Question

Find the base of an isosceles triangle whose area is 60 cm ^ 2 and the length of one of its equal sides is 13 cm.​

Answer 1

Answer:

Let ABC be the given isosceles triangle in which AB=AC=13cm. Draw AD perpendicular from A on BC

Let BC=2x cm. Then BD=DC=x cm

In △ABD, we have

AB

2

+AD

2

+BD

2

(By pythagoras theorem)

⇒13

2

=AD

2

+x

2

⇒AD=

13

2

−x

2

=

169−x

2

Now, Area =60cm

2

⇒

2

1

(BC×AD)=60⇒

2

1

{2x×

169−x

2

}=60

⇒x

169−x

2

=60⇒x

2

(169−x

2

)=3600

⇒(x

2

−144)(x

2

−25)=0⇒x

2

=144orx

2

=25

⇒x=12orx=5

Hence, Base 2x=24cm or 10cm