1. Tardigrade
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  4. Let ABCD be a trapezium, in which AB is parallel to CD, AB = 11, BC =4, CD = 6 and DA = 3. The distance between AB and CD is

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Q. Let $ABCD$ be a trapezium, in which $AB$ is parallel to $CD, AB = 11, BC =4, CD = 6$ and $DA = 3$. The distance between $AB$ and $CD$ is

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Solution:

$ABCD$ is a trapezium.

$AB$ is parallel to $CD$.

$AB = 11,BC = 4, CD = 6$ and $DA = 3$

Construct $CE$ is parallel to $DA$.

image

$\therefore CE = 3$

$BC = 4$

$BE = 5$

$\therefore \angle BCE $ is a right angled triangle.

$\therefore $ Area of $\Delta BCE = \frac{1}{2} EC \times BC$

$= \frac{1}{2} \times 3 \times 4 \,...(i)$

Also, area of

$\Delta BCE = \frac{1}{2} \times BE \times h$

$ = \frac{1}{2} \times 5 \times h \,...(ii)$

From Eqs. (i) and (ii),

$\frac{1}{2} \times 3 \times 4 = \frac{1}{2} \times 5 \times h$

$\Rightarrow h = 2.4 $