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
Explanation
Video Explanation
No video available
Post your Contribution
Discussions (11)
Can someone please explain the first line of the solution. I don't understand how the height of the triangle inscribed in the trapezium became 12
Change your perspective . 12 is correct, the height is any point on the line of Q touching any point on the line PS and if you drag the line around to the the point p it will still be the same height -12cm
in (a) you said area of PQR = 1/2*b*h taking 12cm as the base and area is given.
In my view it should be A(PQR) = 1/2*a*B*sinQ, where a=PQ and can be obtained using pythagoras. Please check
