Question

The base of an Isosceles triangle is 48cm and its
perimeter is
98 cm. Find its area​

Answer 1

Answer:

Let AB=AC=0

⇒2a+48=108

⇒2a=60+a=30

Perpendicular AD to BC also bisect BC(Isosceles triangle prop)

⇒BD=24cm By Pythagoras theorem,

AD

2

=a

2

−24

2

=900−576=324

⇒AD=18cm

area of triangle =

2

1

×Base×Height=

2

1

×48×18=432cm

2

.

Answer 2

AnSwer -:



Area of triangle = 168 cm^3

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_____________________________

Explanation-:

The area of triangle-:

“A = √s(s-a)(s-b)(s-c)”

S = Semi perimeter

A = Area

A,B,C = three sides of triangle

______________________________

Given,

The bases of isosceles triangle is 48 cm

The perimeter of triangle-: 98 cm

To find ,

The area of triangle

_____________________________

Now ,

Let the congruent sides of the isosceles

triangle =a and the base =b.

B = base = 48 cm

Perimeter of the triangle =Sum of all

sides = a + a + b =98 cm

= 2 a + 48 = 98cm

2a = 98 - 48

2 a = 50

A = 50/2

A = 25

The two sides of isosceles triangle is 25

cm and 25 cm and third side is 48 cm


The area of triangle-:

“A = √s(s-a)(s-b)(s-c)”

S = Semi perimeter = 49 cm

A = Area

A = 25 cm

B = 25 cm

C = 48 cm

Semi perimeter-: a + b + c / 2

-: 25 + 25 + 48 / 2

= 98/2

= 49 cm

Semi perimeter-: 49 cm


= √49(49-25)(49-25)(49-48)

= √49 x 24 x 24 x 1

= √ 28,224

= 168 cm^2

Hence,

Area of triangle = 168 cm^3

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