How Many Square Test 2

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How Many Square Test 2

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Question 1
1 / -0

Find the perimeter of the given figure.

A
42 cm

B
45 cm

C
49 cm

D
53 cm

Solution

Perimeter of the given figure
=(5+ 6 + 4 + 4 +5+8+3+ 2+12)cm
= 49 cm
Question 2
1 / -0

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

A
196 cm²

B
148cm²

C
200 cm²

D
184 cm²

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)cm²
= 240 cm²
Area of shaded part of rectangle
= 240/2 = 120 cm²
Total shaded area = (64+ 120) cm²
=184 cm²
Question 3
1 / -0

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

A
144 cm²

B
216 cm²

C
240 cm²

D
300 cm²

Solution

Area of rectangle ABCD = (18x8) cm²
=144 cm²

Area of rectangle GEFC = (12x6) cm²
= 72 cm²

∴ Total area = (144 + 72) cm² = 216 cm²
Question 4
1 / -0

Find the area of the shaded figure.

A
49.5 cm²

B
36.5 cm²

C
24.5 cm²

D
30 cm²

Solution

Area of 1 small square = 3x3 = 9 cm²
Question 5
1 / -0

Which shape has the largest area?

A
Q

B
R

C
P

D
S

Solution

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
1 / -0

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?

A
39 m

B
42 m

C
35 m

D
53 m

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
1 / -0

Directions For Questions

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

...view full instructions

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

A
716 sq. cm

B
784 sq. cm

C
700 sq. cm

D
768 sq. cm

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
1 / -0

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?

A
50 cm

B
75 cm

C
80 cm

D
95 cm

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
1 / -0

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

A
139 sq. cm

B
118 sq. cm

C
126 sq. cm

D
152 sq. cm

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
1 / -0

The shaded part in the given figure is covered with cement. If it costs Rs 84 to cement an area of 3 cm² , find the total cost of cementing.

A
Rs. 8820

B
Rs. 8125

C
Rs. 8610

D
Rs.8804

Solution

Area of larger rectangle
= (32x25) cm² = 800cm²
Also, area of smaller rectangle = 485 cm²
∴Area of shaded part 22
= (800 - 485) cm = 315cm²
Now,
Cost of cementing of 3cm² = Rs.84
∴ Cost of cementing of 1 cm²
= (84/3)
= Rs.28
∴ Cost of cementing of 315 cm²
= Rs. (28x315) = Rs.8820