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## Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.1

Question 1.

The table given below contains some measures of the rectangle. Find the unknown values.

Solution:

(i) Area of the rectangle = (length × breadth) sq unit.

Perimeter of a rectangle = 2(1 + b) units.

l = 5 cm

b = 8 cm

∴ p = 2 (l + b) cm = 2 (5 + 8) cm = 2 × 13 cm

p = 26 cm

Area = (l × b) cm^{2} = (5 × 8) cm^{2}

A = 40 cm^{2}

(ii) l = 13 cm

p = 54 cm

Perimeter = 2 (l + b) units

54 = 2 (13 + b) cm

\(\frac{54}{2}\) = 13 + b

27 = 13 + b

b = 27 – 13

b = 14 cm

Area = l × b sq. unit = 13 × 14 cm^{2}

A = 182 cm^{2}

(iii) b = 15 cm

p = 60 cm

p = 2 (l + b) units

60 = 2 (l + 15) cm

\(\frac{60}{2}\) = l + 15

30 = l + 15

l = 30 – 15 .

l = 15 cm

Area = l × b unit^{2} = 15 × 15 cm^{2} = 225 cm^{2}

A = 225 cm^{2}

(iv) l = 10 m

Area = 120 sq metre

Area = l × b sq.m

120 = 10 × 6

b = \(\frac{120}{10}\)

b = 12 m

Perimeter =2 (l + b) units = 2(10 + 12) units = 2 × 22 m

A = 44 m

(v) b = 4 feet.

Area = 20 sq. feet

Area = l × b sq .feet

20 = l × 4

l = \(\frac{20}{4}\) feet

l = 5 feet

Perimeter = 2 (l + b) units.

p = 2 (5 + 4) feet = 2 × 9

p = 18 feet

Completing the unknown values in the table.

Question 2.

The table given below contains some measures of the square. Find the unknown values.

Solution:

Perimeter of a square = (4 × side) units

Area of a square = (side × side) unit^{2}

(i) s = 6 cm

Perimeter = 4s units = 4 × 6 cm = 24 cm

P = 24 cm

Area = s × s unit^{2} = 6 × 6 cm^{2} = 36 cm^{2}

A = 36 cm^{2}

(ii) Perimeter = 4 × s unit

100 = (4 × s) m

\(\frac{100}{4}\) = s

s = 25 m

Area = s × s unit^{2}= 25 × 25 m^{2} = 625m^{2}

A = 625m^{2}

(iii) Area = s × s unit^{2}

49 = s × s square feet

s^{2} = 7^{2}

s = 7 feet

Perimeter = 4 × s unit = 4 × 7 feet = 28 feet

Perimeter = 28 feet

Completing the unknown values in the table

Question 3.

The table given below contains some measures of the triangle. Find the unknown values.

Solution:

Area of the right triangle = \(\frac{1}{2}\) × (base × height) unit^{2}

(i) b = 20 cm

h = 40 cm

Area = \(\frac{1}{2}\) (b × h) cm^{2} = \(\frac{1}{2}\) × 20 × 40 = 400 cm^{2}

A = 400 cm^{2}

(ii) b = 5 feet

Area = \(\frac{1}{2}\) × b × h unit^{2}

= 20 = \(\frac{1}{2}\) × 5 × h sq. feet

\(\frac{20 \times 2}{5}\) = h

h = 8 feet

(iii) Area = \(\frac{1}{2}\) × (base × height) unit^{2}

24 = \(\frac{1}{2}\) × b × 12 m^{2}

base = \(\frac{24 \times 2}{12}\) m = 4 m

Base = 4m

Tabulating the unknown values

Question 4.

The table given below contains some measures of the triangles. Find the unknown values.

Solution:

Perimeter of a triangle = sum of three sides.

(i) Perimeter = 6 + 5 + 2 cm = 13 cm

p = 13 cm

(ii) Perimeter = (side 1 + side 2 + side 3) m

17 = (side 1 + 8 + 3) m

17 m = (side 1 + 11) m

side 1 = 17 – 11 = 6m

(iii) Perimeter = side 1 + side 2 + side 3

28 feet = 11 feet + side 2 + 9 feet

28 ft = 20 feet + side 2

28 – 20 = side 2

side = 8 feet

Tabulating the unknowns.

Question 5.

Fill in the blanks.

i) 5 cm^{2} = mm^{2}

Hint: 1 cm^{2} = 100 mm^{2}

ii) 26 m^{2} = cm^{2}

Hint: 1 m^{2} = 10000

iii) 8 km^{2} = m^{2}

Hint 4 1 km^{2}– 1000000 m^{2}

Solution:

(i) 500

(ii) 2,60,000

(iii) 80,00,000

Question 6.

Find the perimeter and area of the following shapes.

Solution:

(i) Perimeter = (4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4) cm = 48 cm

Perimeter = 48 cm

Area of 5 squares of side 4 cm

Area of a square = (side × side) unit^{2}

∴ A = 5 × (4 × 4) cm^{2} = 5 × 16 cm2 = 80 cm^{2}

80 cm^{2}

(ii) Perimeter = (4 + 5 + 4 + 5 + 4 + 5 + 4 + 5)cm = 36cm

Perimeter = 36 cm

Area of a square of side 3cm + Area of 4 right triangles

= (3 × 3) + [4 × \(\frac{1}{2}\) × 4 × 3] cm^{2} = (9 + 24) cm^{2} = 33 cm^{2}

Area = 33 cm^{2}

(iii) Perimeter = (50 + 12 + 13 + 40 + 10 + 10 + 10 + 5) cm = 150 cm

Perimeter = 150 cm

Area = Area of a rectangle + Area of a square + Area of a right triangle.

= (l × b) + (s × s) + ( \(\frac{1}{2}\) × b × h) cm^{2}

= (50 × 5) + (10 × 10) + \(\frac{1}{2}\) × 12 × 5) cm^{2}

= (250 + 100 + 30) cm^{2} = 380 cm^{2}

Area = 380 cm^{2}

Question 7.

Find the perimeter and area of the rectangle whose length is 6 m and breadth 4 m.

Solution:

Perimeter of a rectangle P = 2(1 + b) units.

Length l = 6m

breadth b = 4 m

p = 2 × (6 + 4)m = 2 × 10 m

Perimeter = 20 m

Area = l × b unit^{2} = 6 × 4 m^{2} = 24 m^{2}

Area = 24 m^{2}

Question 8.

Find the perimeter and the area of the square whose side is 8 cm.

Solution:

Perimeter of a square = (4 × side) units

Side = 8 cm

∴ Perimeter = 4 × 8 cm = 32 cm

Perimeter = 32 cm

Area of a square = (side × side) unit^{2} = (8 × 8) cm^{2} = 64 cm^{2}

Area = 64 cm^{2}

Question 9.

Find the perimeter and area of the right angled triangle whose sides are 6 feet, 8 feet and 10 feet.

Solution:

Perimeter of a right angled triangle = sum of three sides

= (6 + 8 + 10) feets = 24 feets.

Area of a right angled triangle = \(\frac{1}{2}\) × (b × h) unit^{2}

Here the longest side is 10 feet which is the hypoteneous

∴ Sides containing right angle are 6 feet and 8 feet

let base = 6 feet and height = 8 feet

∴ Area = \(\frac{1}{2}\) × 6 × 8 feet^{2} = 24 sq. feet

Area = 24 sq feet.

Question 10.

Find the perimeter of

i) A scalene triangle with sides 7 m, 8 m, 10 m

ii) An isosceles triangle with equal sides 10 cm each and third side is 7 cm.

iii) An Equilateral triangle with a side of 6 cm.

Solution:

i) Perimeter of a scalene triangle = (7 + 8 + 10) m = 25 m

ii) The three sides of the isosceles triangle are 10 cm, 10 cm and 7 cm

∴ Perimeter = (10 + 10 + 7) cm = 27 cm

iii) An equilateral triangle with a side of 6 cm.

The sides of equilateral triangle are 6 cm, 6 cm and 6 cm

∴ Perimeter = (6 + 6 + 6) cm = 18 cm

Question 11.

The area of a rectangular shaped photo is 820 sq. cm. and its width is 20 cm. What is its length? Also find its perimeter.

Solution:Area of a rectangle = (length × breadth) unit^{2}

Here width = 20 cm (breadth)

Area = 820 sq. cm

∴ 820 sq.cm = (length × 20) cm^{2}

Length = \(\frac{820}{20}\) =41 cm

Length of the photo = 41 cm

Perimeter of a rectangle = 2(l + b) units = 2 (41 + 20) cm = 2 × 61 = 122 cm

Perimeter of the photo = 122 cm

Question 12.

A square park has 40 m as its perimeter. What is the length of its side? Also find its area.

Solution:

Given perimeter = 40 m

Perimeter of a square = 4 × Length of a side

40 = 4 × Length of a side

∴ Length of its side = \(\frac{40}{4}\) m = 0 m

∴ Side of the park = 10m

Area of a square = (Side × side) unit^{2} = (10 × 10) m^{2} = 100 m^{2}

∴ Area of the Park = 100 m^{2}

Question 13.

The scalene triangle has 40 cm as its perimeter and whose two sides are 13 cm and 15 cm, find the third side.

Solution:

Given two sides of a scalene triangle are 13 cm and 15 cm

The perimeter of the triangle = sum of three sides

40 = 13 + 15 + Third side

40 = 28 + Third side

∴ Third side = 40 – 28 = 12 cm

∴ The third side of the triangle = 12 cm

Question 14.

A field is in the shape of right angled triangle whose base is 25 m and height 20 m. Find the cost of levelling the field at the rate of ₹ 45/- per sq. m.

Solution:

Area of a right angled triangle = \(\frac{1}{2}\) × (base × height) unit^{2}

base = 25 m

height = 20 m

∴ Area = \(\frac{1}{2}\) × (25 × 20)

Area = 250 m^{2}

Cost of levelling per m^{2} = ₹ 45.

∴ Cost of levelling 250 m^{2} = 250 × 45 = ₹ 11,250

Cost of levelling = ₹ 11,250

Question 15.

A square of side 2 cm is joined with a rectangle of length 15 cm and breadth 10 cm. Find the perimeter of the combined shape.

Solution:perimeter of the combined figure

= (2 + 15 + 10 + 15 + 8 + 2 + 2) cm

= 54 cm

Objective Type Questions

Question 16.

The following figures are of equal area. Which figure has the least perimeter?

Solution:

Hint:

(a) 12 units

(b) 10 units

(c) 12 units

(d) 12 units

Question 17.

If two identical rectangles of perimeter 30 cm are joined together, then the perimeter of the new shape will be

(a) equal to 60 cm

(b) less than 60 cm

(c) greater than 60 cm

(d) equal to 45 cm

Solution:

(b) less than 60 cm

Hint:

Question 18.

If every side of a rectangle is doubled, then its area becomes times.

(a) 2

(b) 3

(c) 4

(d) 6

Solution:

(c) 4

Question 19.

The side of a square is 10 cm. If its side is tripled, then by how many times will its perimeter increase?

(a) 2 times

(b) 4 times

(c) 6 times

(d) 3 times

SolutionL

(d) 3 times

30 × 4 = 120 = 3 × 40

Question 20.

The length and breadth of a rectangular sheet of paper are 15 cm and 12 cm respectively. A rectangular piece is cut from one of its corners. Which of the following statement is correct for the remaining sheet?

(a) Perimeter remains the same but the area changes

(b) Area remains the same but the perimeter changes

(c) There will be a change in both area and perimeter

(d) Both the area and perimeter remains the same

Solution:

(c) There will be a change in both area and perimeter

Hint: