Login

find the area of an isosceles triangle whose one side is 10 cm greater than its equal sides and its perimeter is 100 cm

Let the equal side of an isosceles triangle be x cm.According to question, other side=x+10 cmGiven, perimeterp=100 cm⇒ x+x+x+10 =100⇒3x=90⇒x=30 cmThus, equal sides=30 cm And, other side=30+10=40 cmNow, s=p2=1002=50 cmArea of an isosceles triangle=ss-as-bs-c                                                  =5050-3050-3050-40                                                  =50×20×20×10                                                  =2005 cmHence, Area of an isosceles triangle is 2005 cm.

  • 26
View Full Answer

Let the equal side be x

the other side = x + 10

Perimeter = sum of all three sides = 100cm

therefore, perimeter is also = x + x + x+10

= 3x +10

By the problem, 3x+10 = 100

therefore, 3x= 100-10

= 90

therefore, x= 90/3

= 30

the equal sides are 30 cm each n the other side is 40 cm

S=perimeter/2

s=100/2 = 50

area = sq. root of [s(s-a) (s-b) (s-c)] where a, b and c are sides of the triangle.

= sq. root of [ 50 (50-30) (50-30) (50-40)]

= sq. root of [50* 20* 20* 10]

= 200 sq. root 5

  • 12
What are you looking for?