Distance between AB and DC is 17-32 cm-If true then write 1 and false then write 0-

Question

Toppr Admin · Accepted Answer

Given- trapezium ABCD- AD-20- -x2220-A-60-x2218-Draw perpendicular from D on AB- which meets AB at ENow- In -x25B3-ADE-sin60-x2218-DEAD-x221A-32-DEADDE-10-x221A-3DE-17-32cm

Distance between AB and DC is 17.32 cm.
If true then write 1 and if false then write 0.

Given, trapezium ABCD, $A D = 20$ , $∠ A = 60^{\circ}$
Draw perpendicular from D on AB, which meets AB at E
Now, In $△ A D E$ ,
$sin 60^{\circ} = \frac{D E}{A D}$
$\frac{\sqrt{3}}{2} = \frac{D E}{A D}$
$D E = 10 \sqrt{3}$
$D E = 17.32 c m$

Distance between AB and DC is 17.32 cm.If true then write 1 and if false then write 0.

Given, trapezium ABCD, AD=20, ∠A=60∘Draw perpendicular from D on AB, which meets AB at ENow, In △ADE,sin60∘=DEAD√32=DEADDE=10√3DE=17.32cm

Distance between AB and DC is 17.32 cm.
If true then write 1 and if false then write 0.

Given, trapezium ABCD, $A D = 20$ , $∠ A = 60^{\circ}$
Draw perpendicular from D on AB, which meets AB at E
Now, In $△ A D E$ ,
$sin 60^{\circ} = \frac{D E}{A D}$
$\frac{\sqrt{3}}{2} = \frac{D E}{A D}$
$D E = 10 \sqrt{3}$
$D E = 17.32 c m$