# Ncert Solutions For Class 8 Maths Ex 11.2

## Ncert Solutions For Class 8 Maths Chapter 11 Ex 11.2

1. The upper surface of a table is trapezium. Find the area if its parallel sides are 2 m and 2.4 m. Perpendicular distance between two parallel lines is 1 m.

Solution:

Area of trapezium = 12$\frac{1}{2}$ × (Sum of parallel sides) × (distance between parallel sides)

= [12×(1+1.4)×(1)]$[\frac{1}{2} \times (1 + 1.4) \times (1)]$

= 0.7 m2$m^{2}$

2. Area of trapezium is 50 cm2$cm^{2}$. Measurement of one of the parallel sides is 15 cm and distance between them is 5 cm. What is the measurement of other parallel side?

Solution:

Area of trapezium = 50 cm2$cm^{2}$

Height = 5 cm

Let the length of another side be c.

Area of trapezium = 12$\frac{1}{2}$ × (Sum of parallel sides) × (distance between parallel sides)

50 cm2$cm^{2}$ = [12×(15+c)×(5)]$[\frac{1}{2} \times (15 + c) \times (5)]$

100 cm = (15 cm + c cm) × 5 cm

20 cm = 15 + c

C = 5 cm

Therefore, the length of the other side is 5 cm.

3. Length of fence of a field which is trapezium shaped PQRS is 150 m. Length of SR = 50 m, QR = 15 m and PQ = 40 m, the area of the field. Sides PS and SR are perpendicular to each other.

Solution:

Length of fence = Perimeter of trapezium

= PQ + QR + RS + SP

150 m = 40 m + 15 m + 50 m + SP m

150 m = 105 m + SP

SP = 45 m

Area of field PQRS = (12$\frac{1}{2}$ × (PQ + SR ) × PS) m2$m^{2}$

= (12$\frac{1}{2}$ × (40 + 50 ) × 45) m2$m^{2}$

= (12$\frac{1}{2}$ × 90 × 45) m2$m^{2}$

= 2025 m2$m^{2}$

4. The diagonal of quadrilateral field is 26 m and the perpendiculars dropped on it from the remaining opposite vertices are 10 m and 15 m. What is the area of field?

Solution:

d = 26 m

Perpendiculars,

P1 = 10 m and P2 = 15 m

Area of quadrilateral = 12$\frac{1}{2}$ × d × (P1 + P2)

= 12$\frac{1}{2}$ × 26 m× (10 + 15) m

= 13 × 25 m2$m^{2}$

= 325 m2$m^{2}$

5. The length of diagonals of a rhombus are 8 cm and 14 cm. What is the area of the rhombus?

Solution:

Area of rhombus = 12$\frac{1}{2}$ × ( product of its diagonals)

= 12$\frac{1}{2}$ × (8 cm × 14 cm)

= 56 cm2$cm^{2}$

6. Find area of rhombus whose sides are 10 m. Its altitude is 6 m. One of its diagonals is 5 m long. What is the length of the other diagonal?

Solution:

Let the length of other diagonal be d2.

Area of rhombus = Base × Height

= 10 m × 6 m

= 60 m2$m^{2}$

Area of rhombus = 12$\frac{1}{2}$ × ( product of its diagonals)

60 = 12$\frac{1}{2}$ × (5 m × d2 m)

120 = 5 m × d2 m

d2 = 24 m

Therefore, the length of the other diagonal d2 is 24 m.

7. Floor of a house consists of 2000 tiles. These tiles are rhombus in shape and measure of its diagonals are 25 cm and 20 cm. What is the cost of polishing the floor if it costs Rs 5 per m2$m^{2}$?

Solution:

Area of rhombus = 12$\frac{1}{2}$ × ( product of its diagonals)

Area of a tile = (12$\frac{1}{2}$ × (25 cm × 20 cm)

= 250 cm2$cm^{2}$

Area of 2000 tiles = (2000 × 250) cm2$cm^{2}$

= 500000 cm2$cm^{2}$

= 50 m2$m^{2}$

Cost of polishing per m2$m^{2}$ = Rs 5

Cost of polishing 50 m2$m^{2}$ = Rs (5 × 50)

= Rs 250

Therefore, the cost of polishing the floor is Rs 250.

8. Rohan wants a field in the shape of trapezium. Its side along the pond is parallel to and double the side along the road. Area of the field is 10200 m2$m^{2}$. The perpendicular distance between two parallel sides is 200 m. What is the length of the side along the river?

Solution:

Length of the field along the road = a

Length of the field along the pond = 2a

Area of trapezium = 12$\frac{1}{2}$ × (Sum of parallel sides) × (distance between parallel sides)

10200 m2$m^{2}$ = 12$\frac{1}{2}$ × (a + 2a) × (200 m)

3a = (10200×2200$\frac{10200 \times 2}{200}$) m

3a = 102

a = 34 m

9. Top surface of a table is regular octagon in shape as shown in the figure. What is the area of octagonal surface table?

Solution:

Side of the regular octagon = 10 cm

Area of PQRW = Area of STUV

Area of PQRW = [12×(6)×(15+10)]m2$[\frac{1}{2} \times (6) \times (15 + 10)] m^{2}$

= [12×(6)×(25)]m2$[\frac{1}{2} \times (6) \times (25)] m^{2}$

= 75 m2$m^{2}$

Area of rectangle WRSV = 15 × 10

= 150 m2$m^{2}$

Area of octagon = Area of PQRW + Area of STUV + Rectangle WRSV

= (75 + 75 + 150) m2$m^{2}$

= 300 m2$m^{2}$

10. There is a café in the shape of pentagon as shown in the figure. To find the area of the café, Sita and Gita divided it into two parts. Find the area of café using both the ways. Can you think about any other way to find the area of café?

Solution:

Sita

Area of café = 2 (Area of trapezium PQRO)

= [2 × 12×(20+40)×(202)]m2$\frac{1}{2} \times (20 + 40) \times (\frac{20}{2}) ]m^{2}$

= (60 m) × (10 m)

= 600 m2$m^{2}$

Gita

Area of café = Area of ΔPQT$\Delta PQT$ + Area of square QRST

= [12×20×(4020)]$[\frac{1}{2} \times 20 \times (40-20)]$ + [202]$[20^{2}]$ m2$m^{2}$

= [200 + 400] m2$m^{2}$

= 600 m2$m^{2}$

11. Diagram of the adjacent picture has inner dimensions 20 cm × 24 cm and outer dimensions 26 cm × 30 cm. What will be the area of each section of frame if the width is same in every section?

Solution:

Width is same for every section as it is given.

NE = EA = FB = FH = GI = GJ = KD = DM

NH = NE + EF + FH

30 = NE + 24 + FH

NE + FH = 30 cm – 24 cm = 6 cm

NE = FH = 3 cm

Therefore, NE = EA = FB = FH = GI = GJ = KD = DM = 3 cm

Area of EQRF = Area of DGSP

= [12(24+30)(3)]$[\frac{1}{2} (24 + 30)(3)]$ m2$m^{2}$

= 81 m2$m^{2}$

Area of PDEQ = Area of GFRS