Maths Chapter-11: Perimeter and Area

Exercise 1:

Q1. Find

(a) the area of the land

(b) the price of the land,

If the land is of rectangular shape measuring 300 m in length and 400 m in breadth. The cost of 1m² land is Rs. 10000.

Ans: (a) Area = Length (l) x Breadth (b)

= 300 x 400

= 120000 m²

(b) Cost of 1 m² land = Rs 10000

Cost of 120000 m²land = 10000 x 120000 = Rs 1200000000.

Q2. The perimeter of a park is 240m. Determine the area.

Ans: Perimeter = 240 m

4 x Length of a side of the park = 240

Length of a side of park = 240/4 = 60 m

Area = (Length of a side of park)² = (60)² = 3600 m²

Q3. The area and length of a rectangular plot of land are 1200 m² and 40 m respectively. Determine its perimeter.

Ans: Area = Length (l) x Breadth (b) = 1200 m²

= 40 x Breadth = 1200

= Breadth = 1200/40 = 30

Perimeter = 2 (Length (l) + Breadth (b) )

= 2 (30 + 40) = 2 (70) = 140 m

Q4. Find the breadth and area if the perimeter and length of the rectangular sheet is 120 cm and 40 cm respectively.

Ans: Perimeter = 2 (Length + Breadth) = 120 cm

2 (40 + Breadth) = 120

40 + b = 60

b = 6O – 40 =20 cm

Area = Length (l) x Breadth (b) = 40 x 20 = 800 cm²

Q5. The area of the rectangular park is equal to a square park. If the length of rectangular park measures 80m and the side of the square park is 100 m .Determine the breadth of the rectangular park.

Ans: Area of the square park = (side of the park)² = (100)² = 10000 m²

Area of the rectangular park = Length (l) x Breadth (b) = 10000

80 x Breadth = 10000

Breadth = 125 m

Q6. The length and breadth of a wire whose shape is a rectangle are 22 and 40 respectively. The wire is then reshaped into a square. Determine the side of the square and find out which shape encloses more area.

Ans: Perimeter of the rectangle = Perimeter of the square

2 (Length (l) + Breadth (b)) = 4 x Side

2 (22 + 40) = 4 x Side

2 x 62 = 4 x Side

Side =124/4 = 31 cm

Area of the rectangle = 22 x 40 = 880 cm²

Area of the square = (Side)² = 31 x 31 = 961 cm²

Therefore, the square shaped wire encloses more area.

Q7. If the breadth and perimeter of a rectangle are 40 cm and 120 cm respectively, determine the length and area of the rectangle.

Ans: Perimeter = 2 (Length (l) + Breadth (b)) = 120

2 (Length + 40) = 120

Length + 40 = 60

Length = 60 – 40 = 20 cm

Area = Length (l) x Breadth (b) = 20 x 40 = 800 cm²

Q8. A door is fitted in a wall whose length measures 1 m and breadth measures 2 m. The length and breadth of the wall are 5 m and 3.5 m (in the given figure). The price of white washing the wall is Rs 10 per m². Find the cost of whitewashing the wall without the door.

Ans: Area of the wall = 5 x 3.5 = 17.5 m²

Area of the door = 2x 1 = 2 m²

Area to be white-washed = 17.5 – 2 = 15.5 m²

Cost of white-washing 1 m² area = Rs 10

Cost of white-washing 15.5 m² area = 15.5 x 10 = Rs 155

Exercise 2:

Q1. Determine the area of the parallelograms in the diagram given below

Ans:

(a) Area of a parallelogram = Height x Base

Base = 8 cm

Height = 5 cm

Area = 8 x 5 = 40 cm²

(b) Area of a parallelogram = Height x Base

Base = 6 cm

Height = 4 cm

Area = 6 x 4 = 24 cm²

Base = 4.5 cm

Height = 7 cm

Area = 4.5 x 7 = 31.5 cm²

(d) Area of a parallelogram = Height x Base

Base = 9.6 cm

Height = 10 cm

Area = 9.6 x 10 = 96 cm²

(e) Area of a parallelogram = Height x Base

Base = 8.8 cm

Height = 4 cm

Area = 8.8 x 4 = 35.2 cm²

Q2. Determine the area of the triangles .

Ans:

(a) Area of a triangle \(= \frac{1}{2}\times Base \times height\)

Height = 6 cm

Base = 8 cm

Area \(= \frac{1}{2}\times 8 \times 6\) = 48 cm²

(b) Area of a triangle \(= \frac{1}{2}\times Base \times height\)

Height = 6.4 cm

Base = 10 cm

Area \(= \frac{1}{2}\times 10 \times 6.4\) = 64 cm²

Height = 6 cm

Base = 8 cm

Area \(= \frac{1}{2}\times 8 \times 6\) = 48 cm²

(d) Area of a triangle \(= \frac{1}{2}\times Base \times height\)

Height = 4 cm

Base = 6 cm

Area \(= \frac{1}{2}\times 4 \times 6\) = 24 cm²

Q3. Fill the empty cells:

S. No	Base	Height	Area of parallelogram
A	10 cm		124 cm²
B		12 cm	96 cm²
C	7 cm		91 cm²
D	11 cm		121 cm²

Ans:

(a) Area of a parallelogram = Height x Base

B = 10 cm

H =?

Area = 124 cm²

10 x h = 124

\(h = \frac{124}{10}=12.4\)

So, the height of the parallelogram is 12.4 cm

(b) Area of a parallelogram = Height x Base

B =?

H = 12 cm

Area = 96 cm²

b x 12 = 96

\(b = \frac{96}{12}=8\)

So, the base of the parallelogram is 8 cm

B = 7 cm

H =?

Area = 91 cm²

7 x h = 91

\(h = \frac{91}{7}=13\)

So, the height of the parallelogram is 13 cm

(d) Area of a parallelogram = Height x Base

B = 11 cm

H =?

Area = 121 cm²

11 x h = 121

\(h = \frac{121}{11}=11\)

So, the height of the parallelogram is 11 cm

Q4. Fill in the blanks:

Base	Height	Area of triangle
12		24 cm²
	30	60 cm²
22		66 cm²

Ans:

(a) Area of a triangle \(= \frac{1}{2}\times Base \times height\)

Height = ?

Base = 12 cm

Area \(= \frac{1}{2}\times Base \times height\) = 24 cm²

\(\frac{1}{2}\times 12 \times h = 24\) \(h = \frac{24\times 2}{12} = 4 cm\)

Therefore, the height of the triangle is 4 cm

(b) Area of a triangle \(= \frac{1}{2}\times Base \times height\)

Height = 30 cm

Base = ?

Area \(= \frac{1}{2}\times Base \times height\) = 60 cm²

\(\frac{1}{2}\times b \times 30 = 60\) \(h = \frac{60\times 2}{30} = 4 cm\)

Therefore, the base of the triangle is 4 cm

Height =?

Base = 22 cm

Area \(= \frac{1}{2}\times Base \times height\) = 66 cm²

\(\frac{1}{2}\times 22 \times h = 24\) \(h = \frac{66\times 2}{22} = 6 cm\)

Therefore, the height of the triangle is 6 cm.

Q5. ABCD is a parallelogram (in the given figure). BX is the height from B to CD and BY is the height from B to AD. If CD = 12 cm and BX = 7.6 cm.

Find: (a) the area of the parallelogram ABCD

(b) BY, if AD = 8 cm.

Ans:

(a) Area of parallelogram = Base x Height = CD x BX

= 7.6 x 12 = 91.2 cm²

(b) Area of parallelogram = Base x Height = AD x BY = 91.2 cm²

BY x 8 = 91.2

BY = 91.2/8 =11.4 cm

Q6. SL and QM are the heights on sides PQ and PS respectively of parallelogram PQRS (in the given figure). If area of parallelogram is 1470 cm², AB = 35 cm and AD = 49 cm, find the length of BM and DL.

`Ans:

Area of a parallelogram = Height x Base = PQ x SL

1470 = 35 x SL

DL = \(\frac{1470}{35}\) = 42 cm

Also, PS x QM = 1470

1470 = 49 x QM

\(QM = \frac{1470}{35} = 30\ cm\)

Q7. ABC is right angled at A (in the given figure). AD Is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, determine the area of triangle ABC. Also find the length of AD.

Ans:

Area of a triangle \(= \frac{1}{2}\times Base \times height\) \(= \frac{1}{2}\times 5 \times 12\) = 30 cm²

Also, area of triangle \(= \frac{1}{2}\times AD \times BC\)

\( 30 = \frac{1}{2}\times AD \times 13\) \(\frac{30\times 2}{13}= AD\)

AD = 4.6 cm

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

Ans:

Area of triangle ABC \(= \frac{1}{2}\times Base \times height\) \(= \frac{1}{2}\times BC \times AD\)

\(= \frac{1}{2}\times 9 \times 6 = 27 cm² \)

Area of triangle ABC \(= \frac{1}{2}\times Base \times height\) \(= \frac{1}{2}\times AB \times CE\)

\(27= \frac{1}{2}\times 7.5 \times CE\)

CE = 7.2