Question

Find the base of an isosceles triangle whose area is 60 cm² and the length of one of its equal sides is 13 cm. ​

Answer 1

Answer:

Base = 24 cm or 10cm

Step-by-step explanation:

REMEMBER:
An isosceles triangle ABC with base BC = ‘b' & height AD = ‘h' & its equal sides =13 cm & area = 60 cm²

Using the formulas

$A=\frac{bh_b}{2}$

$h_b=\sqrt{a^2-\frac{b^2}{4} }$

There are 2 solutions for $b$

$b=2\sqrt{2} \frac{A}{\sqrt{a^2+\sqrt{a^4-4A^2} } } =2*\sqrt{2} *\frac{60}{\sqrt{13^2+\sqrt{13^4-4*60^2} } }$ ≈ $10cm$$b=2\sqrt{2} \frac{A}{\sqrt{a^2+\sqrt{a^4-4A^2} } } =2*\sqrt{2} *\frac{60}{\sqrt{13^2-\sqrt{13^4-4*60^2} } }=24cm$

Less complex:

Area of a triangle = 1/2 * b * h = 60

=> h = 120/b

In right triangle ABD

13² = h² + b² /4 ( by Pythagoras law)

=>169 = 120²/b² + b²/4

=>676 b² = 57600 + b^4

=> b^4 - 676 b² + 57600 = 0

=> b² = 676 +- √(676² - 4*57600) / 2

=> b²= 676 +- √(226576) /2

=> b² = (676 +- 476 )/2

=> b² = 1152/2 , 200 /2

=> b² = 576 , 100

=> b = 24, 10

So, Base = 24 cm or 10cm