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Perimeter of the given figure
=(5+ 6 + 4 + 4 +5+8+3+ 2+12)cm
= 49 cm
Find the total area of the shaded parts of the rectangle.
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}
The given figure is made up of two rectangles Find its total area.
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}
Area of 1 small square = 3x3 = 9 cm^{2}
Number of shaded squares =
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
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?
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
In the given figure. PQRS is a square and TUVW is a rectangle.
Find the remaining area if TUVW was removed from the figure.
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
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?
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
Find the area of the shaded part in the given figure.
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
The shaded part in the given figure is covered with cement. If it costs Rs 84 to cement an area of 3 cm^{2} , find the total cost of cementing.
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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