Question

The base of an isosceles triangle is 24cm and its area is 192sq cm. The length of each of its equal sides is *


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Answer 1

♧QUESTION♧

The base of an isosceles triangle is 24cm and its area is 192sq cm. The length of each of its equal sides is *

☆ANSWER☆ = 64

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 Height21×Base×Height

So, 192=\frac{1}{2} \times 24 \times AD/ 192=12\times AD \frac{192}{12}= AD/te [tex]16= AD/tex] Thus the Altitude = 16 cm Note : The altitude to the base of an isosceles triangle bisects the base. So, BD=DC = BC/2 = 24/2 =12 cm In ΔABD [tex]Hypotenuse^2 = Perpendicular^2 + Base^2192=21×24×AD/tex][tex]192=12×AD/tex][tex]12192=AD/16=AD/tex]ThustheAltitude=16cmNote:Thealtitudetothebaseofanisoscelestrianglebisectsthebase.So,BD=DC=BC/2=24/2=12cmInΔABD[tex]Hypotenuse2=Perpendicular2+Base2

AB^2 =AD^2 + BD^2AB2=AD2+BD2

AB^2 =16^2 + 12^2AB2=162+122

AB^2 =400AB2=400

AB=20AB=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

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