the base of isosceles triangle is 24cm and it's area is 192 find its perimeter
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The base of an isosceles triangle is 24 cm and its area is 192 square cm. Find its perimetet?
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
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
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