiperimeter=\dfrac{\cancel{24}}{\cancel{2}}}}$

$\large\therefore\boxed{\bf{Semiperimeter=12\:m.}}$

Now by Heron's Formula,

$\large{\green{\underline{\boxed{\bf{Area=\sqrt{s(s-a)(s-b)(s-c)}}}}}}$

where,

• s is the semiperimeter = 12 m.
• a is first side = 6 m.
• b is the second side = 8 m.
• c is the third side = 10 m.

Putting the values,

$\large\implies{\sf{Area=\sqrt{12(12-6)(12-8)(12-10)}}}$

$\large\implies{\sf{Area=\sqrt{12\times6\times4\times2}}}$

$\large\implies{\sf{Area=\sqrt{576}}}$

$\large\implies{\sf{Area=24\:m^2.}}$

So, the area of the triangle is 24 m².