9. Perimeter and Area

Solutions to Ex. 9.1 

Q1. Find the area of each of the following parallelograms:

Ans:

(a) Given: Base (b) = 7 cm

Height (h) = 4 cm

Area = b X h = 7 X 4

Area = 28 cm²

(b) Given: Base (b) = 5 cm

Height (h) = 3 cm

Area = b X h = 5 X 3

Area = 15 cm²

(c) Given: Base (b) = 2.5 cm

Height (h) = 3.5 cm

Area = b X h = 2.5 X 3.5

Area = 8.75 cm²

(d) Given: Base (b) = 5 cm

Height (h) = 4.8 cm

Area = b X h = 5 X 4.8

Area = 24 cm²

(e) Given: Base (b) = 2 cm

Height (h) = 4.4 cm

Area = b X h = 2 X 4.4

Area = 8.8 cm²

Q2. Find the area of each of the following triangles:

Ans:

(a) Given: Base (b) = 4 cm

Height (h) = 3 cm

Area = 1/2 X b X h = 1/2 X 4 X 3

Area = 6 cm²

(b) Given: Base (b) = 5 cm

Height (h) = 3.2 cm

Area = 1/2 X b X h = 1/2 X 5 X 3.2

Area = 8 cm²

(c) Given: Base (b) = 3 cm

Height (h) = 4 cm

Area = 1/2 X b X h = 1/2 X 3 X 4

Area = 6 cm²

(d) Given: Base (b) = 3 cm

Height (h) = 2 cm

Area = 1/2 X b X h = 1/2 X 3 X 2

Area = 3 cm²

Q3. Find the missing values:

Ans:

(a) Given: Base (b) = 20 cm

Area = 246 cm²

Area = b X h 

246 = 20 X h

Height = \( \frac{246}{20} \) = 12.3 cm

(b) Given: height (h) = 15 cm

Area = 154.5 cm²

Area = b X h 

154.5 = b X 15

Base = \( \frac{154.5}{15} \) = 10.3 cm

(c) Given: height (h) = 8.4 cm

Area = 48.72 cm²

Area = b X h 

48.72 = b X 8.4

Base = \( \frac{48.72}{8.4} \) = 5.8 cm

(d) Given: Base (b) = 15.6 cm

Area = 16.38 cm²

Area = b X h 

16.38 = 15.6 X h

Height = \( \frac{16.38}{15.6} \) = 1.05 cm

Thus,

Q4. Find the missing values:

Ans:

(a) Given: Base (b) = 15 cm

Area = 87 cm²

Area = 1/2 X b X h 

87 = 1/2 X 15 X h

Height = \( \frac{87 \times 2}{15} \) = 11.6 cm

(b) Given: Height (h) = 31.4 mm

Area = 1256 mm²

Area = 1/2 X b X h 

1256 = 1/2 X b X 31.4 

Base = \( \frac{1256 \times 2}{31.4} \) = 80 mm

(c) Given: Base (b) = 22 cm

Area = 170.5 cm²

Area = 1/2 X b X h 

170.5 = 1/2 X 22 X h

Height = \( \frac{170.5 \times 2}{22} \) = 15.5 cm

Thus,

Q5. PQRS is a parallelogram (Fig 9.14). QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find: (a) the area of the parallelogram PQRS (b) QN, if PS = 8 cm

Ans:

(a) Given: Base SR (b) = 12 cm

        Height QM (h) = 7.6 cm

Area = b X h = 12 X 7.6

Area of parallelogram PQRS = 91.2cm²

(b) Given: Area = 91.2 cm²

        Base PS (b) = 8 cm

Area = b X h 

91.2 = 8 X h

h = \( \frac{91.2}{8} \) 

QN = 11.4cm

Q6. DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (Fig 9.15). If the area of the parallelogram is 1470 cm², AB = 35 cm and AD = 49 cm, find the length of BM and DL.

Ans:

Given: Area = 1470 cm²

   Base AB = 35 cm

   Base AD = 49 cm

Area = base  X height 

Area = AD  X  BM

1470 = 49 X BM

BM = \( \frac{1470}{49} \) 

BM = 30 cm

Area = AB X DL

1470 = 35 X DL

DL = \( \frac{1470}{35} \) 

DL = 42 cm

Q7. ∆ABC is right angled at A (Fig 9.16). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, Find the area of ∆ABC. Also find the length of AD.

Ans:

Given: ∆ABC is right angled ∆

   AB = 5 cm, BC = 13 cm,  AC = 12 cm

So, AC is the height for base AB.

Thus, Area of ∆ABC = 1/2 X base (AB) X height (AC)

Area of ∆ABC = 1/2 X 5 X 12

Area of ∆ABC = 30 cm²

Area of ∆ABC = 1/2 X base (BC) X height (AD)

30 = 1/2 X 13 X AD

AD = \( \frac{30 \times 2}{13} \)

AD = \( \frac{60}{13} \) cm

Q8. ∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 9.17). The height AD from A to BC, is 6 cm. Find the area of ∆ABC. What will be the height from C to AB i.e., CE?

Ans: 

Given: ∆ABC is isosceles ∆

   AB = AC = 7.5 cm,

   BC = 9 cm,  AD = 6 cm

Thus, Area of ∆ABC = 1/2 X base (BC) X height (AD)

Area of ∆ABC = 1/2 X 9 X 6

Area of ∆ABC = 27 cm²

Area of ∆ABC = 1/2 X base (AB) X height (CE)

27 = 1/2 X 7.5 X CE

CE = \( \frac{27 \times 2}{7.5} \)

CE = 7.2 cm

Solutions to Ex. 9.2

Q1. Find the circumference of the circles with the following radius: (Take π = 22/7)

Ans:

(a) Radius (r) = 14 cm

Circumference = 2πr

      = 2 X \( \frac{22}{7} \) X 14

Circumference = 88 cm

(b) Radius (r) = 28 mm

Circumference = 2πr

      = 2 X \( \frac{22}{7} \) X 28

Circumference = 176 mm

(c) Radius (r) = 21 cm

Circumference = 2πr

      = 2 X \( \frac{22}{7} \) X 21

Circumference =132 cm

Q2. Find the area of the following circles, given that: (Take π = 22/7)

Ans:

(a) Radius (r) = 14 mm

Area = πr²

         = \( \frac{22}{7} \) X 14 X 14

Area = 616 mm²

(b) Diameter (d) = 49 cm 

Radius = d / 2 = \( \frac{49}{2} \) cm

Area = πr²

         = \( \frac{22}{7} \) X \( \frac{49}{2} \) X \( \frac{49}{2} \)

Area = 1886.5 cm²

(c) Radius (r) = 5 cm

Area = πr²

         = \( \frac{22}{7} \) X 5 X 5

Area = \( \frac{550}{7} \) cm²

Q3. If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π = 22/7)

Ans:

Given: Circumference = 154 m

Circumference = 2πr

154 = 2 X \( \frac{22}{7} \) X r

r = \( \frac{154 \times 7}{2 \times 22} \)

Radius = 24.5 m

Area = πr²

         = \( \frac{22}{7} \) X 24.5 X 24.5

Area = 1886.5 m²

Q4. A gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs ₹4 per meter. (Take π = 22/7)

Ans:

Given: Diameter (d) = 21 m

Circumference = πd

      =  \( \frac{22}{7} \) X 21

      = 66 m

Length of rope = 2 X 66 = 132 m

Cost of the rope = 132 X 4 = ₹528

Q5. From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)

Ans:

Given: Radius of sheet (r₁) = 4 cm

  Radius of circle (r₂) = 3 cm

Area = πr²

A₁ = 3.14 X 4² = 50.24 cm²

A₂ = 3.14 X 3² = 28.26 cm²

Area of remaining sheet = A1 - A₂

   = 50.24 - 28.26

Area of remaining sheet = 21.98 cm²

Q6. Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs ₹15. (Take π = 3.14)

Ans:

Given: Diameter (d) = 1.5 m

Circumference = πd

      = 3.14 X 1.5

      = 4.71 m

Length of rope = 4.71m

Cost of the rope = 4.71 X 15 = ₹70.65

Q7. Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Ans:

Given: Diameter (d) = 10 cm

Circumference of semicircle = πd / 2

= 3.14 x 10 / 2 = 15.7 cm

Perimeter = circumference of semicircle + diameter

Perimeter = 15.7 + 10 = 25.7 cm

Q8. Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹15/m². (Take π = 22/7)

Ans:

Given: Diameter (d) = 1.6 m

Radius = d / 2 = \( \frac{1.6}{2} \) = 0.8 m

Area = πr²

        = \( \frac{22}{7} \) X 0.8²

Area of table-top = 2.01 m²

Cost of polishing = 2.01 X 15 = ₹30.15

Q9. Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take π = 22/7)

Ans:

Given: Perimeter = Circumference = 44 cm

Circumference = 2πr

44 = 2 X \( \frac{22}{7} \) X r

r = \( \frac{44 \times 7}{2 \times 22} \)

Radius of the circle = 7 cm

Area = πr²

         = \( \frac{22}{7} \) X 7 X 7

Area of the circle = 154 cm²

Perimeter of square = 44 cm = 4 X length

So, length of each side of square = 44 / 4 = 11cm

Area of square = length² = 11² = 121 cm²

Thus, the circle encloses more area than the square.

Q10. From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. (Take π = 22/7)

Ans:

Given: Radius = 14 cm

Radius of 2 circles = 3.5 cm

Length of rectangle = 3 cm

Breadth of rectangle = 1 cm

Area of circle = πr²

                     = \( \frac{22}{7} \) X 14 X 14

Area of the circular sheet = 616 cm²

Area of 2 circles = 2 X  \( \frac{22}{7} \) X 3.5 X 3.5

Area of 2 circles = 77 cm²

Area of rectangle = l X b = 3 X 1 = 3 cm²

Area of remaining sheet = Area of circular sheet - (area of rectangle + area of 2 circles)

   = 616 - (77 + 3) = 616 -80

Area of remaining sheet  = 536 cm²

Q11. A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take π = 3.14)

Ans:

Given: Side of square sheet = 6 cm

   Radius of circle = 2 cm

Area of square sheet = (side)² = 6²

Area of square sheet = 36 cm²

Area of circle = πr²

                     = 3.14 X 2 X 2

Area of circle = 12.56 cm²

Area of left over sheet = Area of square sheet - Area of circle

= 36 - 12.56

Area of left over sheet = 23.44 cm²

Q12. The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take π = 3.14)

Ans:

Given: Circumference = 31.4 cm

Circumference = 2πr

31.4 = 2 X 3.14 X r

r = \( \frac{31.4}{2 \times 3.14} \)

Radius of the circle = 5 cm

Area = πr²

         = 3.14 X 5 X 5

Area of the circle = 78.5 cm²

Q13. A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)

Ans:

Given: Diameter of the flower bed = 66 m

Radius of the flower bed (r) = d/2 = 66/2

r = 33 m

Area of flower bed = πr² = 3.14 X 33²

Area of flower bed = 3419.46 m²

Radius of flower bed & path = 33 + 4 = 37 m

Area = πr² = 3.14 X 37²

Area of flower bed & path = 4298.66 m²

Area of the path = Area of the flower bed & path – Area of the flower bed

= 4298.66 – 3419.46

Area of the path = 879.20 m²

Q14. A circular flower garden has an area of 314 m². A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden? (Take π = 3.14)

Ans:

Given: Area of the circular flower garden = 314 m²

  Radius of sprinkler = 12 m

Area of the circular flower garden = πr²

314 = 3.14 X r²

r² = \( \frac{314}{3.14} \) = 100

Radius of the circular flower garden = 10 m

As the sprinkler can cover an area of a radius of 12 m.

Hence, the sprinkler will water the whole garden.

Q15. Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)

Ans:

Given: Radius of outer circle = 19 cm

Radius of inner circle = Radius of outer circle – 10

Radius of inner circle = 19 – 10 = 9 cm

Circumference = 2πr

                        = 2 X 3.14 X 9

Circumference of inner circle = 56.52 m

Circumference of the outer circle = 2 X 3.14 X 19

Circumference of the outer circle = 119.32 m

Q16. How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = 22/7)

Ans:

Given: Radius of the wheel = 28 cm

   Distance covered = 352 m = 35200 cm

Circumference = 2πr

      = 2 X \( \frac{22}{7} \) X 28

Circumference of the wheel = 176 cm

Number of rotations = Total distance \( \div \) Circumference of the wheel

                        = 35200 \( \div \) 176 

Number of rotations = 200

thus, the wheel should rotate 200 times.

Q17. The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour. (Take π = 3.14)

Ans:

Given: Length of the minute hand = 15 cm

Distance of the tip of minute hand in 1 hour = Circumference of minute hand

Distance = 2πr

     = 2 × 3.14 × 15 = 94.2 cm

Thus, the tip of the minute hand moves 94.2 cm in 1 hour.