In the diagram, \(\overline{PQ//RS}\) is a trapezium with QR//PS. U and T are points on \(\overline{PS}\) such that \(\overline{|PU|}\) = 5 cm, \(\overline{...

In the diagram, \(\overline{PQ//RS}\) is a trapezium with QR//PS. U and T are points on \(\overline{PS}\) such that \(\overline{|PU|}\) = 5 cm,

\(\overline{|QU|}\) = 12 cm and ?PUQ= ?STR =90°. If the area of PQR = 20 cm\(^2\),

calculate, correct to the nearest whole number, the:

(a) perimeter; (b) area; of the trapezium

Answer Details

(a) Perimeter: We can calculate the perimeter of a trapezium by adding up the lengths of all four sides. If we call the length of side PR "x", then the length of side QR is 5 cm + x. The length of side PS is 12 cm + 5 cm = 17 cm, and the length of side QS is 12 cm.

So the perimeter is: x + 5 + 17 + 12 + x = 2x + 34 cm

(b) Area: To find the area of the trapezium, we first need to find the height. We can use the Pythagorean theorem to find the height of the trapezium.

Let h be the height of the trapezium. Then, h² = PR² - PU² = x² - 25

And, h² = QR² - QU² = (5 + x)² - 144

So, x² - 25 = (5 + x)² - 144

Expanding both sides: x² - 25 = x² + 10x + 25 - 144

10x = -144 + 25 - 25 = -144

x = -14.4 cm

Since x has to be positive, the height of the trapezium is not possible. So, the problem has no solution.

Explanation: In this problem, we are given a trapezium PQRS with parallel sides PQ and RS, and we are asked to find the perimeter and area of the trapezium. To find the perimeter, we simply add up the lengths of all four sides. To find the area, we need to find the height of the trapezium, which we can do using the Pythagorean theorem. However, in this case, the height of the trapezium is not possible, so the problem has no solution.