Question

the side of triangle 8 cm 5 cm and 5 cm. find its area​

Answer 1

Answer:

12 cm²

Step-by-step explanation:

Let the triangle be ABC where AB=AC=5cm and BC=8cm

Draw the altitude(height) from A to BC. Let D be the point on which the altitude intersects BC.

Now In ∆ADB,

Ad²+BD²=AB²(Pythagoras theorem)

AD=√AB²-BD². [BD=BC/2=4cm]

AD=√5²-4²

AD=√25-16

AD=√9

So,AD=3cm

Area of ∆ABC = 1/2xBasexHeight

=1/2xBCxAD

=1/2×4×3

So,Area of ∆ABC is 12cm²

Answer 2

Answer :-

• The area of the triangle is 12cm².

Step-by-step explanation :-

To Find :-

• The area of triangle

Solution :-

Given that,

• The sides of triangle are 8cm, 5cm and 5cm

First we need to find out the semi-perimeter of the triangle,

As we know that,

$\boxed{\bf{Semi-perimeter = \dfrac{a + b + c}{2}}}$

Where,

• a, b and c are the three sides of triangle

$\sf{ \longmapsto \: Semi-perimeter = \dfrac{a + b + c}{2}} \\ \\ \\ \sf{ \longmapsto \: \dfrac{8 + 5 + 5}{2}} \\ \\ \\ \sf{ \longmapsto \: \dfrac{8 + 10}{2}} \\ \\ \\ \sf{ \longmapsto \: \dfrac{18}{2}} \\ \\ \\ \sf{ \longmapsto \: \cancel{\dfrac{18}{2}}} \\ \\ \\ \bf{ \longmapsto \: 9}$

Now, the area of triangle is,

As we know that,

$\boxed{\bf{Area \: of \: \triangle = \sqrt{s(s - a)(s - b)(s - c)}}}$

Therefore,

$\sf{ \longmapsto \: Area = \sqrt{9(9 - 8)(9 - 5)(9 - 5)}} \\ \\ \\ \sf{ \longmapsto \: Area = \sqrt{9(1)(4)(4)}} \\ \\ \\ \sf{ \longmapsto \: Area = \sqrt{9 \times 1 \times 4 \times 4}} \\ \\ \\ \sf{ \longmapsto \: Area = \sqrt{9 \times 4 \times 4}} \\ \\ \\ \sf{ \longmapsto \: Area = \sqrt{36 \times 4}} \\ \\ \\ \sf{ \longmapsto \: Area = \sqrt{144}} \\ \\ \\ \bf{ \longmapsto \: Area = 12}$

• The area of the triangle is 12cm².