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2. Introduction
A closed plane figure, the perimeter is the distance around its
boundary and its area is the region covered by it. We found the
area and perimeter of various plane figures such as triangles,
rectangles, circles etc

3. Shape Figure Area Perimeter
Rectangle Length x Breadth 2 (Length + Breadth)
Square (Side)2 4 x Length
Triangle ½ Base x Height a + b + c, where a, b and c are
three sides of triangle
Parallelogram Base x Height 2 (a + b), where a and b are
two different sides of
parallelogram
Circle π x (Radius)2 2 x π x radius

4. Examples :
Example 1: Find the area of square whose side is 50m.
● Solution: The area of square = (Side) 2
Given, side = 50 m.
Then, area of square = (50)2 = 2500 m2.

5. Example 2: Find the perimeter of a triangle with
isocelles sides of 15 cm and one side of 20 cm.
Solution: We know that,
Perimeter of triangle = sum of sides = (15 + 15 + 20) = 50 cm.
Example 2: Write the perimeter of the shape.
7cm
14cm
Perimeter of rectangle
= 2(l+b) = 2X(7+14)= 42

6. Ncert Exersise 11.1
Q.1 A square and a rectangular field with measurements as given in the
figure have the same perimeter. Which field has a larger area?
Q.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 ` 55 per m2 .
Q. 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.

7. Side of a square = 60 m (Given)
And the length of rectangular field, l = 80
m (Given)
According to question,
Perimeter of rectangular field = Perimeter
of square field
2(l+b) = 4×Side (using formulas)
2(80+b) = 4×60
160+2b = 240 = B = 40
Breadth of the rectangle is 40 m.
Now, Area of Square field
= (side)2
= (60)2 = 3600 m2
And Area of Rectangular field
= length×breadth = 80×40
= 3200 m2
Hence, area of square field is larger.
Answer:1

8. Answer: 2
Side of a square plot = 25 m
Formula: Area of square plot = square of a side =
(side)2
= (25)2 = 625
Therefore the area of a square plot is 625 m2
Length of the house = 20 m and Breadth of the
house = 15 m
Area of the house = length×breadth
= 20×15 = 300 m2
Area of garden = Area of square plot –
Area of house
= 625–300 = 325 m2
∵ Cost of developing the garden per sq. m is
Rs. 55
Cost of developing the garden 325 sq. m =
Rs. 55×325
= Rs. 17,875
Hence total cost of developing a garden
around is Rs. 17,875.

9. Answer:5
(a) Radius = Diameter/2 = 2.8/2 cm = 1.4 cm
Circumference of semi-circle = πr
= (22/7)×1.4 = 4.4
Circumference of semi-circle is 4.4 cm
Total distance covered by the ant= Circumference of semi -
circle+Diameter
= 4.4+2.8 = 7.2 cm
(b) Diameter of semi-circle = 2.8 cm
Radius = Diameter/2 = 2.8/2 = 1.4 cm
Circumference of semi-circle = r
= (22/7)×1.4 = 4.4 cm
Total distance covered by the ant= 1.5+2.8+1.5+4.4 = 10.2
cm
(c) Diameter of semi-circle = 2.8 cm
Radius = Diameter/2 = 2.8/2
= 1.4 cm
Circumference of semi-circle = π r
= (22/7)×1.4
= 4.4 cm
Total distance covered by the ant = 2+2+4.4 = 8.4
cm

10. Area of Trapezium:
For any given trapezium, if a and b are two parallel sides and h is
perpendicular distance between them
Area of trapezium = h x (a +b)/2

11. Examples:
Calculate the area for the following given trapezium.
Solution: As per the formula, Area of trapezium = height x (sum of parallel
sides)/2
Here, height = 4 cm and two sides are 5 cm & 10 cm. Substituting these value
Area of trapezium = 4 x (5 + 10)/2 = 30 cm2.

12. The area of a trapezium is 20 cm2 and the length of one of the parallel sides is
8 cm and its height is 2 cm. Find the length of the other parallel side.
Solution: Given, area = 20 cm, length of one side = a = 8 cm, height = 2 cm.
Now, area of trapezium = height x (a + b)/2
20 = 2 x (8 + b)/2
On solving, b = 12 cm.
Thus, the length of other parallel side for the given trapezium is 12 cm.

13. Area of Quadrilateral:
Any quadrilateral can be divided into two triangles by drawing one of the
diagonals.
Then applying and calculating area of triangle formula on individual
triangles, we obtain the area of quadrilateral.
Suppose the length of diagonal AC is d, then,
Area of quadrilateral ABCD = 1/2 x d x (h1 + h2)
D = Diameter
H1 AND H2 = Height

14. Example 1: Find the area of some quadrilateral whose diagonal length is 8 cm.
And the lengths of two perpendiculars on given diagonal from the vertices are 4
cm and 3 cm.
Solution: Here, d = 8 cm. Let the length of two perpendiculars be h1 = 4 cm
and h2 = 3 cm.
As per the formula of quadrilateral, we can write,
Area of quadrilateral ABCD = 1/2 x d x (h1 + h2)
= 1/2 x 8 x (4 + 3) = 28 cm2.
Examples:

15. Example 2: Find the area of quadrilateral ABCD as shown in figure below.
Solution: Given, d = 5 cm, h1 = 2cm, h2 = 1 cm.
Now, Area of quadrilateral ABCD = 1/2 x d x (h1 + h2) = 1/2 x 5 x (2 + 1)
= 7.5 cm2

16. Area of Rhombus:
In rhombus, the diagonals are perpendicular bisectors of each other. Thus,
total rhombus will be divided into two equal triangles.
Area of rhombus ABCD = area of Δ ABC + area of Δ ACD = (1/2 x d1 x
d2)

17. Example: Find the area of a rhombus in which the length of its two
diagonals is 4cm and 6cm.
Solution: Area of rhombus ABCD = (1/2 x d1 x d2) = 1/2 x 4 x 6 = 12 cm2.
Calculate the area of a rhombus having diagonals equal to 6 cm and 8 cm.
Solution: Area of rhombus PQRS = (1/2 x d1 x d2) = 1/2 x 6 x 8 = 12 cm2.
Examples:

18. Area of a Polygon
To find area of any poloygon . We split a polygon into quadrilaterals
into triangles and find its area.
For example:
Example: Find the area of octagon shown in figure below.

19. Solution: 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) = h x (sum of two parallel sides)/2
= 2 x (10 + 3)/2 = 13 cm2.
Area of part B (Rectangle) = length x breadth
= 10 x 3 = 30 cm2.
Thus, area of Octagon = 2 x area of part A + area of part B
= 2 x 13 + 30 = 56 cm2.

20. Ncert Exersise 11.2
Do it on your own and ask your doubts

21. Solid shapes
Solid Shapes:
The three dimensional objects having width, depth and height are known as solid
shapes.
Example: Cube, cuboid, cylinder, etc.

22. 1. Surface area of Cuboid:
A cuboid is made of six parts out of which four parts are same and remaining
two parts are same.
From figure, we can write surface area of cuboid as sum of area of six individual
parts.
Thus, area of cuboid = l x h + b x h + l x b + b x h + l x h + l x b
= 2 (lb + bh + hl)

23. Examples:
Example 1: Find total surface area of cuboid having length, breadth and
height as 10cm, 5cm and 2cm, respectively.
Solution: Given, l = 10cm, b = 5cm, h = 2cm
We know that, area of cuboid = 2 (lb + bh + hl)
= 2 (10x5 + 5x2 + 2x10) = 160 cm2.
Hence, total cost of white washing entire room = 760 x 10 = 7600 Rs.

24. Example 2: The internal measures of a cuboidal room are 20 m × 10 m × 6 m.
Find the total cost of whitewashing all walls of a room including ceiling, if the
cost of white washing is Rs 10 per m2 .
Solution: Given, length = 20m, breadth = 10m, h = 6 m.
Surface area of room = 2 (lb + bh + hl)
= 2 (20 x 10 + 10 x 6 + 6 x 20) = 760 m2.
Cost of white washing per m2 = Rs 10

25. 2. Surface area of Cube:
A cube is made of six equal parts.
We know that area of square = (side) 2. In cube, there are six equal square sides.
Thus, Total surface area of cube = 6 a2; where a is the length of any side.

26. Examples:
Example 1: Find total surface area of cube having its one length as 2cm.
Solution: Here, a = 2cm (Given)
We know that, area of cube = 6 a2
= 6 x 2 = 24 cm2.2

27. Example 2: Find the side of a cube whose surface area is 864 cm2.
Solution: Given, surface area of cube = 864 cm2. Let ‘a’ be the length of the cube.
We know that, area of cube = 6 a2
864 = 6 a2
Thus, a = 12 cm.

28. 3. Surface area of Cylinder:
From the above figure, we can see that cylinder is made of two equal circles and a
rectangle in circular form.
So, we can write, area of cylinder = area of circular base + area of curved rectangle +
area of circular top
= πr2 + 2πrh + πr = 2πr (r + h)

29. Example 1: Find the total surface area of cylinder given below.
Solution: Here, r = 7cm and h = 4cm. (Given)
We know that, area of cylinder = 2πr (r + h)
= 2 x π x 7 (7 + 4) = 484 cm2.
Examples:

30. Example 2: Find the height of a cylinder whose radius is 7 cm and the total surface area
is 968 cm2 .
Solution: Given, r = 7cm, area = 968 cm2.
We know that, area of cylinder = 2πr (r + h)
968 = 2 x π x 7 (7 + h)
On solving, h = 15 cm.

31. Volume and Capacity
Volume:
The amount of space occupied by an object three dimensionally is termed as its volume.
volume of a solid is measured by counting the number of unit cubes it contains. Cubic
units which we generally use to measure volume are:

32. Volume of Cuboid:
The volume of cuboid = lbh.

33. Example:
Example 1: A godown is in the form of a cuboid of measures 50 m × 40 m × 30 m. How
many cuboidal boxes can be stored in it if the volume of one box is 10 m3?
Solution: Given, volume of one box = 10 m3
Volume of godown = 50 × 40 × 30 = 60000 m3
Thus, number of boxes that can be stored in the godown = 60000 / 10 = 6000

34. Volume of Cube
The cube is a special case of a cuboid, where l = b = h.
Hence, volume of cube = l × l × l = l
3

35. Example:
Example 1: Find volume of cube having its one length as 5cm.
Solution: Here, a = 5cm (Given)
We know that, volume of cube = L
= 5 = 125 cm
3
3 3

36. Volume of Cylinder
Volume of cylinder = area of base × height
= πr2 × h = πr2h

37. Example:
Example 2: A rectangular piece of paper 10 cm × 5 cm is folded without overlapping to
make a cylinder of height 5 cm. Find the volume of the cylinder.
Solution: Here, the length of the paper will become the perimeter of the base of the
cylinder and width becomes height.
Now, perimeter of base of cylinder = 2πr = 10
Thus, r = 1.6 cm. Now, volume of cylinder = πr2h
= π x 1.6 x 1.6 x 5 = 40.23 cm
3