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Question-187819




Question Number 187819 by normans last updated on 22/Feb/23
Answered by mr W last updated on 22/Feb/23
Commented by mr W last updated on 22/Feb/23
a=10  b=8  cos α=((2c)/(a+b))  ED^2 =a^2 +c^2 −2ac×((2c)/(a+b))  BD^2 =a^2 +4c^2 −4ac×((2c)/(a+b))  ED^2 +BD^2 =c^2   a^2 +c^2 −2ac×((2c)/(a+b))+a^2 +4c^2 −4ac×((2c)/(a+b))=c^2   c^2 =((a^2 (a+b))/(2(2a−b)))  ⇒c=a(√((a+b)/(2(2a−b))))  BC=(√((a+b)^2 −(2c)^2 ))=(√((b(a^2 −b^2 ))/(2a−b)))  area of triangle ABC:  Δ=(((2c)×BC)/2)=a(√((a+b)/(2(2a−b))))(√((b(a^2 −b^2 ))/(2a−b)))  Δ=((a(a+b))/(2a−b))(√((b(a−b))/2))  Δ=((10×18)/(12))(√((8×2)/2))=30(√2) ✓
a=10b=8cosα=2ca+bED2=a2+c22ac×2ca+bBD2=a2+4c24ac×2ca+bED2+BD2=c2a2+c22ac×2ca+b+a2+4c24ac×2ca+b=c2c2=a2(a+b)2(2ab)c=aa+b2(2ab)BC=(a+b)2(2c)2=b(a2b2)2abareaoftriangleABC:Δ=(2c)×BC2=aa+b2(2ab)b(a2b2)2abΔ=a(a+b)2abb(ab)2Δ=10×18128×22=302
Commented by normans last updated on 22/Feb/23
nice solution, thank you sir
nicesolution,thankyousir

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