Question

abcd is a trapezium in which ab is parallel to dc. if p is the point of intersection of diagonals ac and bd such that area triangle dpc = 50 cm2 and area of triangle apb 32 cm2, then what is the area of the area of trapezium abcd ?

Answer 1

Given: ab c d is a tpzium  ab║dc.Point p is intersection of diagonals ac and b d such that Ar(d p c)=50 cm² and Ar(a p b)=32 cm²

To find: Area(Trpzm a b c d)

Solution:In Δ a p b  and Δ c p d,  a b║c d

    ∠1=∠2,[Alternate angle]

∠3=∠4,[Alternate angle]

∠5=∠6 [Vertically opposite angles]

   Δ a p b  ~ Δ c p d, [AA similarity]

As we know when triangles are similar , their corresponding sides are proportional.

$\frac{ap}{pc}=\frac{bp}{pd}=\frac{ab}{cd}$

Also,when triangles are similar, ratio of square of corresponding sides is equal to ratio of their areas.

$\frac{Ar(a p b)}{Ar(c p d)}=\frac{32}{50}$

$\frac{Ar(a p b)}{Ar(c p d)}=\frac{16}{25}$

$\frac{ap^{2}}{pc^{2}}=\frac{16}{25}$

$\frac{ap}{pc}=\frac{4}{5}$

Now consider, Δ a p d and Δ b p c

$\frac{ap}{pc}=\frac{bp}{pd}$  [proved above]

and , ∠ a p d =∠ Δ b p c   [ Vertically opposite angles]

Δ a p d ~ Δ b p c [ SAS similarity]

$\frac{Ar( a p d)}{Ar( b p c)}=\frac{ap^{2}}{pc^{2}}$  [when triangles are similar, ratio of square of corresponding sides is equal to ratio of their areas.]

$\frac{Ar( a p d)}{Ar( b p c)}=\frac{ap^{2}}{pc^{2}}=\frac{16}{25}$

As you can see the trpzm , the two triangles  Δa p d and Δ b p c are smaller in area than Δ a p b and Δ d p c.

So, Ar(Δa p d)= 16 cm²,Ar( Δb p c)=25 cm²

Ar[ Tr p z m ( a b c d)]=32+25+50+16=123 cm²

Answer 2

Answer:

123 is answer dear friend