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Question

Triangle ABC is isosceles with AB=AC and BC=65 cm. P is a point on BC such that the perpendicular distance from P to AB and AC are 24 cm and 36 cm respectively.
The area of triangle ABC in sq-cm is?

A
1254
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B
1950
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C
2535
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D
5070
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Solution

The correct option is C 2535
From the fig:
AB=AC; B=C
BC=BP+PC=65 ...(1)
and PD=24; PE=36

In PBD
BP=PDsinB=24sinB ....(2)

In PCE
PC=PEsinC=36sinB ....(3)

From (2) & (3), we get BP=2PC3 ....(4)
from (1) & (4),
BP=39 and PC=26

From (2), sinB=2439=1213
cosB=513

A=πBC=π2B
sinA=sin2B=2sinBcosB=120169

using sine rule in ΔABC:

AB=AC=BCsinBsinA=65sinBsin2B=652cosB=1692

Therefore, =(AB)(AC)sinA2=2535

194270_116966_ans_1fe2a633ee6d497286e6ec45b75f8c87.png

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