FIND AREA WHOLE AND HALF UNITS OF LENGTH

Find the area of the following figures.

Each square stand for 1 square unit and each triangle stands for a half square unit.

Problem 1 :

Solution :

Total number of square = 4

Area of complete squares

= number of squares x area of each square

= 4 × 1

Area of squares = 4

Total number of triangles = 4

Area of triangles = 4 × 0.5

= 2

Required area = 4 + 2

= 6 square units

Problem 2 :

Solution :

By observing the figure,

Total number of square = 6

= 6 × 1

= 6

Total number of triangle = 4

= 4 × 0.5

= 2

Required area = 6 + 2

= 8 square units

Problem 3 :

Solution :

Total number of square = 6

Area of squares = 6 × 1  ==> 6

Total number of triangle = 2

Area of triangles = 2 × 0.5  ==> 1

Area of the shaded portion = 6 + 1

= 7 square units

Problem 4 :

Solution :

Total number of square = 12

Area of shaded squares = 12 × 1 ==> 12

Total number of triangle = 4

Area of shaded triangles = 4 × 0.5 ==> = 2

Total number of area = 12 + 2

= 14 square units

Problem 5 :

Solution :

Total number of square = 6

Area of squares = 6 × 1 ==> 6

Total number of triangle = 2

Area of triangles = 2 × 0.5 ==> 1

Area of shaded portion = 6 + 1

= 7 square units

Problem 6 :

Solution :

Total number of square = 6

Area of shaded squares = 6 × 1 ==> 6

Total number of triangle = 2

Area of shaded triangles = 2 × 0.5 ==> 1

Required area = 6 + 1

= 7 square units

Problem 7 :

Solution :

Total number of square = 4

Area of shaded squares = 4 × 1 ==> 4

Total number of triangle = 8

Area of shaded triangles = 8 × 0.5 ==> 4

Required area = 4 + 4

= 8 square units

Problem 8 :

Solution :

Total number of square = 6

Area of shaded squares = 6 × 1 ==> 6

Total number of triangle = 4

Area of shaded triangles = 4 × 0.5 ==> 2

Required area = 6 + 2

= 8 square units

Problem 9 :

Solution :

By observing the figure,

Total number of square = 8

Area of shaded squares = 8 × 1 ==> 8

Total number of triangle = 6

Area of shaded triangles = 6 × 0.5 ==> 3

Required area = 8 + 3

= 11 sq.units

Problem 10 :

Find the perimeter of square whose side is 13 cm.

Solution :

Perimeter of square = 4 (side length)

Side length = 13 cm

Perimeter of the square = 4 x 13

= 52 cm

Problem 11 :

Area of the square whose side length is 1 unit.

Solution :

Side length of square = 1 unit

Area of square = side x side

= 1 x 1

= 1 square units.

Problem 12 :

Find the perimeter of the regular hexagon with each side measuring 8 cm.

a)  48 cm    b)  40 cm     c) 32 cm   d) 24 cm

Solution :

Since it is regular hexagon all sides will be equal.

Perimeter of regular hexagon = 6 (side length)

Side length of hexagon = 8 cm

= 6(8)

= 48 cm

So, option a is correct.

Problem 13 :

The area of the rectangle is 225 cm², its width is 25 cm. Find its length.

a)  7 cm    b)  15 cm     c) 13 cm   d) 9 cm

Solution :

Area of rectangle = 225 cm²

Length x width = 225

Length = ?, width = 25 cm

length x 25 = 225

length = 225/25

= 9 cm

So, option d is correct.

Problem 14 :

The perimeter of the closed rectangular figure is

a)  sum of all sides   b)  sum of any two sides    c) Sum of three sides 

 d) none

Solution :

Rectangle will have four sides and opposite sides will be equal in length. To find the perimeter of the rectangle, will be adding sum of all sides of the rectangle.

Problem 15 :

Fidn the perimeter of an isosceles triangle with equal sides 8 cm and third side is 6 cm

a)  25 cm      b)  18 cm     c) 20 cm     d) 22 cm

Solution :

Since it is isosceles triangle, any two sides will be equal in length.

Perimeter of triangle = 8 + 8 + 6

= 22 cm

So, option d is correct.

Problem 16 :

What will happen to the area of a rectangle if its length is doubled and width is tripled ?

a)  6 times       b)  4 times     c) 2 times     d) 8 times

Solution :

Let l and w be the length and width of the rectangle

length is doubled, then 2l

Width is tripled, then 3 w

Area = 2l(3w)

= 6 lw

So, new area of the rectangle will be 6 times area of the old rectangle. Option a is correct.