RBSE Solutions for Class 7 Maths Chapter 16 Perimeter and Area In Text Exercise is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 16 Perimeter and Area In Text Exercise.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 7 |
| Subject | Maths |
| Chapter | Chapter 16 |
| Chapter Name | Perimeter and Area |
| Exercise | In Text Exercise |
| Number of Questions | 9 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 7 Maths Chapter 16 Perimeter and Area In Text Exercise
(Page 182)
Question 1
Take a graph paper and draw different (RBSESolutions.com) triangles on it by taking base and height of the same measure.
Take three triangles of measure ABC, A’BC and A”BC and observe.
Number of squares surrounded by three squares are same, which means that area of all three triangles are same.
Solution:
Number of boxes covered by the (RBSESolutions.com) triangles are as follows:
Number of boxes covered by ∆A’BC = 155 sq. unit
Number of boxes covered by ∆A’BC = 155 sq. unit
Number of boxes covered by ∆A”BC = 155 sq. unit
So, if is clear that the area of all three triangles are equal.
Question 2
Can they completely cover each other? Cut and see them.
Solution:
When we cut the triangles and keep on each other then they do not cover each other. If BA and CA” would have same length, then ∆ABC and A”BC will cover each other.
Activity 1
(i) Take is transparent (RBSESolutions.com) paper/sheet.
(ii) Cut the parallelograms of different sizes on it.
(iii) Find the area of these on putting squared block sheet or graph paper.
(iv) Separate a triangular figure by cutting perpendicularly from vertex on the side opposite to the base of parallelogram.
(v) Make a rectangle by putting the separated figure on the other side.
(vi) Find the area of rectangle thus formed from the graph paper/squared block sheet.
(vii) Compare the areas of parallelogram and rectangle.
(viii) Here both the area are same.
Process of Activity
The number of covered boxes on graph (RBSESolutions.com) paper by the figures A and B are as follows :
Area of parallelogram (A) = 12 square units
Area of parallelogram (B) = 30 square units
Area of figures A and B are as follows :
Area of rectangle (A) = 12 square units
Area of rectangle (B) = 30 square units
Ratio of area of rectangle and parallelogram
It is clear that area of parallelogram and rectangle is equal.
Activity 2
(Page 190)
Compare the area of circle and to find (RBSESolutions.com) out the area of rectangle made by circle.
Steps of Activity :
Step 1. Draw a circle on paper whose radius is 6.4 cm.
Step 2. Shaded it’s half portion with the pencil.
Step 3. It is in folder for next 6 times.
Step 4. It is cut in 64 pieces in form of strips.
Step 5. Arrange these strips as shown in figure.
Process of Activity
Radius of circle = 6.4 cm
Circumference (RBSESolutions.com) of circle = 2π x radius
= 2 x 3.14 x 6.4 = 40.19 cm ≃ 40 cm
Area of circle = π x (radius)2
= 3.14 x (6.4)2 = 3.14 x 40.96 = 128.61 cm2
So length of rectangle obtained such a way = 19.8 cm ≃ 20 cm
Breadth of rectangle = 6.4 cm
Question
Can you tell the formula of area (RBSESolutions.com) of this circle?
Solution:
For obtained rectangle
Length of rectangle = \(\frac { 1 }{ 2 }\) x circumference
⇒ Area of rectangle = length x breadth
= \(\frac { 1 }{ 2 }\) x circumference x breadth
If radius of circle is r, then
Area = \(\frac { 1 }{ 2 }\) x 2πr x r = πr2.
Do and Learn
Question 1
The figures of plates showing registration number is given below. Calculate (RBSESolutions.com) the perimeter by measuring length and breadth of plates of bus, taxi and private vehicles around
Solution:
Length of number plate in front of bus = 3 cm
Breadth = 1.6 cm
Perimeter of rectangular (RBSESolutions.com) size number plate of bus = 2 (length + breadth)
= 2 (3 + 1.6) cm = 2 (4.6) = 2 x 4.6
= 9.2 cm
Taxi number plate is also in same size as bus so
perimeter of taxi number plate is also 9.2 cm
Length of number plate of private vehicle is = 6.2 cm
Breadth = 1.6 cm
Perimeter of private vehicle number plate is
= 2 (length + breadth) = 2(6.2 + 1.6) 2(7.8)
= 15.6 cm
Question 2
In the following case what do we (RBSESolutions.com) require to find perimeter or area?
(i) Stitching the lace on the edges of dupatta.
(ii) Putting the black soil in hockey ground.
(iii) Filling the ceiling of room.
(iv) Fencing around the farm
Solution:
(i) Perimeter
(ii) Area
(iii) Area
(iv) Perimeter
(Page 182)
Question
Draw parallelograms of different measures. Cut them (RBSESolutions.com) along any diagonal and make two triangles.
- Are both the triangles congruent in every situation?
- Area of two congruent triangles always equal?
- Is the converse of it always true ?
Solution:
Let ABCD is a parallelogram by cutting the parallelogram in the (RBSESolutions.com) direction of diagonal AC or BC it is divided into two triangles.
- These triangles are congruent.
- Two congruent triangles are having same area.
- Yes, the converse of it is always true.
(Page 193)
Question
To show traffic symbols of 5 circular discs are prepared by (RBSESolutions.com) cutting and iron sheet Radius of all discs is 21 cm.
Find out the meaning of all these symbols with the help of your (RBSESolutions.com) teacher and find out circumference and area of discs.
Solution:
The first symbol represents the restriction on foot movement and second symbol represents the restriction of smoking.
Third symbol represents the restriction of truck. Fourth symbol represents the restriction on horn. Fifth symbol represents the maximum speed limit of 50 km/hour.
Radius of circle = 21 cm
Circumference of the circle = 2πr ( where π = \(\frac { 22 }{ 7 }\) )
= 2 x \(\frac { 22 }{ 7 }\) x 21 = 2 x 22 x 3
= 44 x 3 = 132 cm.
Area of the circle = πr2
= \(\frac { 22 }{ 7 }\) x 21 x 21 = 22 x 3 x 21
= 1386 cm2.
We hope the RBSE Solutions for Class 7 Maths Chapter 16 Perimeter and Area In Text Exercise will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 16 Perimeter and Area In Text Exercise, drop a comment below and we will get back to you at the earliest.
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