In a trapezium ABCD, AB and DC are parallel with a perpendicular distance of $8 ext{ m}$ between them. The non-parallel sides are equal: $AD = BC = 10 ext{ m}$, and the shorter parallel side is $AB = 15 ext{ m}$. Find the perimeter of the trapezium ABCD.

Question

In a trapezium ABCD, AB and DC are parallel with a perpendicular distance of $8 	ext{ m}$ between them. The non-parallel sides are equal: $AD = BC = 10 	ext{ m}$, and the shorter parallel side is $AB = 15 	ext{ m}$. Find the perimeter of the trapezium ABCD.

Accepted Answer

**Solution:**

1. Given trapezium ABCD with $AB \parallel DC$, height $AE = BF = 8 \text{ m}$, and equal non-parallel sides $AD = BC = 10 \text{ m}$. The shorter parallel side is $AB = 15 \text{ m}$.

2. Draw perpendiculars from points A and B to side DC, meeting at points E and F respectively. This forms two right triangles $\triangle ADE$ and $\triangle BCF$ and a rectangle $ABFE$.

3. Use the Pythagorean theorem in $\triangle ADE$:
   $$
   AD^2 = AE^2 + DE^2 \implies 10^2 = 8^2 + DE^2 \implies 100 = 64 + DE^2 \implies DE^2 = 36 \implies DE = 6 \text{ m}
   $$

4. Since trapezium is isosceles, $FC = DE = 6 \text{ m}$.

5. Length of longer parallel side $DC$ is sum of $DE + EF + FC$ where $EF = AB = 15 \text{ m}$:
   $$
   DC = 6 + 15 + 6 = 27 \text{ m}
   $$

6. Calculate perimeter $P$:
   $$
   P = AB + BC + CD + DA = 15 + 10 + 27 + 10 = 62 \text{ m}
   $$

**Answer:** The perimeter of trapezium ABCD is $62 \text{ m}$.

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