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Practice each question, then open the answer or worked solution.
The following figures are formed by joining six-unit squares. Which figure has the smallest perimeter in Fig. 6.4
(A) (ii)
(B) (iii)
(C) (iv)
(D) (i)
Ans: Correct option - D
Perimeter=sum of all sides
For the option (i) = 10 Units:
For the option (ii) = 12 Units
For the option (iii) = 14 Units
For the option (vi) = 14 Units
So, figure (i) has the smallest perimeter.
A square-shaped park ABCD of side 100m has two equal rectangular flower beds each of size 10m $\times$ 5m. Fig.6.5 Length of the boundary of the remaining park is
(A) 360m
(B) 400m
(C) 340m
(D) 460m
Ans: Correct option - B
Given ,length =10m
Breadth = 5 m
Than remaining park’s length=90m
Breadth =95m
So, perimeter of remaining park $=90+5+10+95+90+5+10+95=400$m
The side of a square is 10cm. How many times will the new perimeter become if the side of the square is doubled?
(A) 2 times
(B) 4 times
(C) 6 times
(D) 8 times
Ans: Correct option - A
Given,side of square = 10 cm
Than perimeter of square= $4 \times$ Sides $=4\times10cm=40cm$
If the side of the square is doubled.
New side of square $=2\times10 cm = 20cm$
Perimeter of new square $=4\times \text{Sides}=4\times20cm=80cm$
So,the new perimeter has doubled from the old perimeter.
The length and breadth of a rectangular sheet of paper are 20 cm and 10 cm, respectively. A rectangular piece is cut from the sheet as shown in Fig.6.6 . Which of the following statements is correct for the remaining sheet?
(A) Perimeter remains the same but area changes.
(B) The area remains the same but the perimeter changes.
(C) Both area and perimeter are changing.
(D) Both area and perimeter remain the same.
Ans: Correct option - A
Perimeter remains the same as the given figure but the area changes after cutting the piece.
Old perimeter $=2(L+B)=2(20+10)=60m$
New perimeter $=10+20+8+5+2+15=60m$
Old Area $=L \times B =20 \times 10=200{m}^{2}$
New Area $=(20 \times 10)-(5 \times 2)=190{m}^{2}$
Two regular Hexagons of perimeter 30cm each are joined as shown in Fig. 6.7. The perimeter of the new figure is
(A) 65cm
(B) 60cm
(C) 55cm
(D) 50cm
Ans: Correct option - D
Given,Perimeter of hexagon=30cm
Perimeter of hexagon $= 6 \times \text{side}$
$30= 6 \times \text{side}$
$\dfrac{30}{6}$=sides
5cm=side.
So,perimeter of given Figure=$10 \times \text{side}$
$=10\times 5cm=50cm$ $\text{[total number of sides of the new figure is 10]}$
In Fig. 6.8 which of the following is a regular polygon? All have equal side except (i)
(A) (i)
(B) (ii)
(C) (iii)
(D) (iv)
Ans: Correct option - B
Because for a regular polygon, all sides and angles must be equal. Only (ii) is satisfying this condition.
Match the shapes (each side's measures 2cm) in column I with the corresponding perimeters in column II:
Column I | Column II |
(A)
(B)
(C )
(D)
| (i) 16 cm (ii) 20 cm (iii) 24 cm (iv) 28 cm (v) 32 cm |
Ans: Perimeter of Given figure = sum of all sides
Given ,each side= 2cm
(A) No. of sides=14
Perimeter =14$\times $ length of sides
$=14\times2cm = 28cm$
(B) No. of sides=8
Perimeter =$8\times$ length of sides
$=8\times2cm=16cm$
(C) No. of sides=10
Perimeter =$10\times $ length of side
$=10\times2cm = 20cm$
(D) No. of sides=12
Perimeter$=12\times$ length of sides
$=12\times2cm= 24cm$
Column I | Column II |
(A)
| (iv) 28 cm |
(B)
| (i) 16 cm |
(C )
| (ii) 20 cm |
(D) | (iii) 24 cm |
Match the following.
Shapes | Perimeter |
(A)
| (i) 10 |
(B)
| (ii) 18 |
(C )
| (iii) 20 |
(D)
| (iv) 25 |
Ans: Perimeter of Given figure = sum of all sides
(A) We know it's a rectangle.
Perimeter of rectangle$=2\times (l+b)$
$=2\times(6+4)= 2\times 10 = 20 unit$
(B)We know it's a square
Perimeter of square $= 4\times $ length of sides
$=4\times 5$unit $= 20$ unit
(C) We know it's a equilateral triangle
Perimeter of equilateral triangle $=3\times$ length of side
$=3\times 6$unit $=18 $unit
(D)We know it's a Isosceles triangle
Perimeter Isosceles triangle=sum of all sides
$=4+4+2= 10$ unit
Perimeter of the shaded portion in Fig. 6.9 is
AB+_+_+_+_+_+_+HA
Ans: We know Perimeter = sum of all sides
The sides of given figure AB, BM, MD, DE, EN, NG, GH and HA
So,the perimeter of the given figure = AB + BM + MD + DE + EN + NG + GH + HA.
In questions 9 to 13, fill in the blanks to make the statements true.
The amount of region enclosed by a plane closed figure is called its……….
Ans: The amount of region enclosed by a plane closed figure is called its Area.