Advertisements
Advertisements
प्रश्न
In trapezium ABCD, as shown, AB // DC, AD = DC = BC = 20 cm and ∠ A = 60°. Find: length of AB
Advertisements
उत्तर
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
संबंधित प्रश्न
Find ‘x’, if:
Find 'x', if:
Find angle 'x' if :
Find AD, if :
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: AC
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: AE.
Find: AD
Find: AC
Find AB and BC, if:
Find PQ, if AB = 150 m, ∠P = 30° and ∠Q = 45°.
.
