This set of Class 8 Maths Chapter 11 Multiple Choice Questions & Answers (MCQs) focuses on “Mensuration – Area of General Quadrilaterals”.
1. Find the area of the given quadrilateral.
a) 26cm2
b) 27cm2
c) 29cm2
d) 25cm2
View Answer
Explanation: 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
View Answer
Explanation: 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
View Answer
Explanation: 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
View Answer
Explanation: 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
View Answer
Explanation: 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.
Sanfoundry Global Education & Learning Series – Mathematics – Class 8.
To practice all chapters and topics of class 8 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.
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