Question

In a trapezium ABCD, as shown, AB ‖ DC, AD = DC = BC = 24 cm and ∠A = 30°. Find: length of AB

Answer 1

Construction: Draw DP ⊥ AB and CM ⊥ AB

a. In right ΔADP,

cos30° = `"AP"/"AD"`

⇒ `sqrt(3)/(2) = "AP"/(24)`
⇒ AP = `12sqrt(13)"cm"`
Similarly, from right ΔBCM, we have MB = `12sqrt(3)"cm"`
Now, in rectangle PMCD, we have CD = PM = 24cm
∴ Length of AB
= AP + PM + MB
= `12sqrt(3) + 24 + 12sqrt(3)`
= `24(sqrt(3) + 1)"cm"`.
b. In right ΔADP,

sin30° = `"PD"/"AD"`

⇒ `(1)/(2) = "PD"/(24)`
⇒ PD = 12cm
Similarly, from right ΔBCM, we have MB = `12sqrt(3)"cm"`
Now, in rectangle PMCD, we have CD = PM = 24cm
∴ Length of AB
= AP + PM + MB
= `12sqrt(3) + 24 + 12sqrt(3)`
= `24(sqrt(3) + 1)"cm"`.