Question

find the area of the triangle whose perimeter is 42 cm and two sides are 20 cm and 10 cm​

Answer 1

Answer:

a+b+c=42

20+10+x=42

x=42-30

x=12cm

then s=a+b+c/2

42/2=21

so s(s-a) (s-b) (s-c)

after putting the values and solving this we get area=3root231cm square.

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Answer 2

• Given

• Perimeter = 42 cm
• Two sides = 20 cm and 10 cm

• To find

• Area of triangle

• Procedure

• We have the value of perimeter of the triangle and the measure of its two sides. Firstly, we will let the third side be x. And by using the formula of perimeter of triangle we will find the third side (x).

• And then we will use Heron's formula and find the area of triangle.

Formulae to be used -

• Perimeter = Sum of all sides.
• Heron's formula = $\sf{\sqrt{s(s - a)(s - b)(s - c)}}$
• s = (a + b + c)/2

• Solution

Let the third side be x.

⇒ Perimeter = Sum of all sides

⇒ 42 = 20 + 10 + x

⇒ 42 = 30 + x

⇒ 42 - 30 = x

⇒ 12 = x

•°• The value of x is 12.

_____________________________

Let's verify -

⇒ 20 + 10 + 12 = 42

⇒ 42 = 42

•°• LHS = RHS

Hence, verified.

_______________________________

• Third side = 12 cm

Let the sides of triangle be a,b and c.

• a = 20 cm
• b = 10 cm
• c = 12 cm

⇒ s = (a + b + c)/2

Substituting the given values,

⇒ s = (20 + 10 + 12)/2

⇒ s = 42/2

⇒ s = 21

⇒ Heron's formula = $\sf{\sqrt{s(s - a)(s - b)(s - c)}}$

⇒ $\sf{\sqrt{21(21 - 20)(21 - 10)(21 - 12)}}$

⇒ $\sf{\sqrt{21(1)(11)(9)}}$

⇒ $\sf{\sqrt{21(1)(11)(9)}}$

⇒ $\sf{66.24~cm^2}$

• Area of triangle = 66.24 cm²