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Practice each question, then open the answer or worked solution.
Observe the shapes 1, 2, 3 and 4 in the figures. Which of the following statements is not correct?
Solution:-
(a) Shapes 1, 3 and 4 have different areas and different perimeters.
Shapes 1, 3 and 4 have the same area and same perimeter.
A rectangular piece of dimensions 3 cm × 2 cm was cut from a rectangular sheet of paper of dimensions 6 cm × 5 cm (Fig. 9.14).
Area of remaining sheet of paper is
(a) 30 cm2 (b) 36 cm2 (c) 24 cm2 (d) 22 cm2
Solution:-
(c) 24 cm2
First we have to calculate the area of rectangular piece = length × breadth
= 2 cm × 3 cm
= 6 cm2
So, area of sheet of paper of dimensions 6 cm × 5 cm
= 30 cm2
Then, Area of remaining sheet of paper is = 30 cm – 6 cm
= 24 cm2
36 unit squares are joined to form a rectangle with the least perimeter. The perimeter of the rectangle is
(a) 12 units (b) 26 units (c) 24 units (d) 36 units
Solution:-
(b) 26 units
From the question, it is given that, the area of rectangle = 36 units2
36 can be written as = 6 × 6
= (2 × 3) × (2 × 3)
= 22 × 32
= 4 × 9
Then, the sides of the rectangle are 4 cm and 9 cm.
We know that, perimeter of the rectangle = 2 (length + breadth)
= 2 (4 + 9)
= 2 (13)
= 26 units
A wire is bent to form a square of side 22 cm. If the wire is rebent to form a circle, its radius is
(a) 22 cm (b) 14 cm (c) 11 cm (d) 7 cm
Solution:-
(b) 14 cm
From the question it is given that, side of square is 22 cm.
And also, perimeter of a square and circumference of circle are equal, because the length of the wire is same.
Perimeter of square = circumference of circle
4 × side = 2 × π × r
4 × 22 = 2 × (22/7) × r
r = (4 × 22 × 7)/(2 × 22)
r = 14 cm
Therefore, radius of circle is 14 cm.
Area of the circle obtained in Question 4 is
(a) 196 cm2 (b) 212 cm2 (c) 616 cm2 (d) 644 cm2
Solution:-
(c) 616 cm2
We know that, area of circle = πr2
= (22/7) × 14 × 14
= 22 × 14 × 2
= 616 cm2
Area of a rectangle and the area of a circle are equal. If the dimensions of the rectangle are 14cm × 11 cm, then radius of the circle is
(a) 21 cm (b) 10.5 cm (c) 14 cm (d) 7 cm.
Solution:-
(d) 7 cm
From the question it is given that, dimensions of rectangle length = 14 cm, breadth = 11 cm
As area of rectangle = area of circle
length × breadth = πr2
14 × 11 = (22/7) × r2
r2 = (14 × 11 × 7)/22
r2 = 49
r = √49
r = 7 cm
Area of shaded portion in Fig. 9.15 is
(a) 25 cm2 (b) 15 cm2 (c) 14 cm2 (d) 10 cm2
Solution:-
(d) 10 cm2
From the figure, length of rectangle = 5 cm and,
breadth of the rectangle = 3 cm + 1 cm = 4 cm
So, area of rectangle = length × breadth
= 5 × 4
= 20 cm2
By observing the figure, the shaded part covered exactly half of the rectangle,
Therefore, area of shaded part is = area of rectangle/2
= 20/2
= 10 cm2
Area of parallelogram ABCD (Fig. 9.16) is not equal to
(a) DE × DC (b) BE × AD (c) BF × DC (d) BE × BC
Solution:-
(a) DE × DC
We know that, area of parallelogram = base × corresponding height
Then, area of parallelogram ABCD = DC × BF
AD × BE = BC × BE … [because AD = BC]
Area of triangle MNO of Fig. 9.17 is
(a) ½ MN × NO (b) ½ NO × MO
(c) ½ MN × OQ (d) ½ NO × OQ
Solution:-
(d) ½ NO × OQ
MNO is a triangle.
We know that, area of triangle = ½ (base × height)
= ½ × NO × OQ
Ratio of area of ∆MNO to the area of parallelogram MNOP in the same figure 9.17 is (a) 2 : 3 (b) 1 : 1 (c) 1 : 2 (d) 2 : 1
Solution:-
(c) 1 : 2
From the figure,
Area of ∆MNO = ½ × base × height
= ½ × NO × OQ
Area of parallelogram MNOP = base × corresponding height
= MP × OQ
= NO × OQ … [from the figure MP = NO]
Then, ratio of parallelogram and triangle = (½ × NO × OQ)/(NO × OQ)
= ½
= 1: 2