Question: (a) Match the following figures with their respective areas in the box.
(b) Write the perimeter of each shape.
(b) (i) The given figure is a rectangle in which
Length = 14 cm
Breadth = 7 cm
∵ Perimeter of a rectangle = 2 * [Length + Breadth]
∴ Perimeter of the given figure = 2 * [14 cm + 7 cm]
= 2 * 21 cm = 42 cm
(ii) The figure is a square housing its side as 7 cm.
∵ Perimeter of a square = 4 * Side
∴ Perimeter of the given figure = 4 * 7 cm = 28 cm
Question 1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?
(a) Side of the square = 60 m
∴ Its perimeter = 4 * Side
= 4 * 60 m = 240 m
Area of the square = Side * Side
= 60 m * 60 m = 3600 m2
(b) ∵ Perimeter of the rectangle = Perimeter of the given square
∴ Perimeter of the rectangle = 240 m
or 2 * [Length + Breadth] = 240 m
or 2 * [80 m + Breadth] = 240 m
or 80 m + Breadth = 240/2 m = 120 m
∴ Breadth = (120 – 80) m = 40 m
Now, Area of the rectangle = Length* Breadth
= 80 m *40 m
= 3200 m2
Since, 3600 m2 > 3200 m2
∴ Area of the square field (a) is greater.
Question 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: ∵ The given plot is a square with side as 25 m.
∴ Area of the plot = Side * Side
= 25 m * 25 m = 625 m2
∵ The constructed portion is a rectangle having length = 20 m and breadth = 15 m.
∴ Area of the constructed portion = 20 m * 15 m = 300 m2
Now area of the garden = [Total plot area] – [Total constructed area]
= (625 – 300) m2 = 325 m2
∴ Cost of developing the garden = Rs 55 * 325 = Rs 17,875
Question 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 this garden. [Length of rectangle is 20 – (3.5 + 3.5) metres.]
Solution: For the semi-circular part:
Diameter of the semi-circle = 7 m
∴ Radius of the semi-circle = 7/2 = 3.5 m
Area of the 2 semi-circles =
= πr2 = 22/7 *(3.5 m)2
Also, perimeter of the 2 semicircles
For rectangular part:
Length of the rectangle = 20 – (3.5 + 3.5) m = 13 m
Breadth of the rectangle = 7 m
∴ Area = Length * Breadth
= 13 m * 7 m = 91 m2
Perimeter = 2 * [Length + Breadth]
= 2 * [13 m + 0 m]
= 2 * 13 m = 26 m
Now, Area of the garden = (38.5 + 91) m2
= 129.5 m2
Perimeter of the garden = 22 m + 26 m = 48 m
Question 4. A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles area required to cover a floor of area 1080 m2? (If required you can split the tiles in whatever way want to fill up the corners.)
Solution: Area of a parallelograms = Base * Corresponding height
Area of a tile =
∴ Area of the floor = 1080 m2
Now, number of tiles = Total area/Area of one tiles
= 1080/0.024 = 45000 tiles
Question 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 = 2pr, where r is the radius of the circle.
Solution: (a) Diameter = 2.8 cm ﬁ Radius = 2.8/2 cm = 1.4 cm
Perimeter of a circle = 2pr ⇒ Perimeter of a semicircle = 2πR/2 = πr
∴ Perimeter of the figure = πr + Diameter
= 22/7 * 1.4 cm + 2.8 cm
= 22/7* 14/10 cm + 2.8 cm
= 4.4 cm + 2.8 cm = 7.2 cm
(b) Perimeter of the semi-circular part = πr
= 22/7 * 1.4 cm
= 4.4 cm
Perimeter of the remaining part = 1.5 cm + 2.8 cm + 1.5 cm= 5.8 cm
∴ Perimeter of the figure = 4.4 cm + 5.8 cm = 10.2 cm
(c) Perimeter of the semi-circular part = πd/z = 22/7 * 2.8/2 = 4.4 cm
∴ Perimeter of the figure = 4.4 cm + 2 cm + 2 cm
= 8.4 cm
Since, 7.2 cm < 8.4 cm < 10.2 cm
∴ Perimeter of figure ‘b’ has the longest round.
Note: In figures ‘b’ and ‘c’, the diameters are not part of the figures.
Area of Trapezium
The area of a trapezium is the half of the sum of the length of parallel sides and then multiply the perpendicular distances between then.
Here ABCD is a trapezium.
Here AB and CD are parallel sides.
Also AE and BF are perpendicular distance between them.
∴ Area of trapezium of ABCD = 1/2 (AB + CD) * AE
Question 1. Nazma’s sister also has a trapezium shaped plot. Divide it into three parts as shown. Show that the area of trapezium WXYZ =
Area of Δ PWZ =1/2 * Base * Altitude
= 1/2 * c * h = 1/2 ch
Area of the rectangle PQYZ = b * h = bh
Area of Δ QXY = 1/2 * d * h = 1/2 dh
∴ Area of the trapezium XYZW = 1/2 ch + bh +1/2 dh [∵ c + d + b = a]
= 1/2 (c+d)h + bh [∵ c + d = a – b]
Question 2. If h = 10 cm, c = 6 cm, b = 12 cm, d = 4 cm, find the values of each of its parts separately and add to find the area WXYZ. Verify it by putting the values of h, a and b in the expression
Solution: Area of Δ PWZ = 1/2 ch =1/2 * 6 * 10 cm2 = 30 cm2
Area of Δ QXY = 1/2 dh = 1/2 * 4 * 10 cm2 = 20 cm2
Area of rectangle PQYZ = Length * Breadth = b * h
= 12 * 10 cm2 = 120 cm2
\ Area of trapezium WXYZ = Area of Δ PWZ + Area of Δ QXY + Area of rectangle PQYZ
= 30 cm2 + 20 cm2 + 120 cm2 = 170 cm2
Also, area of the trapezium WXYZ =
Hence, the area of trapezium is verified.
Question: Find the area of the following trapeziums.
Note: Area of a trapezium = 1/2* [sum of the parallel sides]*[Perpendicular distance between the parallel sides]
(i) Area of the given trapezium
= 1/2 * (Sum of parallel sides) * (Perpendicular distance between the parallel sides)
= 1/2 * (9 + 7) cm * 3 cm
= 1/2 * 16 cm * 3 cm = 24 cm2
(ii) Area of the given trapezium
= 1/2 * (Sum of the parallel sides) * (Perpendicular distance between the parallel sides)
= 1/2 * (10 + 5) cm * 6 cm
= 1/2 * 15 cm * 6 cm = 45 cm2
Area of a General Quadrilateral
Let ABCD be a quadrilateral. Join its vertices A and C such that AC is a diagonal.
Draw BP AC and DQ AC.
Let BP = h1 and DQ = h2
Now, area of quadrilateral ABCD
[Area of a quadrilateral] = 1/2 * (One of the diagonals) *(Sum of the perpendicular drawn on the diagonal from the opposite vertices)
Example 1. Find the area the quadrilateral LMNO (as shown in the figure).
Solution: Diagonal LN = 8 cm
Perpendicular MX = 4.5 cm
Perpendicular OY = 3.5 cm
∵ Sum of the perpendiculars = MX + OY = 4.5 cm + 3.5 cm = 8 cm
∴ Area of the quadrilateral =1/2 * (A diagonal) * (Sum of the lengths of the perpendiculars on it from opposite vertices)