Question

In trapezium ABCD, as shown, AB || DC, AD = DC = BC = 20 cm and ∠A = ${60}^{\circ }$. Find:
(i) length of AB
(ii) distance between AB and DC

Answer 1

Constructing two perpendiculars to AB from the point D and C respectively.
Now, since AB||CD we have PMCD as a rectangle
Considering the figure,


(i) From right ∆ADP, we have
$\cos {60}^{\circ }=\frac{AP}{AD}\frac{1}{2}=\frac{AP}{20}$
AP = 10
Similarly,
In right ∆BMC, we have
BM = 10 cm
Now, from the rectangle PMCD we have
CD = PM = 20 cm
Therefore,
AB = AP + PM + MB
= 10 + 20 + 10
= 40 cm
(ii) Again, from the right ∆APD, we have
$\sin {60}^{\circ }=\frac{PD}{20}\frac{\sqrt{3}}{2}=\frac{PD}{20}PD=10\sqrt{3}$
Hence, the distance between AB and CD is $10\sqrt{3}$ cm.