Question

the base of an isosceles triangle is 24cm and it's area is 192sq.cm. find its perimeter

Answer 1

1/2*b*h=192
h=2*192/24=16
equal side^2=(16^2)+(12^2)
Eq. side =20
perimeter =2(20)+24
=64

Answer 2

Answer:

64 cm.

Step-by-step explanation:

Refer the attached figure

In ΔABC

AB = AC since the given triangle is isosceles.

Base = BC = 24 cm

Altitude = AD

Area = 192 sq.cm.

Area of triangle = $\frac{1}{2} \times Base \times Height$

So, $192=\frac{1}{2} \times 24 \times AD/tex]</p><p> [tex]192=12\times AD/tex]</p><p> [tex]\frac{192}{12}= AD/tex]</p><p> [tex]16= AD/tex]</p><p>Thus the Altitude = 16 cm</p><p> Note : The altitude to the base of an isosceles triangle bisects the base.</p><p>So, BD=DC = BC/2 = 24/2 =12 cm</p><p>In ΔABD</p><p>[tex]Hypotenuse^2 = Perpendicular^2 + Base^2$

$AB^2 =AD^2 + BD^2$

$AB^2 =16^2 + 12^2$

$AB^2 =400$

$AB=20$

Thus AB =AC = 20 cm

BC = 24 cm

Perimeter of triangle = Sum of all sides

                                  = AB+BC+AC

                                  = 20+24+20

                                  = 64 cm.

Hence the perimeter of triangle is 64 cm.