## RS Aggarwal Class 6 Math Twenty-one Chapter Concept Of Perimeter And Area Exercise 21A Solution

## EXERCISE 21A

**PERIMETER OF A PLANE FIGURE** – *The sum of the lengths of all side of all sides of a plane figure, or the length of its boundary, is called the perimeter of the figure.*

**PERIMETER OF A RECTANGLE** –* Perimeter of a rectangle = 2(length + breadth)*

*=2 (l + b) units*

**PERIMETER OF A SQUARE** – *Perimeter of a square = (4 x side) = 4a units*

**(1) Find the perimeter of a rectangle in which:**

(i) Length = 16.8 cm and breadth = 6.2 cm

Solution: We know, 2(l + b) = 2(16.8 + 6.2) = 2 x 23 = 46 cm

(ii) Length = 2 m 25 cm and breadth = 1 m 50 cm

Solution: Here 2 m 25 cm = 225 cm and 1 m 50 cm = 150 cm

We know, 2(l + b) = 2(225 + 150) = 2 x 375 = 750 cm = 7 m 50 cm

(iii) Length = 8 m 5 dm and breadth = 6 m 8 dm

Solution: Here, 8 m 5 dm = 85 dm and 6 m 8 dm = 68 dm

We know, 2(l + b) =2(85 + 68) = 2 x 153 = 306 dm = 30m 6 dm

**(2) Find the cost of fencing a rectangular field 62 m long and 33 m wide at Rs 16 per meter.**

Solution: Length of the field = 62 m

Breadth of the field = 33 m

Perimeter of the rectangle = 2(l + b) = 2(62 + 33) = 2 x 95 = 190 m

Cost of fencing per meter = Rs 16

Total cost of fencing = Rs (190 x 16) = Rs 3040.

**(3) The length and the breadth of a rectangular field are in the ratio 5 : 3. If its perimeter is 128 m, find the dimensions of the field.**

Solution: Let the length of the rectangle be 5x m.

Then, its breadth = 3x m.

**(4) The cost of fencing a rectangular field at Rs 18 per meter is Rs 1980. If the width of the field is 23 m, find its length.**

Solution: Total cost of fencing = Rs 1980

Rate of fencing = Rs 18 per meter

Hence, the length of the field is 32 m.

**(5) The length and the breadth of a rectangular field are in the ratio 7 : 4. The cost of fencing the field at Rs 25 per meter is Rs 3300. Find the dimensions of the field.**

Solution: Total cost of fencing = Rs 3300

Rate of fencing = Rs 25 per meter

**(6) Find the perimeter of square, each of whose side measures:**

(i) 3.8 cm

Solution: Each side of the square = 3.8 cm

Perimeter of the square = ( 4 x 3.8) = 15.2 cm

(ii) 4.6 m

Solution: Each side of the square = 4.6 cm

Perimeter of the square = ( 4 x 4.6) = 18.4 cm

(iii) 2 m 5 dm

Solution: Here, 2 m 5 dm = 25 dm

Each side of the square = 25 dm

Perimeter of the square = ( 4 x 25) = 100 dm =10 m

**(7) The cost of putting a fence around a square field at Rs 35 per meter is 4480. Find the length of each side of the field.**

Solution: Total cost of fencing = Rs 4480

Rate of fencing = Rs 35 per meter

Hence, the length of each side of the field is 32 m.

**(8) Each side of a square field measures 21 m. Adjacent of this field, there is a rectangular field having its in the ratio 4 : 3. If the perimeters of both the fields are equal, find the dimensions of the rectangular field.**

Solution: Each side of the square = 21 m

Perimeter of the square field = (4 x 21) = 84 m

Let the length of the rectangular field be 4x m.

Then breadth = 3x m.

Perimeter of the rectangular field = 2(4x + 3x) =14x

Perimeter of the rectangular field = Perimeter of the square field

**(9) Find the perimeter of **

(i) a triangle of sides 7.8 cm, 6.5 cm and 5.9 cm,

Solution: Perimeter of the triangle = (7.8 + 6.5 + 5.9) = 20.2 cm

(ii) an equilateral triangle of side 9.4 cm,

Solution: Perimeter of the triangle = (3 x 9.4) = 28.2 cm

(iii) an isosceles triangle with equal sides 8.5 cm each and third side 7 cm,

Solution: Length of the equal side of the triangle = 8.5 cm

Length of the third side of the triangle = 7 cm

Perimeter of the triangle = (8.5 + 8.5 +7) = 24 cm

**(10) Find the perimeter of **

(i) a regular pentagon of side 8 cm,

Solution: Perimeter of the pentagon = 5 x 8 = 40 cm.

(ii) a regular octagon of side 4.5 cm,

Solution: Perimeter of the octagon = 8 x 4.5 = 36 cm.

(iii) a regular decagon of side 3.6 cm.

Solution: Perimeter of the decagon = 10 x 3.6 = 36 cm.

**(11) Find the perimeter of each of the following figures:**

(i) Perimeter of the given figure = (27 + 35 + 35 + 45) = 142 cm

(ii) Perimeter of the given figure = (18 + 18 + 18 +18) = 72 cm

(iii) Perimeter of the given figure = (8 + 16 + 16 + 4 + 4 + 12 + 12) = 72 cm

**For more exercise solutions, Click Below –**

**Exercise – 21B****Exercise – 21C****Exercise – 21D****Exercise – 21E**

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