Mensuration - 8th math

Mensuration Class Eighth Math NCERT Exercise 11.2 Question (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 m² is ₹ 4.

Solution

Given,

Total number of rhombus shaped tiles the floor building have = 3000 tiles

One diagonal of tile = 45 cm

= 45/100 m

Thus, one diagonal (d₁) of tile = 0.45 m

And second diagonal of tile (d₂) = 30 cm

Thus, another diagonal (d₂) of tile = 0.3 m

Rate of polishing the floor = ₹ 4 per square meter

Thus, total cost of polishing of the floor = ?

Now, we know that, Area of a rhombus = 1/2 × product of diagonals

Thus, area of given one rhombus shaped tile = 1/2 × 0.45 m × 0.3 m

= 0.45 m × 0.15 m

= 0.0675 m²

Thus, area of one tile = 0.0675 m²

Thus, area of floor which consists total of 3000 tiles

= Area of one tile × 3000

= 0.0675 m² × 3000

= 202.5 m²

Now,

∵ Cost of polishing of 1 square meter = ₹ 4

∴ Cost of polishing of 202.5 m² = ₹4 × 202.5

= ₹ 810

Thus, cost of polishing of given floor = ₹ 810 Answer

Mensuration Class Eighth Math NCERT Exercise 11.2 Question (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 m² and the perpendicular distance between the two parallel sides is 100m, find the length of the side along the river.

Solution

Given, the area of trapezium shaped field = 10500 m²

Side along the river is twice the side along the road.

Perpendicular distance between parallel sides (h) = 100m

Then, length of the side along the river = ?

Let, the length of side along the road (a) = a

Therefore, according to question, the length of side along with river (b) = 2a

Now, we know that, Area of trapezium = 1/2 × (a + b) × h

Thus, area of given trapezium shaped field = 1/2 (a + 2a) 100m

⇒ 10500 = (a + 2a) 50

⇒ (a + 2a) 50 = 10500

⇒ 3a × 50 = 10500

⇒ 150 a = 10500

⇒ a = 70 m

Thus, side along with the river = 2a

= 2 × 70 m = 140 m

Thus, side of trapezium shaped field along with river = 140 m Answer

Mensuration Class Eighth Math NCERT Exercise 11.2 Question (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.

Solution

Given, Side of octagon shaped platform = 5m

And, AF = 11m

And HN = 4m

Thus, area of this octagon shaped platform = ?

Calculation of Area of rectangle ABEF formed in the octagon platform

Length AF of rectangle ABEF = 11m

And, width EF = 5m

Now, we know that, Area of a rectangle = Length × Width

Thus, area of rectangle ABEF = 11m × 5m

= 55 m²

Calculation of Area of trapezium AFGH

One parallel side AF (a) = 11m

And second parallel side HG (b) = 5m

And Perpendicular height between these parallel sides (h) = 4m

Now, we know that, Area of a trapezium = 1/2 × (a + b) × h

Thus, area of trapezium AFGH = 1/2 × (11m + 5m) 4m

= (11m + 5m) 2m

= 16m × 2m

Thus, area of AFGH = 32 m²

Now, since BCDE = AFGH

Thus, area of BCDE = 32 m²

Now, Total area of octagon = Area of AFGH + Area of BCDE + Area of ABEF

= 32 m² + 32 m² + 55 m²

= 119 m²

Thus, area of given octagon shaped platform = 119 m² Answer

Mensuration Class Eighth Math NCERT Exercise 11.2 Question (10) There is a pentagonal shaped park 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?

Solution

(a) Finding Area using Jyoti's Diagram

Given, AB = (a) = 15m

Thus, AF = FB = 15/2 = (h) = 7.5 m

And, AE = BC = 15m

DF = (b) = 30 m

Thus, Area of given pentagon shaped diagram = ?

Here, AFDE forms a trapezium.

Now, we know that Area of a trapezium = 1/2 × (a + b) h

Thus, area of trapezium AFDE = 1/2 (15m + 30m) 7.5m

= 22.5m × 7.5m

Thus, area of AFDE = 168.75 m²

Now, since AFDE = FBCD

Thus, area of FBCD = 16.75 m²

Thus, area of given pentagon = 2 × Area of AFDE

= 2 × 168.75 m²

= 337.50 m²

Thus, area of given pentagon = 337.50 m² Answer

(b) Finding Area using Kavita’s Diagram

Given, AE = BC = 15m

AB = EC = 15m

And, OD = 30m

Thus, DF = FC = 1/2 EC = 7.5m

And, FD = OD – OF

= 30m – 15m

⇒ FD = 15m

In square ABCE

We know that, Area of a square = side²

Here, Side of square ABCE = 15m

Thus, area of square ABCE = (15 m)²

Thus, area of square ABCE = 225 m²

Now, in triangle DEC

We know that, area of a triangle = 1/2 × Base × Height

Here, Base (EC) = 15m

And, Height (DF) = 15m

Thus, area of triangle DEC = 1/2 × 15m × 15m

= 1/2 × 225m²

= 112.5 m²

Thus, area of triangle DEC = 112.5 m²

Now, Area of given pentagon = Area of square ABCE + Area of triangle DEC

= 225 m² + 112.5 m²

= 337.50 m²

Thus, Area of given pentagon using Kavita's diagram = 337.50 m² Answer

Mensuration Class Eighth Math NCERT Exercise 11.2 Question (11) Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.

Solution

Given, AB = DC = 24cm

And, AD = BC = 28 cm

And, EF = HG = 16 cm

And, HE = FG = 20 cm

And width of each section is same.

Thus, area of each section =?

Now, since DC = 24cm

And, HG = 16 cm

Thus, DC – HG = PH + GM

= 24cm – 16 cm

PH + GM = 8cm

And now, since PH = GM [As per question width of each section are same]

Thus, PH + PH = 8cm

⇒ 2 PH = 8cm

Thus, PH = 8cm/2 = 4 cm

Thus, PH = 4 cm = GM

Similarly, BC = 28 cm

And, FG = 20cm

Thus, BC – FG = NF + GO

= 28cm – 20 cm = 8 cm

⇒ NF + GO = 8cm

Now, since with of each section are same

Thus, NF = GO

Thus, NF + NF = 8cm

⇒ 2 NF = 8cm

⇒ NF = 8cm/2 = 4cm

Thus, NF = GO = 8cm

Calculation of area of trapezium BCGF

Here, one parallel side BC = (a) = 28cm

And, other parallel side FG = (b) = 20cm

And, perpendicular distance GM (h) = 4cm

Now, we know that, Area of a trapezium = 1/2 (a + b) h

Thus, area of trapezium BCGF = 1/2 (28cm + 20cm) 4cm

= (28cm + 20cm) 2cm

= 48 cm × 2cm

= 96 cm²

Now, since trapezium AEHD = trapezium BCGF

Thus, area of trapezium AEHD = Area of trapezium BCGF

Thus, area of trapezium AEHD = 96 cm²

Calculation of Area of trapezium ABFE

Here, one parallel side AB = a = 24cm

And other parallel side EF = HG = b = 16 cm

And perpendicular height FN (h) = 4cm

Now, we know that, Area of a trapezium = 1/2 (a + b) h

Thus, area of trapezium ABFE = 1/2 (24cm + 16cm) 4cm

= (24 cm + 16cm) 2cm

= 40cm × 2cm

Thus, area of trapezium ABFE = 80 cm²

Now, since area of trapezium ABFE = area of trapezium DHGC

Thus, area of trapezium DHGC = 80 cm²

Thus, area of each trapezium are 96 cm², 80 cm², 96 cm² and 80 cm² respectively Answer