Mathematics Multiple Choice Questions & Answers on “Mensuration – Area of General Quadrilaterals”.
1. Find the area of the given quadrilateral.
a) 26cm2
b) 27cm2
c) 29cm2
d) 25cm2
Answer: b
Clarification: Area of general quadrilateral = (frac{1}{2}) × (h1 + h2) × d [d – diagonal of quadrilateral]
= (frac{1}{2}) × (4 + 2) × 9
= 27cm2.
2. Find the height of the quadrilateral if the area is 144cm2.
a) 24 cm
b) 12 cm
c) 22 cm
d) 11 cm
Answer: a
Clarification: Area of general quadrilateral = (frac{1}{2}) × (h1 + h2) × d (d – diagonal of quadrilateral)
⇒ 144 = (frac{1}{2}) × (8 + 4) × d
⇒ d = 24 cm.
3. The area of quadrilateral PQRS is 228cm2. Find length of QA.
a) 16 cm
b) 9 cm
c) 8 cm
d) 18 cm
Answer: d
Clarification: Area of general quadrilateral = (frac{1}{2}) × (h1 + h2) × d (d – diagonal of quadrilateral)
⇒ 228 = (frac{1}{2}) × (h1 + 20) × 12
⇒ l(QA) = 18 cm.
4. Find the area of the shaded region. The length of the diagonal DB is 8 cm. The perpendicular distances from the vertices to the diagonal DB are 4 cm and 3 cm. The length of segment BC is 6 cm. (π = 3.14)
a) 21.87cm2
b) 9.87cm2
c) 13.87cm2
d) 17.87cm2
Answer: c
Clarification: Area of shaded region = Area of general quadrilateral – Area of semicircle
= (frac{1}{2}) × (h1 + h2) × d – (frac{pi}{2}) × (r)2 = (frac{1}{2}) × (4 + 3) × 8 – (frac{pi}{2}) × (3)2 = 13.87cm2.
5. Find the area of pentagon PQRSA. The perpendicular distances from the vertices to the diagonal PR are 5 cm and 2 cm.
Given: l(PR) = 10 cm, l(PA) = 8 cm, l(SA) = 7cm.
a) 7cm2
b) 6cm2
c) 8cm2
d) 5cm2
Answer: a
Clarification: Area of pentagon PQRSA = Area of Quadrilateral PQRS – Area of Triangle PAS
= (frac{1}{2}) × (h1 + h2) × d – (frac{1}{2}) × b × h = (frac{1}{2}) × (5 + 2) × 10 – (frac{1}{2}) × 8 × 7 = 7cm2.