Question

The base of an isosceles triangle is 48 cm and its perimeter is 98 Finds area please give answer with explanation​

Answer 1

Since two sides of isosceles triangle is equal,

So, let them be ' x cm ' each.

Now,

48 + x + x = 98

2x = 98 - 48 = 50

x = 50/2 = 25cm

s = 98 / 2 = 49 cm

So, Area =

$= \sqrt{s(s - a)(s - b)(s - c)}$

$= \sqrt{49(49 - 48)(49 - 25)(49 - 25)}$

$= \sqrt{49 \times 1 \times 24 \times 24}$

$= 168 \: {cm}^{2}$

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Answer 2

Answer:

The area of the isosceles triangle is 168 cm².

Step-by-step-explanation:

NOTE: Refer to the attachment for the diagram.

In figure, △ABC is an isosceles triangle.

BC is the base having length of 48 cm.

The perimeter of the triangle is 98 cm.

We have to find the area of the triangle.

Now, we know that,

Two sides of an isosceles triangle are congruent.

AB = AC = x cm

Now, we know that,

Perimeter of triangle = Sum of all sides

⇒ P = AB + BC + AC

⇒ 98 = x + 48 + x

⇒ 98 = x + x + 48

⇒ 98 = 2x + 48

⇒ 2x = 98 - 48

⇒ 2x = 50

⇒ x = 50 ÷ 2

⇒ x = 25 cm

∴ AB = AC = 25 cm

In figure, AD ⊥ BC,

We know that,

A perpendicular drawn to the base of an isosceles triangle bisects the base.

∴ BD = DC = ½ * BC

⇒ DC = ½ * BC

⇒ DC = ½ * 48

⇒ DC = 24 cm

Now, in △ADC, m∟ADC = 90°,

∴ ( AC )² = ( AD )² + ( DC )² - - - [ Pythagoras theorem ]

⇒ ( 25 )² = ( AD )² + ( 24 )²

⇒ 625 = AD² + 576

⇒ AD² = 625 - 576

⇒ AD² = 49

⇒ AD = 7 cm - - - [ Taking square roots ]

Now, we know that,

Area of isosceles triangle = ( Base * Height ) / 2

⇒ A ( △ABC ) = ( BC * AD ) / 2

⇒ A ( △ABC ) = ( 48 * 7 ) / 2

⇒ A ( △ABC ) = 48 ÷ 2 * 7

⇒ A ( △ABC ) = 24 * 7

⇒ A ( △ABC ) = 168 cm²

∴ The area of the isosceles triangle is 168 cm².