Question 327.
The fig. shows a triangle ABC of area S with a line DE. If S₁ is area of triangle BDE , prove that S₁/ S = BD * BE / AB * BC.
Answer 327.
Refer to the figure in the following figure.
Let CM and EN be perpendiculars from C and E on AB.
Right Δs BCM and BEN are similar
=> CM/EN = BC/BE ... (1)
S = Area of Δ ABC = (1/2) AB * CM
S₁ = Area of Δ BDE = (1/2) BD * EN
=> S₁/S
= [(1/2) BD * EN] / [(1/2) AB * CM]
= (BD/AB) * (EN/CM)
= (BD/AB) * (BE/BC) ... [Plugging (EN/CM) = (BE/BC) from (1).
= BD * BE / AB * BC.
Link to YA!
The fig. shows a triangle ABC of area S with a line DE. If S₁ is area of triangle BDE , prove that S₁/ S = BD * BE / AB * BC.
Answer 327.
Refer to the figure in the following figure.
Right Δs BCM and BEN are similar
=> CM/EN = BC/BE ... (1)
S = Area of Δ ABC = (1/2) AB * CM
S₁ = Area of Δ BDE = (1/2) BD * EN
=> S₁/S
= [(1/2) BD * EN] / [(1/2) AB * CM]
= (BD/AB) * (EN/CM)
= (BD/AB) * (BE/BC) ... [Plugging (EN/CM) = (BE/BC) from (1).
= BD * BE / AB * BC.
Link to YA!
No comments:
Post a Comment