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 $7\frac{5}{7}$ and 18.

Step-by-step 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

⇒ $\frac{3+x}{18} = \frac{x}{11}$

⇒  33 +11x = 18x

⇒ x = 33/7= $4\frac{5}{7}$

Thus, PD= $4\frac{5}{7}$

But, AP= PD + DA

AP= $4\frac{5}{7}+3 =7\frac{5}{7}$

Now, let PC =y,

⇒ $\frac{7+y}{18} = \frac{y}{11}$

⇒ 77 + 11y = 18y

⇒ y = 77/7 = 11

Thus, PC= 11

But, PB= PC + CB

PB= 11+7 = 18