Question

In a trapezoid the lengths of bases are 11 and 18. The lengths of legs are 3 and 7. The extensions of the legs meet at some point. Find the length of segments between this point and the vertices of the greater base.

Answer 1

Answer: The length of segments between this point and the vertices of greater base are
`7(5)/(7)` and 18.

Explanation:

Let ABCD is the trapezoid, ( shown in below diagram)

In which AB is the greater base and AB = 18 DC= 11, AD= 3 and BC = 7

Let P is the point where The extended legs meet,

So, according to the question, we have to find out : AP and BP

In Δ APB and Δ DPC,

∠ DPC ≅ ∠APB ( reflexive)

∠ PDC ≅ ∠ PAB ( By alternative interior angle theorem)

And, ∠ PCD ≅ ∠ PBA ( By alternative interior angle theorem)

Therefore, By AAA similarity postulate,

`\triangle APB\sim \triangle D PC`

Let, DP =x

⇒
`(3+x)/(18) = (x)/(11)`

⇒ 33 +11x = 18x

⇒ x = 33/7=
`4(5)/(7)`

Thus, PD=
`4(5)/(7)`

But, AP= PD + DA

AP=
`4(5)/(7)+3 =7(5)/(7)`

Now, let PC =y,

⇒
`(7+y)/(18) = (y)/(11)`

⇒ 77 + 11y = 18y

⇒ y = 77/7 = 11

Thus, PC= 11

But, PB= PC + CB

PB= 11+7 = 18