Advertisements
Advertisements
Question
In trapezium ABCD, as shown, AB // DC, AD = DC = BC = 20 cm and ∠ A = 60°. Find: length of AB
Advertisements
Solution
First, draw two perpendiculars to AB from point D and C respectively. Since AB || CD therefore PMCD will be a rectangle.
Consider the figure,
From right triangle ADP we have
cos 60° = `"AP"/"AD"`
`(1)/(2) = "AP"/(20)`
AP = 10
Similarly from the right triangle 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.
APPEARS IN
RELATED QUESTIONS
Find 'x', if :
Find 'x', if:
Find angle 'A' if :
Find angle 'x' if :
In trapezium ABCD, as shown, AB // DC, AD = DC = BC = 20 cm and A = 60°. Find: distance between AB and DC.
Find the length of AB.
In the given figure, AB and EC are parallel to each other. Sides AD and BC are 2 cm each and are perpendicular to AB.
Given that ∠ AED = 60° and ∠ ACD = 45°. Calculate: AB.
In the given figure, ∠B = 60°, AB = 16 cm and BC = 23 cm,
Calculate:
- BE
- AC
Find: AD
In right-angled triangle ABC; ∠ B = 90°. Find the magnitude of angle A, if: AB is √3 times of BC.
