NCERT Solutions For Class 7 Maths Chapter 11

NCERT Solutions Class 7 Maths Perimeter and Area

NCERT Solutions For Class 7 Maths Chapter 11 Exercises

Exercise 11.1

 

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 m2 land = 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.

8

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 11.2

 

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

Ans:

9

(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

(c) Area of a parallelogram = Height x Base

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:

10

(a) Area of a triangle =12×Base×height

Height = 6 cm

Base = 8 cm

Area =12×8×6 = 48 cm2

(b) Area of a triangle =12×Base×height

Height = 6.4 cm

Base = 10 cm

Area =12×10×6.4 = 64 cm2

(c) Area of a triangle =12×Base×height

Height = 6 cm

Base = 8 cm

Area =12×8×6 = 48 cm2

(d) Area of a triangle =12×Base×height

Height = 4 cm

Base = 6 cm

Area =12×4×6 = 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=12410=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=9612=8

So, the base of the parallelogram is 8 cm

(c) Area of a parallelogram = Height x Base

B = 7 cm

H =?

Area = 91 cm2

7 x h = 91

h=917=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=12111=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 =12×Base×height

Height = ?

Base = 12 cm

Area =12×Base×height = 24 cm2

12×12×h=24 h=24×212=4cm

Therefore, the height of the triangle is 4 cm

(b) Area of a triangle =12×Base×height

Height = 30 cm

Base = ?

Area =12×Base×height = 60 cm2

12×b×30=60 h=60×230=4cm

Therefore, the base of the triangle is 4 cm

(c) Area of a triangle =12×Base×height

Height =?

Base = 22 cm

Area =12×Base×height = 66 cm2

12×22×h=24 h=66×222=6cm

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:

11

(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:

12

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

1470 = 35 x SL

DL = 147035 = 42 cm

Also, PS x QM = 1470

1470 = 49 x QM

QM=147035=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:

13

Area of a triangle =12×Base×height  =12×5×12 = 30 cm2

Also, area of triangle =12×AD×BC

30=12×AD×13 30×213=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:

14

Area of triangle ABC =12×Base×height =12×BC×AD

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

Area of triangle ABC =12×Base×height =12×AB×CE

27=12×7.5×CE

CE = 7.2

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