Question

abcd is a trapezium in which ab parallel to dc. a line through a parallel to bc meets diagonal bd at p. if ar(∆bpc) = 5cm^2, then ar(∆abd) is equal to?​

Answer 1

abcd is a trapezium in which ab parallel to dc. a line through a parallel to bc meets diagonal bd at p

Given,

ar(∆bpc) = 5cm^2

Consider the attached figure, while going through the following steps:

The diagonal of a trapezium divides the trapezium into 2 equal triangles.

∴ Δ bcd = Δ abd

as, dp = pb

line pc cuts the Δ bcd into 2 equal triangles Δ bpc and Δ pdc

Δ bpc = Δ pdc

ar(∆ bpc) = ar(∆ pdc) = 5 cm^2

Now consider in ∆bcd,

∆ bcd = ∆ bpc + ∆ pdc

⇒ ar(∆ bcd) = ar(∆ bpc) + ar(∆ pdc)

= 2 ×  ar(∆bpc)

= 2 × 5

∴ ar(∆ bcd) = 10 cm^2

Since we have proved that,

Δ bcd = Δ abd

⇒ ar(∆ bcd)  = ar(∆ abd)

∴ ar(∆ abd) = 10 cm^2