Lesson 8 Area and Volume
Exercise 8.1
Find the perimeter of the following triangles.
Triangle (a):Sides: 4 cm, 2 cm, 4.5 cm
Perimeter: 4 + 2 + 4.5 = 10.5 cm
Triangle (b):Sides: 3 cm, 6 cm, 6.5 cm
Perimeter: 3 + 6 + 6.5 = 15.5 cm
Triangle (c):Sides: 5 cm, 5 cm, 5√2 cm (right triangle)
Perimeter: 5 + 5 + 5√2 ≈ 18.07 cm
Triangle (d):Sides: AB = 3 cm, BC = 8 cm, AC = 6 cm
Perimeter: 3 + 8 + 6 = 17 cm
Triangle (e):Sides: 5 cm, 8 cm, 10 cm (right triangle)
Perimeter: 5 + 8 + 10 = 23 cm
Triangle (f):Sides: 4 ft, 3 ft, 5 ft (right triangle)
Perimeter: 4 + 3 + 5 = 12 ft
2. Find the area of following triangles:
(a)Solution:
In ∆EFG,
Base = 12 cm, Height = 4 cm
Area = ?
We know that,
Area = (1/2) × Base × Height
Area = (1/2) × 12 × 4 = 24 cm²
(b)Solution:
In ∆KLM,
Base = 6.4 cm + 3.6 cm = 10 cm, Height = 3.6 cm
Area = ?
We know that,
Area = (1/2) × Base × Height
Area = (1/2) × 10 × 3.6 = 18 cm²
(c)Solution:
In equilateral triangle PQR,
Side = 8 cm
Area = ?
We know that,
Area = (√3 / 4) × Side²
Area = (√3 / 4) × 8² ≈ 27.71 cm²
(d)Solution:
In equilateral triangle ABC,
Side = 5 cm
Area = ?
We know that,
Area = (√3 / 4) × Side²
Area = (√3 / 4) × 5² ≈ 10.83 cm²
(e)Solution:
In equilateral triangle XYZ,
Side = 12 cm
Area = ?
Using Heron's formula:
Semi-perimeter (s) = (12 + 12 + 12) / 2 = 18 cm
Area = √[s × (s - a) × (s - b) × (s - c)]
Area = √[18 × (18 - 12) × (18 - 12) × (18 - 12)] ≈ 62.35 cm²
(f)Solution:
In isosceles triangle PQR,
Sides = 10 cm, Base = 12 cm
Area = ?
Using Heron's formula:
Semi-perimeter (s) = (10 + 10 + 12) / 2 = 16 cm
Area = √[s × (s - a) × (s - b) × (s - c)]
Area = √[16 × (16 - 10) × (16 - 10) × (16 - 12)] = 48 cm²
(g)Solution:
Isosceles triangle with sides = 17 cm, Base = 16 cm,
Area = ?
We know that,
Height = √(17² - (16/2)²) = √(289 - 64) = √225 = 15 cm
Area = (1/2) × Base × Height
Area = (1/2) × 16 × 15 = 120 cm²
(h)Solution:
Isosceles triangle with sides = 20 cm, Base = 32 cm,
Area = ?
We know that,
Height = √(20² - (32/2)²) = √(400 - 256) = √144 = 12 cm
Area = (1/2) × Base × Height
Area = (1/2) × 32 × 12 = 192 cm²
(i)Solution:
In ∆XYZ,
Base = YZ = 8 ft, Height = 12 ft
Area = ?
We know that,
Area = (1/2) × Base × Height
Area = (1/2) × 8 × 12 = 48 ft²
(j)Solution:
In ∆STU,
Base = 18 cm, Height = 15 cm
Area = ?
We know that,
Area = (1/2) × Base × Height
Area = (1/2) × 18 × 15 = 135 cm²
(k)Solution:
In ∆ABC,
Base = 4 cm, Height = 3 cm
Area = ?
We know that,
Area = (1/2) × Base × Height
Area = (1/2) × 4 × 3 = 6 cm²
(l)Solution:
In ∆ABC,
Base = Height = 16 cm
Area = ?
We know that,
Area = (1/2) × Base × Height
Area = (1/2) × 16 × 16 = 128 cm²
3. Find the area of given quadrilaterals.
(a) RectangleGiven:
Length = 5 cm, Height = 3.6 cm
Area Formula:
Area = Length × Height
= 5 × 3.6 = 18 cm2
(b) SquareGiven:
Side = 1.4 ft
Area Formula:
Area = Side2
= 1.4 × 1.4 = 1.96 ft2
(c) ParallelogramGiven:
Base = 9 m, Height = 4 m
Area Formula:
Area = Base × Height
= 9 × 4 = 36 m2
(d) ParallelogramGiven:
Base = 11 mm, Height = 7 mm
Area Formula:
Area = Base × Height
= 11 × 7 = 77 mm2
(e) TrapeziumGiven:
Parallel sides: 4.5 cm and 7.5 cm, Height = 6 cm
Area Formula:
Area = 1/2 × (Sum of Parallel Sides) × Height
= 1/2 × (4.5 + 7.5) × 6
= 1/2 × 12 × 6 = 36 cm2
(f) TrapeziumGiven:
Parallel sides: 9 cm and 13 cm, Height = 9 cm
Area Formula:
Area = 1/2 × (Sum of Parallel Sides) × Height
= 1/2 × (9 + 13) × 9
= 1/2 × 22 × 9 = 99 cm2
(g) RhombusGiven:
Diagonals: 12 cm and 8 cm
Area Formula:
Area = 1/2 × (Product of Diagonals)
= 1/2 × (12 × 8)
= 1/2 × 96 = 48 cm2
(h) RhombusGiven:
Diagonals: 8 cm and 6 cm
Area Formula:
Area = 1/2 × (Product of Diagonals)
= 1/2 × (8 × 6)
= 1/2 × 48 = 24 cm2
(i) KiteGiven:
Diagonals: 3 cm and 5 cm
Area Formula:
Area = 1/2 × (Product of Diagonals)
= 1/2 × (3 × 5)
= 1/2 × 15 = 7.5 cm2
(j) KiteGiven:
WY= 20 cm + 12 cm = 32cm
XZ= 12 cm + 12 cm= 24cm
Area = ?
Area Formula:
Area = 1/2 × (Product of Diagonals)
Area = 1/2 × (20 × 12)
= 1/2 × 240 = 120 cm2
(k) KiteGiven:
WY = 20 cm + 12 cm = 32 cm
XZ = 12 cm + 12 cm = 24 cm
Area = ?
Area Formula:
Area = 1/2 × (Product of Diagonals)
= 1/2 × (32 × 24)
= 1/2 × 768 = 384 cm2
(k) Quadrilateral with Perpendicular DiagonalsGiven:
Diagonals: AC = 8 cm, BM = 4.5 cm, DN = 6.5 cm
Area Formula:
Area = 1/2 × AC × (BM + DN)
= 1/2 × 8 × (4.5 + 6.5)
= 1/2 × 8 × 11 = 44 cm2
(l) Rhombus or KiteGiven:
Diagonals: AC = 10 cm, BD = 8 cm
Area Formula:
Area = 1/2 × (Product of Diagonals)
= 1/2 × (10 × 8)
= 1/2 × 80 = 40 cm2
4. Find the area of following geometrical shape:
(a)
For Rectangle ABCD:
Length (l) = 18 cm
Breadth (b) = 12 cm
Area of Rectangle (A₁):
A₁ = l × b
A₁ = 18 × 12 = 216 cm²
For Triangle BCD,
BC= CD= 10 cm
BD = 12 cm
Area of Triangle (A₂):
A₂ = (1/2) × b × h
A₂ = (1/2) × 10 × 12 = 60 cm²
For isosceles triangle BCD:
Using Heron's Formula:
Semi-perimeter (s):
s = (a + b + c) / 2
s = (10 + 10 + 12) / 2 = 16 cm
Area of Triangle (A₂):
A₂ = √[s(s-a)(s-b)(s-c)]
A₂ = √[16(16-10)(16-10)(16-12)]
A₂ = √[16 × 6 × 6 × 4]
A₂ = √[2304]
A₂ = 48 cm²
Total Area of the Shape:
Total Area = A₁ + A₂
Total Area = 216 + 48 = 264 cm²
(b)
For the rectangle ABCD,
Length (AB) = 9 cm
Width (AD) = 5 cm
Area = Length × Width
= 9 cm × 5 cm
= 45 cm²
For the triangle MAD,
Base (MD) = 3 cm
Height (MA) = 4 cm
Areatriangle = 1/2 × Base × Height
= 1/2 × 3 cm × 4 cm
= 6 cm²
Total Area = 45 cm² + 6 cm²
= 51 cm²
(c)
1. Area of Triangle PQR:
Base (PQ) = 5 cm
Height (QR) = 12 cm
Area of PQR = 1/2 × Base × Height = 1/2 × 5 cm × 12 cm = 30 cm²
2. Area of Triangle PQS:
Base (PS) = 4 cm
Height (SQ) = 3 cm
Area of PQS = 1/2 × Base × Height = 1/2 × 4 cm × 3 cm = 6 cm²
Total Area:
Total Area = Area of PQR + Area of PQS = 30 cm² + 6 cm² = 36 cm²
(d)
For ΔPMO,
Base (PM) = 5 cm
Height (PO) = 6 cm
Area ΔPMO = 1/2 × Base × Height = 1/2 × 5 cm × 6 cm = 15 cm²
For ΔMON,
Base (MN) = 7 cm
Height (ON) = 4 cm
Area ΔMON = 1/2 × Base × Height = 1/2 × 7 cm × 4 cm = 14 cm²
Total Area PMON = Area ΔPMO + Area ΔMON = 15 cm² + 14 cm² = 29 cm²
So, the total area of quadrilateral PMON is 29 cm².
(e)
For Rectangle ABCG:
l = 9 cm and b = 4 cm
A₁ = l × b = 9 cm × 4 cm = 36 cm²
For Rectangle CDEF:
l = 4 cm and b = 6 cm
A₂ = l × b = 4 cm × 6 cm = 24 cm²
Total Area of the L-shaped figure:
Total Area = A₁ + A₂ = 36 cm² + 24 cm² = 60 cm²
So, the total area of the L-shaped figure is 60 cm². 😊
(f)
Rectangle 1 (top rectangle):
Length: 24 meters
Breadth: 24 meters
A₁ = Length × Breadth = 24 m × 24 m = 576 m²
Rectangle 2 (bottom rectangle):
Length: 40 meters
Breadth: 24 meters
A₂= Length × Breadth = 40 m × 24 m = 960 m²
Total Area:
Total Area = Area 1 + Area 2 = 576 m² + 960 m² = 1536 m²
So, the area of the shape is 1536 m².
(g)
Rectangle 1:
Length: 12 meters
Breadth: 3 meters
Area 1 = Length × Breadth = 12 m × 3 m = 36 m²
Rectangle 2:
Length: 5 meters
Breadth: 12 meters
Area 2 = Length × Breadth = 6 m × 5 m = 30 m²
Rectangle 3:
Length: 15 meters
Breadth: 3 meters
Area 3 = Length × Breadth = 15 m × 3 m = 45 m²
Total Area = Area 1 + Area 2 + Area 3 = 36 m² + 30 m² + 45 m² = 111 m²
So, the area of the composite shape is 111 m².
(h)
For rectangle,Length: 10 inches
Breadth: 24 inches
Area (A) = Length × Breadth = 10 in × 24 in = 240 square inches
For triangle,
Base: 24 inches
Height: 9 inches
Area (A) = 1/2 × Base × Height = 1/2 × 24 in × 9 in = 108 square inches
Total Area = Area rectangle + Area triangle = 240 square inches + 108 square inches = 348 square inches
So, the total area of the composite figure is 348 square inches.
(i)
Rectangle 1 :
Length: 4 meters
Breadth: 12 meters
Area 1 = Length × Breadth = 4 m × 12 m = 48 m²
Rectangle 2:
Length: 5 meters
Breadth: 3 meters
Area 2 = Length × Breadth = 5 m × 3 m = 15 m²
Rectangle 3 :
Length: 3 meters
Breadth: 12 meters
Area 3 = Length × Breadth = 3 m × 12 m = 36 m²
Total Area = Area 1 + Area 2 + Area 3 = 48 m² + 15 m² + 36 m² = 99 m²
The total area of the composite shape is 99 m².
5. Find the area of the shaded part from the given figure:
(a)In Parallelogram EFGH,
Base: 22 cm
Height: 12 cm
Area = Base x Height
Area = 22 cm x 12 cm = 264 cm²
In Triangle GHI,
Base: 22 cm
Height: 12 cm
Area = (1/2) x Base x Height
Area = (1/2) x 22 cm x 12 cm = 132 cm²
Shaded Area = Area of Parallelogram - Area of Triangle
= 264 cm² - 132 cm² = 132 cm²
(b)In trapezium ABCD,
Base (AB) = 4.8 cm
Height (BC) = 4.6 cm
Area = (1/2) x (Base1 + Base2) x Height
Area = (1/2) x (4.8 + 8.2) x 4.6 = 29.9 cm²
In parallelogram CDEF,
Base (EF) = 8.2 cm
Height (CF) = 3.5 cm
Area = Base x Height
Area = 8.2 x 3.5 = 28.7 cm²
3. Total Area
Total Area = Area of Trapezium + Area of Parallelogram
Total Area = 29.9 cm² + 28.7 cm² = 58.6 cm²
(c)Given:
In trapezium EFGH,
a = 12 cm
b = 6 cm
h = 7 cm
Area (A) = ?
Area = (1/2) × (Base1 + Base2) × Height
= (1/2) × (12 + 6) × 7
= (1/2) × 18 × 7
= 63 cm²
In Triangle GHJ,
Area = (1/2) × Base × Height
= (1/2) × 6 × 7
= 21 cm²
Shaded Area = Area of Trapezium - Total Area of Triangle
= 63 cm² - 21 cm²
= 42 cm²
(d)Given:
In trapezium ABCD,
a = 25 inch
b = 15 inch
h = 8 inch
Area (A) = ?
Area = (1/2) × (Base1 + Base2) × Height
= (1/2) × (25 + 15) × 8
= (1/2) × 40 × 8
= 160 inch²
In Triangle,
Base = 25 inch
Height = 8 inch
Area = (1/2) × Base × Height
= (1/2) × 25 × 8
= 100 inch²
Shaded Area = Area of Trapezium - Total Area of Triangle
= 160 inch² - 100 inch²
= 60 inch²
Here,
Length of park (l) = 120 m
Breadth of park (b) = 110 m
Area (A)=?
We know that,
Area of park = l × b = 120 × 110 = 13,200 m²
Now,
Length of volleyball court (l) = 18 m
Breadth of volleyball court (b) = 9 m
Area (A)=?
Area of court = l × b = 18 × 9 = 162 m²
Therefore, remaining area = Area of park - Area of court = 13,200 - 162 = 13,038 m²
7. (a) A rectangular park of length 120m and breadth 110m has a volleyball court of length 18m and breadth 9m. Find the area of the park excluding the volleyball court.
Here,
Length of park (l) = 120 m
Breadth of park (b) = 110 m
Area of park = l × b = 120 × 110 = 13,200 m²
Now,
Length of volleyball court (l) = 18 m
Breadth of volleyball court (b) = 9 m
Area of court = l × b = 18 × 9 = 162 m²
Therefore, remaining area = Area of park - Area of court = 13,200 - 162 = 13,038 m²
7. (b) A square fish pond of length 50 m is constructed in the middle of a rectangular field of length 125 m and breadth 100 m. Find the area of the field excluding the fish pond.
Here,
Length of field (l) = 125 m
Breadth of field (b) = 100 m
Area of field = l × b = 125 × 100 = 12,500 m²
Now,
Side of square fish pond (s) = 50 m
Area of fish pond = s × s = 50 × 50 = 2,500 m²
Therefore, remaining area = Area of field - Area of pond = 12,500 - 2,500 = 10,000 m²
8. (a) How many boards of area 1 sq. ft. required to be pasted on a wall having 9 ft length and 7 ft wide without overlapping?
Here,
Length of the wall = 9 ft
Breadth of the wall = 7 ft
Area of the wall = length × breadth = 9 × 7 = 63 sq. ft.
As the area of each board is 1 sq. ft.,
Number of boards required = Area of the wall / Area of each board = 63 / 1 = 63 boards
8. (b) In a wall of 25 m length and 3 m wide, how many 1 sq. m. ply can be pasted without overplacing?
Here,
Length of the wall = 25 m
Breadth of the wall = 3 m
Area of the wall = length × breadth = 25 × 3 = 75 sq. m.
As the area of each ply is 1 sq. m.,
Number of plies required = Area of the wall / Area of each ply = 75 / 1 = 75 plies
9. (a) If the perimeter of a square field is 200 feet,
(i) Find the length of the field.
We know that the perimeter of a square is the total length of all four sides. The formula for the perimeter of a square is:
Perimeter = 4 × side
We are given that the perimeter is 200 feet, so we can substitute this into the formula:
200 = 4 × side
or, side = 200 / 4 = 50 feet
The length of the field is 50 feet.
(ii) Find the area of the field.
To find the area of the square field, we use the formula:
Area of the square = side × side
We know that the side is 50 feet, so:
Area = 50 × 50 = 2500 sq. ft.
The area of the field is 2500 square feet.
9. (b) If the length of the wall surrounding a square ground is 80 m,
(i) Find the length of the ground.
The perimeter of a square is the total length of all four sides. The formula for the perimeter is:
Perimeter = 4 × side
We are given that the perimeter is 80 meters, so we can substitute this into the formula:
80 = 4 × side
or, side = 80 / 4 = 20 m
The length of the ground is 20 meters.
(ii) Find the area of the ground.
To find the area of the square ground, we use the formula:
Area of the square = side × side
We know that the side is 20 meters, so:
Area = 20 × 20 = 400 sq. m.
The area of the ground is 400 square meters.
9. (c) If the length of rectangular room is 15 ft and its perimeter is 54 ft,
(i) What is breadth of the room?
We know that the perimeter of a rectangle is given by the formula:
Perimeter = 2 × (length + breadth)
We are given that the perimeter is 54 feet and the length is 15 feet. We can substitute these values into the formula:
54 = 2 × (15 + breadth)
or, 54 = 2 × (15 + breadth) → 54 = 30 + 2 × breadth
or, 54 - 30 = 2 × breadth → 24 = 2 × breadth
or, breadth = 24 / 2 = 12 feet
The breadth of the room is 12 feet.
(ii) Find the area of room.
To find the area of the room, we multiply the length and the breadth:
Area of the room = length × breadth = 15 × 12 = 180 sq. ft.
The area of the room is 180 square feet.
9. (d) The length of a rectangular room is twice the breadth. If the perimeter is 60 ft,
(i) What is the length and breadth of the room?
We are given that the perimeter of the room is 60 feet and the length is twice the breadth.
Let the breadth be represented by 'b'. Then, the length will be 2b. Now, we can use the perimeter formula:
Perimeter = 2 × (length + breadth)
60 = 2 × (2b + b)
60 = 2 × 3b → 60 = 6b
Now, divide both sides by 6:
b = 60 / 6 = 10 feet
The breadth of the room is 10 feet, and the length of the room is twice the breadth, so:
Length = 2 × 10 = 20 feet
(ii) How many square feet of carpet is required to cover the floor of room?
To find the area of the room, we multiply the length and breadth:
Area of the room = length × breadth = 20 × 10 = 200 sq. ft.
So, 200 square feet of carpet is required to cover the floor.
Solution of selected questions from Excel in Mathematics - Book 8
9. The perimeter of a square ground is 100 m. Find the cost of plastering the ground at Rs 80 per sq. meter.
Solution:
Here,
The perimeter of a square ground = 100 m
or, 4l = 100 m → l = 25 m
Now, area of the garden = l² = (25 m)² = 625 m²
Hence, the cost of plastering the ground = Area × Rate of cost
= 625 × Rs 80
= Rs 50,000
10. The length of the floor of a room is two times its breadth and the perimeter of the floor is 48 m. Find the cost of paving marbles on the floor at Rs 125 per sq. meter.
Solution:
Let the breadth of the room (b) = x m
Then the length of the room (l) = 2x m
Now,
The perimeter of the room = 2(l + b)
or, 48 m = 2(2x + x)
Breath of the room (b) = 8 m and the length (l) = 2x m = 2 × 8 m = 16 m
Also, area of the floor = l × b = 16 m × 8 m = 128 m²
Hence, the cost of paving marbles on the floor = Area × Rate of cost
= 128 × Rs 125
= Rs 16,000
11. The length of the rectangular ground is three times its breadth. If the perimeter of the ground is 96 m, find the cost of growing grass on it at Rs 65 per sq. meter.
Solution:
Let the breadth of the room (b) = x m
Then the length of the room (l) = 3x m
Now,
The perimeter of the room = 2(l + b)
or, 96 m = 2(3x + x)
or, 96 m = 8x → x = 12 m
Thus, Breath of the room (b) = x = 12 m and the length (l) = 3x m = 3 × 12 m = 36 m
Also, area of the floor = l × b = 36 m × 12 m = 432 m²
Hence, the cost of growing grass on the floor = Area × Rate of cost
= 432 × Rs 65
= Rs 28,080
12. A rectangular park of length 60 m and breadth 50 m encloses a volleyball court of length 18 m and breadth 10 m. Find the area of the park excluding the court. Also, find the cost of paving stone in the park excluding the court at the rate of Rs 110 per square meter.
Solution:
Area of the park = l × b = 60 m × 50 m = 3000 m²
Area of the volleyball court = l × b = 18 m × 10 m = 180 m²
Area of the park excluding the court = 3000 m² - 180 m² = 2820 m²
Hence, the cost of paving stone in the park excluding the court = Area × Rate of cost
= 2820 × Rs 110
= Rs 3,10,200
13. Mr. Thapa builds a circular pond of diameter 28 ft for fish farming in his rectangular field of length 70 ft and breadth 40 ft. Find the area of the field excluding the pond. Also, estimate the cost of planting vegetables at Rs 11 per square foot.
Solution:
Here,
Area of the rectangular field = l × b = 70 ft × 40 ft = 2800 ft²
Radius of the pond = 14 ft.
Area of the pond = Ï€r²
= 22/7 × (14 ft)² = 616 ft²
∴ Area of the field excluding the pond = 2800 ft² - 616 ft² = 2184 ft²
The cost of planting vegetables in the field excluding the pond = Area × Rate of cost
= 2184 × Rs 11
= Rs 24,024
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