Answer: [tex]7\frac{5}{7}[/tex] unit and 18 unit
Step-by-step explanation:
Let ABCD is a trapezoid where AB and CD are the bases. ( In which AB is greatest base which shown in below figure)
AD and BC are the legs of the trapezoid ABCD.
Now, we have ( According to the question ),
AB = 18 unit, BC = 7 unit, AD = 3 unit and DC = 11 unit.
Here the leg AD extends from point D.
Similarly leg BC extends from point C.
Let they meet at point P ( shown in below diagram)
Since In triangles PAB and PDC,
∠PDC ≅ ∠PAB ( because DC ║ AB )
And, ∠ PAB ≅ ∠ PBA
∠DPC ≅ ∠ APB ( reflexive)
Therefore, By AAA similarity postulate,
[tex]\triangle PDC \sim \triangle PAB[/tex]
Thus, By the definition of similarity,
[tex]\frac{PD}{PA} = \frac{DC}{AB}[/tex]
[tex]\frac{PD}{PD+3} = \frac{11}{18}[/tex] ( because PA = PD+DA)
⇒ 18 PD = 11 PD +33
⇒7PD = 33
⇒ PD = 33/7
Again by the definition of similarity,
[tex]\frac{PC}{PB} = \frac{DC}{AB}[/tex]
[tex]\frac{PC}{PC+7} = \frac{11}{18}[/tex] ( because PB = PC + CB)
⇒ 18 PC = 11 PD +77
⇒7PC = 77
⇒ PC = 11
Thus, PA = PD+DA = 33/7 + 3 = [tex]7\frac{5}{7}[/tex]
And, PB = PC + CB = 11 + 7 = 18
