Question

Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is
42cm
​

Answer 1

Given :

• Perimeter of triangle = 42 cm
• Length of first side = 18 cm
• Length of second side = 10 cm

To find :

• Area of triangle

Solution :

We know that :-

=> perimeter = sum of all sides

Let length of third side be x .

Now , according to the question :-

=> 18+10+ x = 42

=> 28+ x = 42

=> x = 42-28

=> x = 14 cm

Therefore , length of third side = 14 cm

Now , to find area of triangle we will use heron's formula.

So, first we will use heron's formula :-

$\implies \sf s = \dfrac{18 + 10 + 14}{2}$

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

$\implies \sf s =21$

Now, area is :-

$\sf \implies A = \sqrt{21(21 - 18)(21 - 10)(21 - 14)}$

$\sf \implies A = \sqrt{21 \times 3 \times11 \times 7}$

$\sf \implies A = \sqrt{3 \times 7 \times 3 \times11 \times 7}$

$\sf \implies A = 3 \times 7 \times \sqrt{ 11 }$

$\sf \implies A = 21 \times \sqrt{ 11 }$

$\sf \implies A = 21 \sqrt{ 11 }$

Answer :

• Area of triangle = 21√(11) cm²

Answer 2

1 st side = 18 cm

2 nd side = 10 cm

therefore,sum of these = 18+10

= 28

perimeter = 42 cm

1st and 2nd side sum= 28 cm

therefore, 3rd side = 42 - 28 cm

= 14 cm ans

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