# Grade 7 Perimeter and Area Worksheet (For CBSE, ICSE, IAS, NET, NRA 2022)

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## (10) the Adjacent Sides of a Parallelogram Are 20 M and 8 M. If the Distance between Its Longer Sides is 5 M, Find the Distance between Its Shorter Sides

(a) What is the area of triangle whose base is 8cm and altitude is 18 cm?

(b) What is the side length of equilateral triangle having area of ?

## (15) Find the Circumference of the Circle Whose

(b) Diameter is 3.5m

(d) Diameter is 42cm

## (16) Find the Area of Circle Whose

(b) Diameter 2.1m

## (17) Find the Radius of the Circle Whose

(a) Circumference is 1.76m

(b) Area is 154 cm2.

## (19) if the Area of Circle is 169 Times the Area of Another Circle, What is Ratio of Their Circumference?

• Perimeter means length of outline of given figure.
• To find out perimeter of given figure first we have to find out missing dimension of figure as below,
• Missing dimension represent if red color in above figure.
• Now perimeter of given figure

• Therefore Perimeter of given figure is

• Perimeter means length of outline of given figure.
• To find out perimeter of given figure first we have to find out missing dimension of figure as below,
• Missing dimension represent if red color in above figure.
• Now perimeter of given figure

• Therefore Perimeter of given figure is

• To find out the area of given figure, first divide the figure in part of rectangle, than find out area of individual rectangle and then do summation area of all rectangle.
• Given figure divide in part of rectangle as below;
• Now Area of Rectangle-1

• Area of Rectangle-2

• Area of Rectangle-3

• Now area of Figure

• Therefore Area of given figure is

• To find out the area of given figure, first divide the figure in part of rectangle, than find out area of individual rectangle and then do summation area of all rectangle.
• Given figure divide in part of rectangle as below;
• Area of Rectangle-1

• Area of Rectangle-2

• Area of Rectangle-3

• Now area of Figure

• Therefore Area of given figure is

• Area of shaded region can be find out by calculating the area of all rectangle and then substitute the area of unshaded region from this.

• Now as shown in figure;

• Now,

• Area of shaded region can be find out by calculating the area of full figure and then substitute the area of unshaded region from this.
• To find out area of full figure consider below figure;

• Now,

• From figure Area of unshaded rectangle

• Now,

• Therefore Area of shaded region is

• Length of rectangular farm house is 60m
• Breadth of rectangular farm house is 40m

Therefore

• Now,

• Therefore, cost of fencing of farm house is ₹ 8000

• Given question can be represent as below.
• Where the door is installed; not able to paint the wall, hence can paint the wall accept door area.
• Now;

• Now,

• Therefore,

• Now,

• Therefore Cost of painting the wall is ₹ 810

• Base (b) of parallelogram
• Height (h) of parallelogram

• Therefore Area of parallelogram is

• Base (b) of parallelogram
• Height (h) of parallelogram

• Therefore Area of parallelogram is

• Base (b) of parallelogram
• Height (h) of parallelogram

• Therefore Area of parallelogram is

• Base (b) of parallelogram
• Height (h) of parallelogram

• Therefore Area of parallelogram is

• Area of parallelogram
• Height of parallelogram
• Now,

• Therefore Base of parallelogram is 23 cm

• Given problem represented in figure as below,
• From figure
• Base of parallelogram
• Height of parallelogram
• Now,

• Now, for other side
• Base of parallelogram
• Height of parallelogram
• Area of parallelogram
• Now,

• Therefore, Distance between shorter sides

• Given that Base of triangle
• Altitude of triangle height of triangle
• Now

• Therefore, Area of triangle

• Suppose Side length of equilateral triangle
• Area of equilateral triangle
• Formula for area of equilateral triangle is;

• Therefore, The equilateral triangle whose area is have side length

• Given figure is made of two figure as shown in below figure
• One is rectangle and other is triangle

• For rectangle
• Length of rectangle

• Now for triangle
• Base of triangle
• Height of triangle

• Now,

• Therefore Area of figure

• Given figure have two part, the one full figure out line form a parallelogram and inside parallelogram there is triangle.
• Therefore,

• Now for area of parallelogram
• Base of parallelogram
• Height of parallelogram

• Now for area of triangle
• Base of triangle
• Height of triangle

• Now,

• Therefore, Area of shaded region of figure

• Given that the base of a triangular field is two times its altitude
• Suppose altitude of triangle
• Than Base of triangle
• Rate of cultivating the field is 18 per sq. m
• And total cost of cultivating the triangle shape field is ₹ 216.
• Now,

• Given that field is of triangular shape
• Therefore Area of triangle Cultivating area

• Therefore;
• Altitude of triangle
• Base of triangle

• Given that Radius of circle is 14 cm

• Therefore Circumference of circle having radius 14 cm is 88cm

• Given that Diameter of circle is 3.5 m

• Therefore Circumference of circle having diameter 3.5 m is 11m

• Given that Radius of circle is 8.4 cm

• Therefore Circumference of circle having radius 8.4 cm is 52.8cm

• Given that Diameter of circle is 42 cm

• Therefore Circumference of circle having diameter 42cm is 132 cm

• Given that radius of circle is 14cm
• Diameter of circle
• Diameter of circle

• Therefore area of circle having radius 14 cm is

• Given that Diameter of circle is 2.1 m

• Therefore area of circle having diameter 2.1 m is

• Circumference of circle is 1.76m

• Therefore Circle whose circumference is 1.76 m have radius 1.12m

• Area of circle is

• Now,
• Therefore Circle whose Area is 154 cm have radius 7 cm

• Diameter of wheel is 42 cm
• In one revolution wheel cover distance equals to its circumference.

• Therefore,

• Therefore wheel cover 15840cm distance in 120 revolution.