Question

Find the area of an isosceles triangle whose 1 side is 4m greater than its equal side and perimeter is 40m

Answer 1

Hey User!!!

according to the question, the non-equal side of the isosceles triangle is 4m more than the length of it's equal sides.

let the length of equal sides be x and y.

x = y so we can take x as y also.

therefore the no-equal side of the isosceles triangle is x + 4

perimeter of the triangle is given 40m

>> x + x + x + 4 = 40m

>> 3x + 4 = 40m

>> 3x = 40 - 4

>> 3x = 36

>> x= 36/3

>> x = 12m

hence, the length of the sides are :-

equal sides = x and y

x = y

therefore length of it's equal sides are 12cm each.

non-equal side of the triangle = x + 4 = 12 + 4 = 16cm

now, we have to find it's area.

by heron's formula :-

s (semi-perimeter) = 40/2 = 20cm

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

area of the triangle =

$\sqrt{20(20-a)(20-b)(20-c)}   \\ = \sqrt{20(20-12)(20-12)(20-16)}  \\ =\sqrt{20*8*8*4}  \\ =\sqrt{5120}  \\ = 71.55cm^{2}  (approximately)$

Cheers!!!

Answer 2

Answer:

Given ΔABE is isosceles with AB=AC and BC=AB+10

Perimeter of ΔABC=100cm & [

5

=2.23]

Perimeter =AB+BC+AC

100=2AB+AB+10

100=3AB+10

90=3AB

[AB=30cm]

now BC=30+10=40cm

and BO=

2

BC

=20cm

using Pythaoras is ΔAOB

AB

2

=AO

2

+OB

2

(30)

2

=(AO)

2

+(20

2

(AO)

2

=900−400=500

[AO=10

5

cm]

area of ΔABC=

2

1

×BC×AO

=

2

1

×40×10

5

=200

5

=200×2.23

area of ΔABC=446cm

2