Question

The Base of an isosceles triangle is 24 cm and area is 192 cm² find its perimeter

Answer 1

GIVEN: ABC an isosceles triangle, AB=AC. So angle B = angle C. Base BC = 24cm,  TO FIND: The perimeter(AB + BC + AC) =?  CONSTRUCTION: AM perpendicular to BC.  Since triangle AMB is congruent to triangle AMC( by RHS congruence criterion)  So, M is the mid point of BC. So, BM = 12cm  Since area(triangle ABC) = 1/2 * BC * AM  => ar(tri ABC) = 1/2 * 24 * AM = 192  => AM = 192/ 12 = 16cm.  Now in right triangle AMB  AB² = AM² + BM²  => AB² = 16² + 12²  => AB² = 256 + 144 = 400  => AB = √400 = 20cm  So, AC = 20cm  So, perimeter = 20+20+24 = 64cm ……….ANS

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assume a ∆PQR with median T

$\textbf{\underline{Area\;of\;isosceles\;triangle}}$

= 192 cm²

QR = 24 cm

$\textbf{\underline{QT = TR = 12\;cm}}$

${\boxed{\sf\:{Using\;Formula\;we\;have}}}$

$\textbf{\underline{Area\;of\;Triangle}}$

$\tt{\rightarrow\dfrac{1}{2}\times Base\times Height}$

$\tt{\rightarrow 192=\dfrac{1}{2}\times 24\times h}$

192 = 12 × h

$\tt{\rightarrow h=\dfrac{192}{12}}$

h = 16 cm

${\boxed{\sf\:{Using\;Pythagoras\;theorem :-}}}$

PQ² = QT² + PT²

PQ² = (16² + 12²)

PQ² = 256 + 144

PQ² = 400

PQ = √400

PQ = 20 cm

Therefore,

$\textbf{\underline{Perimeter = PQ + QR + RP}}$

= 20 + 24 + 20

= 44 + 20

= 64 cm

So,

$\Large{\boxed{\sf\:{Perimeter\;of\;triangle = 64 cm}}}$