Question

The side of a triangle is 21 cm, 17cm and 10 cm find the area

Answer 1

Answer :

• Area of triangle is 84cm²

Given :

• The side of a triangle is 21cm , 17cm and 10cm

To find :

• Area

Solution :

Given , the side of a triangle is 21cm , 17cm and 10cm then,

Let,

• a = 21cm
• b = 17cm
• c = 10cm

》a + b + c

》21 + 17 + 10

》48cm

As we know that,

• s = a + b + c / 2

》s = 21 + 17 + 10 / 2

》s = 48/2

》s = 24cm

Now , we have to find the area of triangle

As we know that,

• A = √s(s - a) (s - b) (s - c)

where , s is 24cm and a is 21cm , 17cm and 10cm

》A = √s(s - a) (s - b) (s - c)

》A = √24(24 - 21) (24 - 17) (24 - 10)

》A = √24 × 3 × 7 × 14

》A = √2 × 2 × 2 × 3 × 3 × 7 × 2 × 7

》A = 2 × 2 × 3 × 7

》A = 4 × 21

》 A = 84cm²

Hence , Area of triangle is 84cm²

Answer 2

Given:

• Sides of triangle are 21cm,17cm and 10cm respectively

To find:

• Area of given triangle?

Solution:

Here, given that sides of a triangle are 21cm, 17cm and 10cm respectively and are asked to find area of this triangle.

Now,

To find area of triangle, we know that:

• $\large{\boxed{\bf{\blue{Area_{(triangle)}\:=\:\sqrt{s(s-a)(s-b)(s-c)}}}}}$

Where,

• s = semi perimeter of triangle
• a = first side of triangle ••••21cm (given)
• b = second side of triangle ••••17cm(given)
• c = third side of triangle ••••10cm(given)

According to find semi perimeter, we know that :

• $\large{\boxed{\sf{\green{Semi~ perimeter~ of triangle \:=\: \dfrac{sum \: of \: sides\:}{2}}}}}$

Let put given values in this formula :-

$\large{\sf{Semi~ perimeter\: of \: triangle\:=\: \dfrac{sum\: of\:sides}{2}}}$

$\large{\implies{\sf{Semi\: perimeter\: triangle\:=\:\dfrac{21+17+10}{2}}}}$

$\large{\implies{\sf{Semi\: perimeter\: of\: triangle\:=\:\dfrac{48}{2}}}}$

$\large{\boxed{\implies{\bf{Semi\: perimeter\:of\: triangle\:=\:24cm}}}}$

_________________

Now, we found that semi perimeter of triangle is 24cm

Now, find area of triangle :-

Putting given value in above mentioned formula :-

$\large{\sf{Area_{(triangle)}\:=\:\sqrt{s(s-a)(s-b)(s-c)}}}$

$\large{\implies{\sf{Area_{(triangle)}\:=\:\sqrt{24(24-21)(24-17)(24-10)}}}}$

$\large{\implies{\sf{Area_{(triangle)}\:=\:\sqrt{24(3×7×14)}}}}$

$\large{\implies{\sf{Area_{(triangle)}\:=\:\sqrt{(24×3×7×14)}}}}$

$\large{\implies{\sf{Area_{(triangle)}\:=\:\sqrt{7056}}}}$

$\large{\boxed{\implies{\sf{\pink{Area_{(triangle)}\:=\:84cm²}}}}}$