Class 5 Maths Chapter 11 HOTS Questions - Chapter 11 - Area and its Boundary

Q1: The side of the small squares is 1 cm. Find the total area covered by the shaded region.
Sol: There are 2 rectangles, each of these rectangles contains 8 square boxes in total
The side of each square box is 1 cm
thus area of each square = 1 sq cm
Then the area of each rectangle will be (8 × 1) sq.cm = 8 sq.cm
Thus the area of 2 rectangles = (2 × 8) sq.cm = 16 sq.cm
Hence, the area of the shaded region is 16 sq.cm

Q2: Let there be a rectangle of area 24 cm² . Then find the total number of square grids made in this rectangle.(Each square grid measures 1 cm²)
(a) 3
(b) 6
(c) 12
(d) 24
Ans: (d)
Sol: Number of square grids = Area of Rectangle / Area of square grid
= 24/1
= 24

Q3: The figure is made up of 11 squares of the same size. The area of the figure is 99 cm2 . Find the perimeter of the figure.

(a) 12 cm
(b) 24 cm
(c) 48 cm
(d) 36 cm
Ans: (c)
Sol: Area of figure= 99 cm²
⇒ Area of 11 small squares = 99 cm²
⇒ Area of 1 small square = 99 ÷ 11= 9 cm²
⇒ Side of 1 small square = 3 cm
∴ Perimeter of the figure = 16 × 3 = 48 cm

Q4: Which stamp has biggest area?________ ?
Ans: A is the answer which has a biggest area.

Q5: What is the area of the triangle?
Sol: 
Area of the triangles in the first figure will be  2 ÷ 2 = 1sqcm
Since the figure consists of two square boxes each having area of 1 sq cm
Area of triangles in second figure will be  6 ÷ 2 = 3sqcm
Since the figure contains 6 square boxes
Area of triangles in third figure will be  12 ÷ 2 = 6sqcm
Since the figure contains 12 square boxes

Q6: A wire is in shape of a rectangle. Its length is 40cm and breadth is 22cm. If the same wire is reshaped into a square, what will be the measure of each side. Also find which shape encloses more area.
Sol: Given,
Length of rectangle(l) = 40 cm
Breadth of rectangle(b) = 22 cm
Perimeter of rectangle = 2(l + b)
= 2(40 + 22) cm
= 124 cm
It is reshaped into square
∴ Perimeter remains same
Let, the edge of square be a
∴4a = 124
⇒ a = 31
Hence, the side of square is 31 cm .
Area of rectangle = l × b
= 40 × 22 sq.cm
= 880 sq.cm
Area of square = Side × Side
= 31 × 31 sq.cm
= 961 sq.cm
Hence, the square encloses more area.

Q7: Out of two figures if one has larger area, then its perimeter needs not to be larger than the other figure.
(a) True
(b) False
Sol: It is true.
Because the perimeter is the sum of all sides of closed shapes or polygons while the area is just bounded space insides.
Let's take example of rectangle and square
Example:
Rectangle
length = 6, breadth = 4,
So, Area = 6 × 4 = 24
And perimeter = 2 × 6 + 2 × 4 = 12 + 8 = 20
Let, square
side = 5
So,  Area = 5²
= 25
And perimeter = 4 × 5 = 20
Here, area of square is larger than rectangle but its perimeter is not larger than rectangles.
hence, given statement holds true.

Q8: 10 cm² = _______m².
Sol: We know that
1 cm = 1/100 m


Q9: Find the areas of the following figures by counting the square.
Sol: The figure contains 5 fully filled squares only. Therefore, the area of this figure will be 5 square units.

Q10: The region given in figure is measured by taking each rectangular box as a unit. What is the area of the region?
Sol: Area of 1 small rectangle = 1 square unit
Total number of small rectangles = 13
∴ Area of region =  Total number of small rectangles  ×  Area of 1 small rectangle
= (13 × 1)
= 13 square units
∴ Area of the region is 13 square units

Q11: Find the areas of the following figures by counting the square.
Sol: The figure contains 2 fully filled squares and 4 half-filled squares.
Total squares = 2 full squares + 4 x half squares = 2 full squares + 2 full squares = 4 full squares
Therefore, the area of this figure will be 4 square units.

Q12: Rectangular wall MNOP of a kitchen is covered with square tiles of 15 cm length (Figure below). Find the area of the wall.
Sol: It can be observed that MN and MP are the length and breadth of rectangle MNOP.
Number of squares along MN = 7
Number of squares along MN = 4
∴ Length of rectangle MNOP (MN) = Number of squares along MN × length of each square
= (7 × 15)cm
= 105 cm
Breadth of rectangle MNOP (MP) = Number of squares along MP × length of each square
= (4 × 15)cm
= 60 cm
Area of rectangle MNOP = Length × Breadth
= MN × MP
= (105 × 60)cm²
= 6300 cm²
∴ Area of wall is 6300cm²

Q13: Find the areas of the following figures by counting square.
Sol: The figure contains 9 fully filled squares only. Therefore, the area of this figure will be 9 square units.

Q14: Find the areas of the following figures by counting square.
Sol: The figure contains 2 fully filled squares and 4 half-filled squares. Therefore, the area of this figure will be 4 square units.

Q15: Consider the rectangle A with perimeter 30 cm. If rectangle B is having same perimeter as rectangle A and length of rectangle B is 9 cm, then what is the breadth of rectangle B?
(a) 8 cm
(b) 6 cm
(c) 9 cm
(d) 5 cm
Ans: (b)
Sol: Given Perimeter of rectangle A = 30m = perimeter of rectangle B
We know Perimeter = 2 × (length + breadth)
length = 9cm
⇒ 18 + 2b=30
⇒ 2b = 12
⇒ b = 6cm