Q21 of 83 Page 367

In the adjoining figure, the point D divides the side BC of ∆ABC in the ratio m:n. Prove that ar(∆ABD): ar(∆ADC) = m:n.

Given: D divides the side BC of ∆ABC in the ratio m:n


Area (Δ ABD) = 1/2 × BD × AL


Area (Δ ADC) = 1/2 × CD × AL


Area (∆ABD): Area (∆ADC) = 1/2 × BD × AL: 1/2 × CD × AL


Area (∆ABD): Area (∆ADC) = BD: CD


Area (∆ABD): Area (∆ADC) = m: n (∵ BD:CD = m:n)


More from this chapter

All 83 →
18

Show that the diagonals of a ‖ gm divide into four triangles of equal area.

 19

In the given figure, BD ‖ CA, E is the midpoint of CA and BD = CA.

Prove that ar(∆ABC) = 2×ar(∆DBC).


 20

The given figure shows a pentagon ABCDE in which EG, drawn parallel to DA, meets BA produced at G and CF drawn parallel to DB meets AB produced at F.

Show that ar(pentagon ABCDE) = ar(∆DGF).


 22

In the give figure, X and Y are the midpoints of AC and AB respectively, QP ‖ BC and CYQ and BXP are straight lines. Prove that ar(∆ABP) = ar(∆ACQ).