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

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Answer 2

The length of segments between the point P and the vertices of the greater base is 7$\frac{5}{7}$ unit and 18 unit.

Step-by-step explanation:

Referring to the figure attached below, we have

ABCD is a trapezoid where AB = 11 unit is the greater base and CD = 18 unit is the smaller base  

AD = 3 unit and BC = 7 unit, are the legs of the trapezoid ABCD

Let both AD and BC extend and meet at point P.

Now, consider in ∆ PDC and ∆ PAB,

∠PDC = ∠PAB …….. [corresponding angles because DC // AB]

∠ P = ∠ P …….. [common angle]

∴ By AA similarity, ∆ PDC ~ ∆ PAB

Since the corresponding sides of two similar triangles are proportional to each other.

∴ PD/PA = DC/AB = PC/PB

1. PD/PA = DC/AB

⇒ PD/(PD+3) = 11/18

⇒ 18PD = 11(PD + 3)

⇒ 18PD – 11PD = 33

⇒ 7PD = 33

⇒ PD = 33/7  

And,

2. DC/AB = PC/PB

⇒ 11/18 = PC/(PC + 7)

⇒ 11PC + 77 = 18PC

⇒ 7 PC = 77  

⇒ PC = 11  

Thus,  

The length of segments between the point P and the vertices of the greater base, i.e., PA and PB are given by,

PA =  PD + DA = $\frac{33}{7} + 3 = \frac{33 + 21}{7} = \frac{54}{7}$ = $7\frac{5}{7}$ unit

And,  

PB = PC + CB = 11 + 7 = 18 unit

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