Question

The area of an isosceles triangle is 240cm^2 and the length of each one of the equal side is 26cm,find its base?

Answer 1

Area=240cm^2

a=26 cm

b=?

Area of an isosceles triangle

= (b/4)√(4a^2-b^2)

240×4=b√(4×26×26-b^2)

960/b=√(2704-b^2)

960×960/b×b=2704-b^2

921600=2704b^2-b^4

b^4-2704b^2+921600=0

b^4-2304b^2-400b^2+921600=0

b^2(b^2-2304)-400(b^2-2304)=0

(b^2-2304)(b^2-400)=0

therefore, b=48 cm or 20 cm

Answer 2

Answer: The base of an isosceles triangle is 48cm or 20cm when its area is 240cm² and the length of each one of the equal side is 26cm.

Step-by-step explanation:

Area of ΔABC = 240cm²
In an isosceles triangle, two sides are always equal and the length of each equal side is 26cm
AB = AC = 26cm
To find the base, we draw a AD perpendicular to BC.

Let CD be x and AD be h.
ar(ΔADC) = $\frac{1}{2} CD (AD)$
$\frac{1}{2}$ ar (ΔABC) = $\frac{1}{2} xh$
$h =\frac{240}{x}$
By using pythagoras theorem
AC² = AD²+ CD²
26² = h² + x²

676 = (240/x)² + x²
x⁴ - 676x² + 57600 = 0
Here we will take p which represents  x²
p²- 676p +57600 = 0
p = 576 , 100
x= 26 , 10
Now
BC = BD+ CD
      =24+ 24, 10+10
BC= 48, 20

Hence, the base is 48cm or 20cm

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