TTMUA Lab
TMUA 2022 · Paper 2 · Question 11 of 20

TMUA 2022 Paper 2 Question 11

Coordinate geometry — Kites · condition for a right angle. Try it first; the answer and a full worked solution are below.

TMUA 2022 · Paper 2Coordinate geometryKites · condition for a right angle5 options

The original question includes a diagram: A kite PQRS with diagonals meeting at O, where OP = OR = x, OS = z above O and OQ = y below O.

A kite PQRS has diagonals meeting at O, with OP = x,   OQ = y,   OR = x,   OS = z

Which of the following is necessary and sufficient for angle SPQ to be a right angle?

  1. Ax = y = z
  2. B2x = y + z
  3. Cx2 = yz
  4. Dy = z
  5. Ey2 = x2 + z2
Show the answer and worked solution
answer · C
  1. Ax = y = z
  2. B2x = y + z
  3. Cx2 = yz
  4. Dy = z
  5. Ey2 = x2 + z2
The diagonals of a kite cross at right angles, so triangles POS and POQ both have a right angle at O. Writing α for angle SPO and β for angle OPQ, we have tanα = zx and tanβ = yx. Angle SPQ = α + β is a right angle exactly when tanα = 1tanβ, that is zx = xy, which rearranges to x2 = yz.