Home » The area of trapezium PQRS is 60(cm^|2|) PQ // RS, /PQ/ = 15 cm

# The area of trapezium PQRS is 60(cm^|2|) PQ // RS, /PQ/ = 15 cm

(a) The area of trapezium PQRS is 60(cm^{2}). PQ // RS, /PQ/ = 15 cm, /RS/ = 25 cm and < PSR = 60°. Calculate the: (i) perpendicular height of PQRS; (ii) |PS|.

(b) Ade received (frac{3}{5}) of a sum of money, Nelly (frac{1}{3}) of the remainder while Austin took the rest. If Austin’s share is greater than Nelly’s share by N3,000, how much did Ade get?

##### Explanation

(a) (theta = frac{60°}{2} = 30°)

Area of trapezium = (frac{a + b}{2} h)

(60 = frac{15 + 25}{2} h implies 60 = 20h)

(h = frac{60}{20} = 3 cm)

(ii) Considering (Delta PSR),

(sin theta = frac{Opp}{Hyp})

(sin 30 = frac{3}{|PS|})

(|PS| = frac{3}{sin 30})

= (frac{3}{0.5} = 6 cm)

(b) Let the total amount to be shared = k.

Ade’s share = (frac{3}{5} k )

Remainder = (k – frac{3}{5}k )

= (frac{2}{5}k)

Nelly’s share : (frac{1}{3} times frac{2}{5}k)

= (frac{2}{15}k)

Austin’s share : (frac{2}{5}k – frac{2}{15}k)

= (frac{4}{15}k)

(frac{4}{15}k – frac{2}{15}k = frac{2}{15}k)

(frac{2}{15}k = N3,000)

(implies k = frac{3,000 times 15}{2})

= (N22,500)