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This mock test of Test : Perimeter And Area - 2 for Class 5 helps you for every Class 5 entrance exam.
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QUESTION: 1

Find the perimeter of the given figure.

Solution:

Perimeter of the given figure

=(5+ 6 + 4 + 4 +5+8+3+ 2+12)cm

= 49 cm

QUESTION: 2

Find the total area of the shaded parts of the rectangle.

Solution:

We have,

AF = FE = ED = GD = GF = HG = AH

Now, AF + FE= 16 cm

AF + AF= 16 cm

AF = 8 cm

Area of square AFGH = 8 ´ 8 = 64 cm2

Now, area of rectangle HBCD = (15x 16)cm^{2}

= 240 cm^{2}

Area of shaded part of rectangle

= 240/2 = 120 cm^{2 }

Total shaded area = (64+ 120) cm^{2}

=184 cm^{2}

QUESTION: 3

The given figure is made up of two rectangles Find its total area.

Solution:

Area of rectangle ABCD = (18x8) cm^{2}

=144 cm^{2}

Area of rectangle GEFC = (12x6) cm^{2}

= 72 cm^{2
}

Total area = (144 + 72) cm^{2} = 216 cm^{2}

QUESTION: 4

Find the area of the shaded figure.

Solution:

Area of 1 small square = 3x3 = 9 cm^{2}

Number of shaded squares =

QUESTION: 5

Which shape has the largest area?

Solution:

Area of 1 = 3 sq. units

Area of shape P = 10 sq. units

Area of shape Q = 11 sq. units.

Area of shape R = 10 sq. units.

Area of shape S = 10 sq. units.

So, shape Q has largest area

QUESTION: 6

ABCD is a square of perimeter 56 m. Two triangular corners have been cut away as shown in figure. What is the perimeter of the remaining figure?

Solution:

Perimeter of square = 56 m

4 x side of square = 56 m

=> Side of square = (56/4) = 14 m

Now, AE + ED= 14 m

7 m + ED= 14 m

ED = 7 m

So, AE = AF = £F = FB = BG = FG = 7 m and ED = CG = 7 m

Perimeter of remaining figure

= ED + DC + CG + GF + FE

= (7 +14+ 7 + 7 + 7)m = 42m

QUESTION: 7

In the given figure. PQRS is a square and TUVW is a rectangle.

Find the remaining area if TUVW was removed from the figure.

Solution:

Area of square PQRS = (32x32) sq. cm =1024 sq. cm

Length of rectangle TUVW = (32- 12) cm

= 20 cm

Breadth of rectangle TUVW = (32- 20) cm =12 cm

So, area of rectangle TUVW

= (12x20) sq. cm

= 240 sq. cm

Area of remaining figure

= (1024 - 240) sq. cm

= 784 sq.cm

QUESTION: 8

The figure below is made up of 6 similar squares and 2 triangles. The perimeter of each square is 32 cm. What is the perimeter of the figure?

Solution:

Perimeter of each square = 32 cm

4xside of square = 32 cm

Side of square = (32/4) cm = 8

Perimeter of the figure = 10x8 = 80 cm

QUESTION: 9

|Find the area of the shaded part in the given figure.

Solution:

Area of larger rectangle = (22x8) sq. cm = 176 sq. cm

Area of unshaded rectangle = (7x6) sq. cm

= 42 sq. cm

Area of square = (4 ´ 4) sq. cm = 16 sq. cm

Area of shaded region

= 176 sq. cm - (42 + 16) sq. cm

=118 sq. cm

QUESTION: 10

The shaded part in the given figure is covered with cement. If it costs ` 84 to: cement an area of 2

3 cm , find the total cost of cementing.

Solution:

Area of larger rectangle

= (32x25) cm^{2} = 800cm^{2}

Also, area of smaller rectangle = 485 cm^{2
}Area of shaded part 22

= (800 - 485) cm = 315cm^{2}

Now,

Cost of cementing of 3cm^{2} = Rs.84

Cost of cementing of 1 cm^{2}

= (84/3)

= Rs.28

Cost of cementing of 315 cm^{2}

= Rs. (28x315) = Rs.8820

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