Wo parallel sides of an isosceles trapezium are 31 cm and 15 cm. Its non-parallel sides are each 17 cm. Find the area of the trapezium.

To find the area of the isosceles trapezium, you can use the formula: Area = \( \frac{1}{2} \times (b_1 + b_2) \times h \) where \( b_1 \) and \( b_2 \) are the lengths of the parallel sides, and \( h \) is the height of the trapezium. First, let's determine the height (h). Since the trapezium is isosceles, we can drop perpendiculars from the top vertices to the line containing the bottom base, creating two right triangles. Let the length of the top base be \( b_1 = 15 \) cm and the bottom base \( b_2 = 31 \) cm. The difference in the lengths of the two bases is \( 31 - 15 = 16 \) cm. Because it's isosceles, the horizontal segments formed at the top from the vertices will each be half of this difference, so each of these segments is \( \frac{16}{2} = 8 \) cm. Now, we can apply the Pythagorean theorem in one of the right triangles formed. Let the length of the non-parallel side be \( 17 \) cm and one leg be \( 8 \) cm. The height \( h \) is our other leg: \[ h^2 + 8^2 = 17^2 \] \[ h^2 + 64 = 289 \] \[ h^2 = 225 \] \[ h = 15 \text{ cm} \] Now plug in the height into the area formula: Area = \( \frac{1}{2} \times (31 + 15) \times 15 \) Area = \( \frac{1}{2} \times 46 \times 15 \) Area = \( 23 \times 15 = 345 \text{ cm}^2 \) So, the area of the trapezium is **345 cm²**.