mensuration class 8

Triangle: Triangle is a closed figure formed by three line segments.

Quadrilateral: Quadrilateral is a closed figure formed by four line segments.

Triangles and quadrilaterals are called 2-D figures. 2 dimensional figures have only length and breadth.

Formulae:

1 Perimeter of a triangle=Sum of lengths of all sides.

2 Area of a triangle=(½)base×height sq.unis.

3 Perimeter of a rectangle=2(length+breadth)=2(l+b) units.

4 Area of rectangle= length×breadth=l×b sq.units

5 Perimeter of a squre= 4 ×side units.

6 Area of squre= side ×side=side² sq.units.

7 Area of parallelogram=base ×height sq.units

8 Circumference of circle= 2πr= πd units.

here r=radius of circle, d= diameter of circle.

9 Area of circle=πr² sq.units

r=radius of circle.

Exercise 11.1

1 A square and a rectangular field with measurements as given in the figure (side of squre is 60m and length of rectangle is 80 m) have the same perimeter. which field has a larger area?

Solution: Given length of square=60m

Perimeter of squre=4 ×side=4 ×60=240 m

And Given that length of rectangle =l=80m

According to problem

perimeter of square= perimeter of rectangle.

4 ×side=2(l+b)

240=2(80+b)

On dividing both sides by 2,

120=80+b

On transposing 80 to L.H.S

120-80=b

b=breadth of rectangle=40.

Area of square=side²=(60)²=3600 m²

Area of rectangle=l × b=80 ×40=3200 m²

∴ Area of square field is larger than area of rectangular field.

2 Mrs. Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of Rs 55 per m².

Solution: Area of the square plot=25 ×25 =625 m²

Area of the house=15 m×20m=300 m²

Area of the remaining portion=Area of the square plot-Area of the house=625-300=325 m²

Given cost of developing the garden around the house is Rs 55 per m².

Total cost of the developing the garden of area 325 m²=55 × 325=Rs 17,875

3 The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of the garden. [length of a rectangle is 20 -(3.5 + 3.5)metre].

Solution: Length of rectangle=[120-(3.5+3.5)] m=13m {∵semi circle radius=3.5m}

Circumference of 1 semi-circular part=πr=

$\frac{22}{7} \times 3.5 = 11 m$

Circumference of both semi-circular parts=(2 ×11) m=22m

Perimeter of the garden =AB+length of semi circle +CD+length of semi circle

=13 +22+13=48 m.

Area of garden =Area of rectangle+ 2 × area of semi circulular regions.

$= [(13 \times 7) + 2 \times \frac{1}{2} \times \frac{22}{7} \times (3.5) ²] m ²$

=(91+38.5) m²

=129.5 m²

4 A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many tiles are required to cover the floor of area 1080 m²? (If required you can split the tiles in whatever way you want to fill up the corners)?

Solution: Area of parallelogram = Base ×height

Area of one tile=(Area of the floor)÷(Area of each tile)

=1080 m²÷240 cm²

$= \frac{(1080 \times 10000) c m ²}{240 c m ²} = 45000 t i l e s$

{ ∵ 1m=100cm}

∴ 45000 tiles are required to cover a floor of area 1080 m².

5 An ant is moving around a few food pieces of different shapes scattered on the floor. For which food piece would the ant have to take a longer round? Remember circumference of a circle can be obtained by using the expression c=2πr, where r is the radius of the circle.

Solution: a) Radius of semi-circular part=

$\frac{2.8}{2} c m = 1.4 c m$

Perimeter of given figure= 2.8 cm+πr

$= 2.8 + (\frac{22}{7} \times 1.4)$

=2.8 +4.4= 7.2 cm

b) Perimeter of semi circle=πr

$\frac{22}{7} \times \frac{2.8}{2} = \frac{22}{7} \times 1.4 = 4.4 c m$

Perimeter of given figure=1.5+2.8+1.5+perimeter of semi circle=1.5+2.8+1.5+4.4=10.2 cm

C) Perimeter of semi circle= πr

$\frac{22}{7} \times \frac{2.8}{2} = \frac{22}{7} \times 1.4 = 4.4 c m$

Perimeter of the figure=2 cm+perimeter of semi circle+2 cm=2+4.4+2=8.4 cm.

∴ The ant will take a longer round for the food-piece b), because the perimeter of the figure b) is greater than a) and c)

Exercise 11.2

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 find the perpendicular distance between them is 0.8 m.

Solution: Area of trapezium=(½)(Sum of parallel sides) × (Distance between parallel sides)

=[(½)(1+1.2)(0.8)] m²= 0.88 m²

2 The area of a trapezium is 34 cm² and the length of one of the parallel side is 10 cm and its height is 4 cm. Find the length of the other parallel side.

Solution: Given that area of trapezium=34 cm² and height=4 cm

Let the length of one parallel side=a.

Area of trapezium=½ × (sum of parallel sides) × (Distance between parallel sides)

34 =½(10 + a) ×4

34=2(10+a)

17 =10+a

a=17-10=7 cm.

∴ Length of the other parallel side =7cm.

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

Solution:

Given length of the trapezium ABCD=AB+BC+CD+AD

120=AB+48+17+40

120=AB+105

120-105=AB

15=AB.

From figure AB is perpendicular distance between parallel sides.

Area of the field ABCD=½(AD+BC) × AB

=[½(40+48) ×15] m²

=(½ ×88 × 15) m²

=660 m²

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

Solution: Given that length of the diagonal=d=24 m

Let length of the perpendicular h₁=8m, h₂=13m

Area of Quadrilateral =(½)d(h₁+h₂)

=½(24) × (8+13)

=12 ×21=252 m²

∴ Area of the field =252 m²

5 the diagonals of a rhombus are 7.5 cm and 12 cm find its area.

Solution: Area of rhombus=½(Product of diagonals)

⇒Area of rhombus=½ × 7.5 ×12=7.5 × 6=45 cm²

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

Solution: let the length of the other diagonal of the rhombus b x is a special case of parallelogram

Area of a parallelogram= base × height

Area of rhombus= base × height=6 × 4 =24 cm²

Area of Rhombus=½( Product of diagonals)

⇒24 =½(8 × x)

⇒x=(²⁴ˣ²/₈) = 6 cm.

∴ The length of the other side of rhombus=6 cm.

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 Rs 4.

Solution: Area of rhombus=½(Product of diagonals)

Area of each tile=(½ ×45 × 30) =45 ×15=675 cm².

Area of 3000 tiles= (675 ×3000)=2025000 cm²=202.5 m² { ∵ 1m=100cm ⇒1m²=10000 cm²}

Cost of polishing 202.5 m² area= Rs(4 ×202.5)= Rs 810.

∴ Cost of polishing the floor = Rs 810.

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 100 m, find the length of the side along the river.

Solution: let the length of the field along the road b ‘l’ m. Then length of the field along the river will be ‘2l’ m.

Area of trapezium=½(sum of parallel sides × Distance between the parallel sides)

⇒ 10500 =½(l+2l) ×(100)

⇒ 10500=3l ×50

⇒3l=210

⇒l=70 m.

∴ length of the field along the river=2l=2 ×70=140 m.

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 :

Side of regular octagon=5 cm

Area of trapezium ABCH= Area of trapezium DEFG.

Area of trapezium ABCH=[½(4)(11+5)] m²=(½ ×4 ×16) m²= 32 m².

Area of rectangle HGDC=11 × 5 =55 m².

Area of octogon= Area of trapezium ABCH+ Area of trapezium DEFG+ Area of rectangle HGDC.

=32+32+55=119 m²

10 Ther 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 the park using both ways. Can you suggest some other way of finding its area?

Solution : Jyoti’s way of finding area is

Area of Pentagon=2(Area of trapezium ABCF)=[2 ×½ (15+30)(¹⁵/₂)] =½ ×45 ×15= 337.5 m².

Kavita’s way of finding area is

Area of Pentagon= Area of ∆ABE+Area of square BCDE

=[½ ×15 ×(30-15)+(15)²]

=(½ ×15 ×15+225)

=(112.5+225)

=337.5 m².

11 Diagram of the adjacent picture frame has outer dimensions is 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: In given figure 4 trapezium s are there. Clearly opposite trapezium s areas are equal.

Perpendicular distance between EF and CD =½(28-20)=4

Area of trapezium CDFE= Area of trapezium ABHG=½ ×4 ×(16+24)=2 ×40=80 cm²

Given width of each section is same

∴ perpendicular distance between HF and BD =4 cm .

Area of trapezium DFHB= Area of trapezium AGEC=½ ×4 ×(28+20)=2 ×48=96 cm²