Question

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

Answer 1

Given:

Area of the isosceles triangle (ABC) = 60 cm²
Length of the each equal side (AB=AC) = 13cm.

Draw AD perpendicular from A on BC.
Let BC = 2x cm .

Then BD = DC= x cm.

In ∆ABD

AB² = AD² + BD²   [by Pythagoras theorem]
13² = AD² + x²
169 = AD² + x²
AD² = 169 - x²

AD = √ 169 - x²

Area of ∆ ABC = ½ base × height
60 = ½ × BC × AD
60 = ½ ( 2x × √ 169 - x²
60 = x ×√ 169 - x²

On squaring both sides
60² = (x √ 169 - x²)²
3600 = x²(169 - x²)
3600 = 169x²  -  x⁴
x⁴ - 169x² +3600= 0
x⁴ -144 x² -25x² +3600= 0
x²(x² -144) -25(x² -144)= 0
(x² -25) (x² -144)= 0
(x² -25) = 0  or  (x² -144)= 0

x² = 25 or x² = 144

x = √25 or  x = √144
x = 5  or  x = 12

When x= 12, then base = 2x = 2×12= 24 cm or
When base = 5 , then base = 2x = 2× 5 = 10 cm.

Hence, the base is 24 cm or 10 cm.

HOPE THIS WILL HELP YOU....

Answer 2

Nice Questions! ♥

Let's unlock the answers now…

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Let base x and altitude p
Area=1/2(x*p)=60
p=120/x
Now applying pythagoras theorem 
(x/2)^2+(120/x)^2=13^2
solve 
x=24, x=10

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Hope my answer helped … ☺