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Area of ∆ ABC = 32 ,
1/2 [ -5(-5-k) + 3(k--1) + 5(-1--5) ] = 32,
[25+5k + 3(k+1) +5(-1+5)] = 32×2,
[25+5k + 3k+3 +5×4] = 64,
[8k +25+3+20]=64,
(8k+48) = 64,
8k=64-48,
8k=16,
then
k=16/8 = 2
1/2 [ -5(-5-k) + 3(k--1) + 5(-1--5) ] = 32,
[25+5k + 3(k+1) +5(-1+5)] = 32×2,
[25+5k + 3k+3 +5×4] = 64,
[8k +25+3+20]=64,
(8k+48) = 64,
8k=64-48,
8k=16,
then
k=16/8 = 2
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