Class 8 Maths Chapter 9 Question Answers - Mensuration

Q1: The volume of a box is 13400 cm³. The area of its base is 670 cm². Find the height of the box.
Ans: Volume of the box = 13400 cm³
Area of the box = 670 cm²

Hence, the required height = 20 cm.

Q2: Two cubes are joined end to end. Find the volume of the resulting cuboid, if each side of the cubes is 6 cm.
Ans: 

Length of the resulting cuboid = 6 + 6 = 12 cm
Breadth = 6 cm
Height = 6 cm
Volume of the cuboid = l × b × h = 12 × 6 × 6 = 432 cm³.

Q3: MNOPQR is a hexagon of side 6 cm each. Find the area of the given hexagon in two different methods.
Ans: Method I: Divide the given hexagon into two similar trapezia by joining QN.
Area of the hexagon MNOPQR = 2 × area of trapezium MNQR
= 2 × 1/2 (6 + 11) × 4
= 17 × 4
= 68 cm²
Method II: The hexagon MNOPQR is divided into three parts, 2 similar triangles and 1 rectangle by joining MO, RP.


Q4: Find the area of the hexagon ABCDEF given below. Given that: AD = 8 cm, AJ = 6 cm, AI – 5 cm, AH = 3 cm, AG = 2.5 cm and FG, BH, EI and CJ are perpendiculars on diagonal AD from the vertices F, B, E and C respectively.

Ans:
Given:
AD = 8 cm
FG = 3 cm
AJ = 6 cm
EI = 4 cm
AI = 5 cm
BH = 3 cm
AH = 3 cm
CJ = 2 cm
AG = 2.5 cm

Area of hexagon ABCDEF = Area of ΔAGF + Area of trapezium FGIE + Area of ΔEID + Area of ΔCJD + Area of trapezium HBCJ + Area of ΔAHB
= 3.75 cm² + 8.75 cm² + 6 cm² + 2 cm² + 7.5 cm² + 4.5 cm²
= 32.50 cm².

Q5: The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².
Ans: Given: Diameter of the roller = 84 cm
Radius = 84/2 = 42 cm
Height = 120 cm
Curved surface area of the roller = 2πrh

Area covered by the roller in one complete revolution = 3.168 m²
Area covered in 500 complete revolutions = 500 × 3.168 = 1584 m²
Hence, the required area = 1584 m².

Q6: A rectangular metal sheet of length 44 cm and breadth 11 cm is folded along its length to form a cylinder. Find its volume.
Ans: Circumference of the base = 2πr

= 423.5 cm³
Hence, the required volume = 423.5 cm³.

Q7: 160 m3 of water is to be used to irrigate a rectangular field whose area is 800 m2. What will be the height of the water level in the field?
Ans: Volume of water = 160 m³
Area of rectangular field = 800 m²
Let h be the height of water level in the field.
Now, the volume of water = volume of cuboid formed on the field by water.
160 = Area of base × height = 800 × h
⇒ h = 0.2
So, required height = 0.2 m

Q8: The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.
Ans: Let ABCD be the trapezium such that
AB = 40 cm and CD 20 cm and AD = BC = 26 cm.
Mensuration NCERT Extra Questions for Class 8 Maths Q23
Now, draw CL || AD
Then ALCD is a parallelogram.
So AL = CD = 20 cm
and CL = AD = 26 cm.
In ΔCLB, we have CL = CB = 26 cm
Therefore, ΔCLB is an isosceles triangle.
Draw altitude CM of ΔCLB.
Since ΔCLB is an isosceles triangle. So, CM is also the median.
Then LM = MB = 1/2 BL = 1/2 × 20 cm = 10 cm

[as BL = AB – AL = (40 – 20) cm = 20 cm].
Applying Pythagoras theorem in ΔCLM, we have
CL² = CM² + LM²
26² = CM² + 10²
CM² = 26² – 10² = (26 – 10) (26 + 10) = 16 × 36 = 576
CM = √576 = 24 cm
Hence, the area of the trapezium = 12
 (sum of parallel sides) × height
= 1/2 (20 + 40) × 24
= 30 × 24
= 720 cm².