Past papers › JAMB 1998 · UME › Question 27 Question JAMB General Maths 1998 Objective Plane mensuration Plane mensuration
In the figure, P Q S T PQST P QS T T S R TSR T S R T S = 10 TS = 10 T S = 10 S R = 8 SR = 8 S R = 8 △ Q R S \triangle QRS △ QR S 20 cm 2 20\text{ cm}^2 20 cm 2 P Q R T PQRT P QR T
A 35 cm 2 35\text{ cm}^2 35 cm 2 B 65 cm 2 65\text{ cm}^2 65 cm 2 C 70 cm 2 70\text{ cm}^2 70 cm 2 D 140 cm 2 140\text{ cm}^2 140 cm 2
Worked solution (try it first) △ Q R S \triangle QRS △ QR S has base
S R = 8 SR = 8 S R = 8 cm:
1 2 × 8 × h = 20 \frac12 \times 8 \times h = 20 2 1 × 8 × h = 20 , so
h = 5 h = 5 h = 5 cm.
This is also the height of the trapezium.
P Q S T PQST P QS T is a parallelogram, so
P Q = T S = 10 PQ = TS = 10 P Q = T S = 10 cm.
The other parallel side is
T R = 10 + 8 = 18 TR = 10 + 8 = 18 T R = 10 + 8 = 18 cm.
Area of the trapezium
= 1 2 ( 10 + 18 ) × 5 = \frac12(10 + 18) \times 5 = 2 1 ( 10 + 18 ) × 5 = 70 cm 2 = 70\text{ cm}^2 = 70 cm 2 , option C.
Watch out
Keep the 1 2 \frac12 2 1 ( 10 + 18 ) × 5 = 140 cm 2 (10 + 18) \times 5 = 140\text{ cm}^2 ( 10 + 18 ) × 5 = 140 cm 2 Also set as JAMB 2016 · UTME · Q30
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