In triangle ABC the angle at B is a right angle, AB = 9 cm and AC = 41 cm. Find the area of the triangle in ${\text{cm}}^2$.

The right angle is at B, so AC is the hypotenuse and BC = √$(41^2-9^2)=$ √(1681 − 81) = √1600 = 40 cm. The two sides enclosing the right angle are AB and BC, so the area = ½ × $9\times 40=180{\text{cm}}^2$. (Multiplying the two lengths actually printed in the question, ½ × $9\times 41=184.5{\text{cm}}^2$, treats the hypotenuse as a leg; ½ × $40\times 41=820{\text{cm}}^2$ makes the same slip the other way; $360{\text{cm}}^2$ drops the ½.)