Triangle ABC is isosceles with AB=AC and BC=65cm. P is a point on BC such that the perpendicular distances from P to AB and AC are 24cm and 36cm, respectively.
The area of triangle ABC is sq. cms 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 C2535 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=π−B−C=π−2B ⇒sinA=sin2B=2sinBcosB=120169 using sine rule in ΔABC: AB=AC=BCsinBsinA=65sinBsin2B=652cosB=1692 Therefore, △=(AB)(AC)sinA2=2535