Question

in trapezium abcd, as shown in the figure, ab is parallel to dc, ad=dc=bc=20cm and ‹a=60°. find: (i) length of ab (ii) distance between ab and dc​

Answer 1

AB = 40 cm

Distance =
`10√(3)` cm

Explanation:

Draw two heights DE and CF. These heights together with bases of trapezoid form rectangle DEFC. In this rectangle, DC=EF=20 cm.

Consider right triangle ADE. In this triangle, angle ADE has the measure of 30° (the sum of the measures of all interior angles is always 180°, so m∠ADE=180°-60°-90°=30°)

Leg opposite to 30° angle is equal to half of the hypotenuse, so

`AE=(1)/(2)AD=(1)/(2)\cdot 20=10\ cm`

Similarly, in triangle CBF, BF=10 cm.

Hence,

`AB=AE+EF+FB=10+20+10=40\ cm`

The height of the trapezoid (the second leg of triangle ADE) is the distance between DC and AB. By the Pythagorean theorem,

`AD^2=AE^2+DE^2\\ \\DE^2=20^2-10^2=400-100=300\\ \\DE=√(300)=10√(3)\ cm`