# Mensuration : Exercise 11.2 - Class 8 Maths - NCERT Solution

**Chapter 11 MENSURATION**

**Exercise 11.2**

**Q.1 The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.**

** Sol.** Given, one side of table = 1 m, second side of table = 1.2 m and height of table = 0.8 m

Therefore,

Area of top surface of table = 1/2 x (Sum of parallel sides) x (distance between parallel sides)

= [1/2 x (1 + 1.2) x (0.8)]m^{2Â }= 0.88 m^{2}

Hence, the area of the top surface of table is 0.88 m^{2}.

**Q.2 The area of a trapezium is 34 cm ^{2} and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.**

** Sol.** Let the other parallel side be

*b*.

Given, one parallel side of trapezium (*a*) = 10 cm and height (*h*) = 4 cm

Now, Area of trapezium = 1/2 x (Sum of parallel sides) x (distance between parallel sides)

Area of trapezium = 1/2 x (*a *+* b*) x (*h*)

34 =1/2 x (10 +* b*) x (4)

34 = 2 x (10 + *b*)

17 = 10 + *b*

Therefore, *b *= 17 â€“ 10 = 7 cm

Hence, the length of the other parallel side is 7 cm.

**Q.3 Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.**

** Sol.** Given, BC = 48 m, CD = 17 m, AD = 40 m and the length of trapezium ABCD = 120 m

Here, length of trapezium ABCD = AB + BC + CD + DA

120 m = AB + 48 m +17 m + 40 m

AB = 120 m â€“ 105 m = 15 m

Now, area of trapezium field ABCD = x (AD + BC) x AB

= [1/2 x (40 + 48) x (15)] m^{2}

=Â [1/2 x (88) x (15)] m^{2}

= 660 m^{2}

Hence, the area of the field is 660 m^{2}Â .

**Q.4 The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.**

** Sol.** Given, length of diagonal (

*d*) = 24 m and the perpendiculars from the opposite vertices to the diagonal are h

_{1 }= 8 m and h

_{2}= 13 m

Now, area of quadrilateral = 1/2 d(h_{1} +Â h_{2})

= 1/2 (24m)(8m + 13m)

=Â 1/2 (24m)(21m)

= 252 m^{2}

Hence, the area of the field is 252 m^{2}.

**Q.5 The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.**

** Sol.** Given, diagonals of rhombus are d

_{1}= 7.5 cm and d

_{2}= 12 cm.

Now, area of rhombus =Â 1/2 x d_{1}Â d_{2}

= 1/2 x 7.5 cm x 12 cm

= 45 cm^{2}

Hence, the area of rhombus is 45 cm^{2}.

**Q.6 Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.**

** Sol.** Given, base of rhombus = 6 cm, altitude = 4 cm and one diagonal d

_{1}= 8 cm.

Now, area of rhombus = base x altitude

= 6 cm x 4 cm = 24 cm^{2}

Also, area of rhombus =Â 1/2 x d_{1}Â d_{2}

24 cm^{2}= 1/2 x 8 cm xÂ d_{2}

d_{2 }= 24 x 2 / 8Â cm

d_{2 }= 6 cm

Hence, the length of the other diagonal is 6 cm.

**Q.7 The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m ^{2} is Rs 4.**

** Sol.** Given, diagonals of rhombus are

*d*= 45 cm and

_{1}*d*= 30 cm.

_{2}Now, area of rhombus =Â 1/2 x d_{1}Â d_{2}

= 1/2 x 45cm x 30 cm

= 675 cm^{2}

Therefore, area of 3000 tiles = (675 x 30000) cm^{2} = 2025000 cm^{2} = 202.5 m^{2}.

Given, the cost of polishing is Rs 4 per m^{2}.

Therefore, cost of polishing 202.5 m^{2} = Rs (4 x 202.5) = Rs 810

Hence, the total cost of polishing the floor is Rs 810.

**Q.8 Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m ^{2} and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.**

** Sol.** Given, perpendicular distance (

*h*) = 100 m, area of field = 10500 m

^{2}

Now, let the side along the river be *l* m and so the other side along the road will be 2*l* m.

Area of trapezium = 1/2 x (Sum of parallel sides) x (perpendicular distance between parallel sides)

10500 m^{2 }= 1/2 x (*l *+* 2l*) x (100 m)

*3lÂ *=Â (2 x 10500 / 100 ) = 210 m

l = 70 m

2l = 2 x 70 m = 140 m

Hence, the length of the field along the river is 140 m.

**Q.9 Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.**

** Sol.** The given octagon can be divided into three parts as shown in figure below:

From figure, area of part A & C will be same which resembles a trapezium. And part B is a rectangle.

Area of part A (Trapezium)= 1/2 x (Sum of parallel sides) x (distance between parallel sides)

= 1/2 x (11 + 5) x (4) = 32 m^{2}.

Area of part B (Rectangle) = length x breadth

= 11 x 5 = 55 m^{2}.

Thus, area of Octagon = 2 x area of part A + area of part B

= 2 x 32 m^{2} + 55 m^{2} = 119 m^{2}.

Hence, the area of the octagonal surface is 119 m^{2}.

**Q.10 There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways.**

**Find the area of this park using both ways. Can you suggest some other way of finding its area?**

** Sol.** Finding area by Jyotiâ€™s diagram:

Here, the pentagonal park is divided into two trapeziums.

Given, sides of trapezium are a = 15 m and b = 30 m, distance between two sides = 15/2 = 7.5 m

Therefore, area of pentagon = 2 x (Area of trapezium)

= 2 x [ 1/2 x (Sum of parallel sides) x (distance between parallel sides)]

= 2 x 1/2 x (15 + 20) x 15/2 m^{2}

= 337.5 m^{2}.

Finding area by Kavitaâ€™s diagram:

Here, the pentagonal park is divided into one triangle and one sqaure.

Therefore, area of pentagon = area of triangle + area of square

= [1/2 x base x height] + [(side)^{2}] m^{2}

= [1/2 x 15 x (30 â€“ 15)] + [(15)^{2}] m^{2}

= [1/2 x 15 x 15 + 225] m^{2}

= [112.5 + 225] m^{2}

= 337.5 m^{2}

Hence, the area of pentagonal shaped park is 337.5 m^{2}.

**Q.11 Diagram of the adjacent picture frame has outer dimensions =24 cm Ã— 28 cm and inner dimensions 16 cm Ã— 20 cm. Find the area of each section of the frame, if the width of each section is same.**

*Sol. *

Given, width of each section is same.

Therefore, area of figure I will be equal figure II and area of figure III will be equal to figure IV.

Now, area of figure I = 1/2 x (Sum of parallel sides) x (distance between parallel sides)

= [1/2 x (20 + 28) x (4)] cm^{2}

= [1/2 x (48) x (4)] cm^{2}

= 96 cm^{2}

Here, Area of figure II = area of figure I = 96 cm^{2}

Now, area of figure III = 1/2 x (Sum of parallel sides) x (distance between parallel sides)

= [1/2 x (24 + 16) x (4)] cm^{2}

= [1/2 x (40) x (4)] cm^{2}

= 80 cm^{2}

Area of figure III = area of figure IV = 80 cm^{2}