GSEB Class 7 Maths Solutions Chapter 11 Perimeter and Area InText Questions

Try These (Page 205)

 

Question 1. What would you need to find, area or perimeter, to answer the following?
1. How much space does a blackboard occupy?
2. What is the length of a wire required to fence a rectangular flower bed?
3. What distance would you cover by taking two rounds of a triangular park?
4. How much plastic sheet do you need to cover a rectangular swimming pool?
Answer:
1. Area
2. Perimeter
3. Perimeter
4. Area
Note: Each side of a regular polygon is of the same length, and each angle is of the same measure.
In simple words: To figure out space, you use area. To figure out length around the edge, you use perimeter.

Exam Tip: Remember, 'area' measures the surface inside a shape, while 'perimeter' measures the distance around its boundary. Choose the correct measurement based on the problem's context.

 

Definitions:

• Perimeter of a regular polygon = Length of a side x Number of sides.
• Perimeter of a square = 4 x Side.
• Area of a square = Side x Side.
• Perimeter of a rectangle = 2 x [Length + Breadth]
• Area of a rectangle = Length x Breadth.

 

Try These (Page 206)

 

Question 1. Experiment with several such shapes and cut-outs. You might find it useful to draw these shapes on squared sheets and compute their areas and perimeters. You have seen that increase in perimeter does not mean that area will also increase.
Answer: Students should perform this activity with the help of their teacher. This exercise helps to understand that changes in perimeter do not always mean a corresponding change in area. For example, a long, thin rectangle can have a large perimeter but a small area, while a square with a smaller perimeter can enclose a larger area. Experimenting with different shapes like rectangles, L-shapes, and irregular polygons on graph paper will demonstrate this important concept.
In simple words: You need to try drawing different shapes yourself on graph paper. Then, find how big they are (area) and how long their edges are (perimeter). You will see that just because the edge gets longer, the inside space doesn't always get bigger too.

Exam Tip: Hands-on activities like drawing on graph paper help reinforce abstract geometric concepts. Focus on understanding the distinct nature of area and perimeter, not just formulas.

 

Question 2. Give two examples where the area increases as the perimeter increases.
Answer:1. **Growing Square:** Consider a square with a side length of 2 cm. Its perimeter is \( 4 \times 2 = 8 \) cm, and its area is \( 2 \times 2 = 4 \) cm\(^2\). If the side length increases to 4 cm, the perimeter becomes \( 4 \times 4 = 16 \) cm (increases), and the area becomes \( 4 \times 4 = 16 \) cm\(^2\) (increases). Here, both area and perimeter increased. 2. **Expanding Circle:** A circle with a radius of 1 cm has a circumference (perimeter) of \( 2\pi \times 1 = 2\pi \) cm and an area of \( \pi \times 1^2 = \pi \) cm\(^2\). If the radius grows to 2 cm, the circumference becomes \( 2\pi \times 2 = 4\pi \) cm (increases), and the area becomes \( \pi \times 2^2 = 4\pi \) cm\(^2\) (increases). In this case, both measurements grew larger.
In simple words: When a shape gets bigger in every way, like a square or a circle growing larger, its outside edge (perimeter) gets longer, and its inside space (area) also gets bigger.

Exam Tip: When providing examples, always state the initial and final dimensions and calculate both perimeter and area to clearly demonstrate the relationship or lack thereof.

 

Question 3. Give two examples where the area does not increase when perimeter increases.
Answer:1. **Reshaping a Rectangle:** Consider a square with sides of 4 cm. Its perimeter is \( 4 \times 4 = 16 \) cm, and its area is \( 4 \times 4 = 16 \) cm\(^2\). Now, reshape this square into a long, thin rectangle with length 7 cm and breadth 1 cm. The area remains \( 7 \times 1 = 7 \) cm\(^2\), which is smaller. However, the perimeter becomes \( 2 \times (7 + 1) = 16 \) cm. In this case, the perimeter stayed the same, but the area decreased. If we had a 3x3 square (P=12, A=9) and reshaped to 5x1 rectangle (P=12, A=5), area decreases. For area *not* to increase *when* perimeter increases, consider: a 1x1 square (P=4, A=1). A 0.5x10 rectangle (P=21, A=5). Here, perimeter increased greatly, but area also increased. Let's find an example where perimeter increases, but area stays same or decreases. Let's re-think: the question is "area does NOT increase when perimeter increases". **Example 1: Stretching a Rectangle:** Start with a 3x3 square. Perimeter is 12 cm, area is 9 cm\(^2\). Now, stretch it into a very long, thin rectangle, say 10 cm by 0.5 cm. The perimeter becomes \( 2 \times (10 + 0.5) = 2 \times 10.5 = 21 \) cm. The area becomes \( 10 \times 0.5 = 5 \) cm\(^2\). Here, the perimeter significantly increased (from 12 to 21), but the area decreased (from 9 to 5). 2. **Irregular Shapes with Protrusions:** Imagine a 4x4 square with a perimeter of 16 cm and an area of 16 cm\(^2\). Now, add a small, thin rectangular "antenna" (1 cm by 0.1 cm) sticking out from one side without changing the main 4x4 block. The area will increase slightly by 0.1 cm\(^2\). However, if we make the antenna very thin and long, or add many small protrusions, the perimeter can increase significantly (by adding many small edges), while the total enclosed area might not grow much, or even remain effectively the same for practical purposes if the added area is negligible compared to the original. This is a more complex example. A simpler one: Consider a 2x4 rectangle. Perimeter = \( 2(2+4) = 12 \) cm. Area = \( 2 \times 4 = 8 \) cm\(^2\). Now, take a 1x10 rectangle. Perimeter = \( 2(1+10) = 22 \) cm. Area = \( 1 \times 10 = 10 \) cm\(^2\). Here both increased. The goal is for *area not to increase* when perimeter increases. This implies the area could decrease or stay the same. **Let's use a standard example of a shape changing from compact to elongated:** 1. **Changing a Square to a long Rectangle:** Start with a 4x4 square. Perimeter = \( 4 \times 4 = 16 \) cm. Area = \( 4 \times 4 = 16 \) cm\(^2\). Now, consider a rectangle that is 7 cm long and 1 cm wide. Its perimeter is \( 2 \times (7 + 1) = 16 \) cm. Its area is \( 7 \times 1 = 7 \) cm\(^2\). Here, the perimeter remained the same, but the area decreased. *This doesn't fit the 'perimeter increases' condition.* Let's generate examples that fit the exact condition: perimeter increases, area does not increase (i.e., area stays same or decreases). **Example 1: Creating Indentations:** Imagine a 4x4 square. Perimeter = 16 cm, Area = 16 cm\(^2\). If you push in a small section (e.g., make a small square cut-out inside the boundary of the original square, then fill it back in by making an indent on one side), you can increase the perimeter by creating more "zig-zags" without increasing the total area, or even slightly decreasing it if the indent effectively reduces the usable space. *Consider a 4x4 square (Perimeter 16, Area 16). Now, imagine cutting a 1x1 square out of one corner and reattaching it to the side as an extension. The area would remain 16, but the perimeter could change.* Let's stick to simpler transformations: 1. **Reshaping from Square to "U" Shape:** Consider a 4x4 square. Perimeter = 16 cm, Area = 16 cm\(^2\). Now, imagine an 'L' shape made from 1x1 unit squares, say 5 units long and 3 units wide in total (like 5 squares in a row, then 2 more below the first one). The area could be 7 square units. A complex 'U' shape can have a large perimeter for a relatively small area. For example, a shape covering 16 square units, like a 4x4 square (P=16, A=16). If we cut a 1x1 square from the center of each side, and extend outwards, we can increase the perimeter. A more direct example: 1. **Changing a Rectangle's Aspect Ratio:** Start with a 2x2 square. Perimeter = 8 cm, Area = 4 cm\(^2\). Now, consider a 1x5 rectangle. Perimeter = \( 2 \times (1+5) = 12 \) cm. Area = \( 1 \times 5 = 5 \) cm\(^2\). Here both increase. Let's find one where perimeter increases, but area decreases or stays the same. *Revised Example 1:* Start with a 5x5 square. Perimeter = 20 cm, Area = 25 cm\(^2\). Now, consider a long, thin rectangle, say 12 cm by 1 cm. Its perimeter is \( 2 \times (12 + 1) = 26 \) cm. Its area is \( 12 \times 1 = 12 \) cm\(^2\). Here, the perimeter has increased (from 20 to 26 cm), but the area has significantly decreased (from 25 to 12 cm\(^2\)). *Revised Example 2:* A complex, zig-zag path can have a very long perimeter but enclose a relatively small area. For instance, think of a very winding river contained within a small valley. The length of the river (perimeter) can be extremely long, but the area of the valley it occupies might be quite limited. This shows that increasing the complexity of a boundary often increases its length (perimeter) without necessarily expanding the overall space it takes up (area). These examples clearly demonstrate how perimeter can increase while area decreases or remains constrained.
In simple words: Sometimes, a shape can get a longer outside edge (perimeter) but its inside space (area) can actually get smaller or stay about the same. For example, if you take a square and stretch it into a very thin, long rectangle, its edge gets much longer, but its space might shrink. Also, a very curvy or zig-zag path can have a long edge but fit into a small space.

Exam Tip: To illustrate this concept, think about how to make a shape's boundary longer without making its interior larger. Extreme aspect ratios for rectangles or adding complex, non-area-expanding features are good examples.

 

Try These (Page 210)

 

Question 1. Each of the following rectangles of length 6 cm and breadth 4 cm is composed of congruent polygons. Find the area of each polygon.
Answer:
Length of the rectangle (l) = 6 cm
Breadth of the rectangle (b) = 4 cm
Therefore, Area of the rectangle = \( (l \times b) = 6 \times 4 \, \text{cm}^2 = 24 \, \text{cm}^2 \)
(i) Here, the number of congruent polygons = 6
Therefore, Area of each polygon = \( \frac{24}{6} \, \text{cm}^2 = 4 \, \text{cm}^2 \)
(ii) Here, the number of congruent polygons = 4
Therefore, Area of each polygon = \( \frac{24}{4} \, \text{cm}^2 = 6 \, \text{cm}^2 \)
(iii) Here, the number of congruent polygons = 2
Therefore, Area of each polygon = \( \frac{24}{2} \, \text{cm}^2 = 12 \, \text{cm}^2 \)
(iv) Here, the number of congruent polygons = 2
Therefore, Area of each polygon = \( \frac{24}{2} \, \text{cm}^2 = 12 \, \text{cm}^2 \)
(v) Here, the number of congruent polygons = 8
Therefore, Area of each polygon = \( \frac{24}{8} \, \text{cm}^2 = 3 \, \text{cm}^2 \)
In simple words: First, work out the total space inside the big rectangle by multiplying its length and width. Then, look at each picture and count how many equal-sized smaller shapes (polygons) are inside that big rectangle. To find the space inside one small shape, just divide the big rectangle's total space by the number of small shapes.

Exam Tip: Remember that "congruent polygons" means all the smaller shapes are identical in size and shape. The total area is always the sum of the areas of its parts.

 

Try These (Page 212)

 

Question 1. Find the area of the following parallelograms:
Answer:
(i) Base = 8 cm, Height = 3.5 cm
Therefore, Area of the parallelogram = Base x Height = \( 8 \, \text{cm} \times 3.5 \, \text{cm} = 28 \, \text{cm}^2 \)
(ii) Base = 8 cm, Height = 2.5 cm
Therefore, Area of the parallelogram = Base x Height = \( 8 \, \text{cm} \times 2.5 \, \text{cm} = 20 \, \text{cm}^2 \)
(iii) Base of parallelogram ABCD = (AB) = 7.2 cm
Height of parallelogram ABCD = 4.5 cm
Therefore, Area of parallelogram ABCD = Base x Height = \( 7.2 \, \text{cm} \times 4.5 \, \text{cm} = 32.4 \, \text{cm}^2 \)
In simple words: To get the area of a parallelogram, you multiply its base (the bottom side) by its height (how tall it is, measured straight up from the base).

Exam Tip: The height of a parallelogram must be the perpendicular distance between the base and the opposite side, not the slanted side length.

 

Try These (Page 213)

 

Question 1. Try the activity given on page 213, NCERT Textbook with different types of triangles.
Answer: Students should complete this activity themselves. This exercise involves cutting different types of triangles from paper and manipulating them to understand properties like area calculation. By comparing and measuring, you can discover relationships between the base, height, and area of various triangles, such as right-angled, acute-angled, and obtuse-angled triangles. This hands-on experience will solidify your understanding of geometric concepts.
In simple words: You need to do this activity yourself. It's about playing with different paper triangles to learn how their size and shape relate to their base, height, and total space inside.

Exam Tip: Practical activities improve conceptual understanding. When instructed to do an activity, ensure you understand the goal and record any observations or measurements clearly.

 

Question 2. Take different parallelograms. Divide each of the parallelograms into two triangles by cutting any of its diagonals. Are the triangles congruent?
Answer: Students should perform this activity themselves. When you draw a diagonal across any parallelogram, you will indeed divide it into two triangles. Upon cutting them out and placing one on top of the other, you will observe that these two triangles are congruent. This means they are identical in shape and size. This happens because the diagonal acts as a common side, and the opposite sides of a parallelogram are equal and parallel, satisfying the SSS (Side-Side-Side) or SAS (Side-Angle-Side) congruence criteria.
In simple words: You need to try this yourself. If you cut any parallelogram down the middle using one of its corner-to-corner lines (a diagonal), you will get two triangles. If you check, these two triangles will be exactly the same shape and size.

Exam Tip: The congruence of triangles formed by a diagonal in a parallelogram is a fundamental property. Understanding why they are congruent (e.g., by SSS or SAS criterion) is more important than just observing it.

 

Try These (Page 219)

 

Question 1. Refer to the adjoining figure.

(a) Which has the larger perimeter, the outer square or the smaller square?
(b) Which is larger, the perimeter of the smaller square or the circumference of the circle?
Answer:
(a) The outer square has the larger perimeter. If we assume the side length of the outer square is 's', its perimeter is 4s. The smaller square, formed by connecting the midpoints, has a side length of \( \frac{s}{\sqrt{2}} \). Its perimeter would be \( \frac{4s}{\sqrt{2}} \), which is clearly less than 4s.
(b) The circumference of the circle is larger than the perimeter of the smaller square. If the outer square has side 's', the inscribed circle has a diameter of 's', so its radius is \( \frac{s}{2} \). The circumference is \( 2\pi r = 2\pi (\frac{s}{2}) = \pi s \). The smaller square (whose vertices are midpoints of the outer square) has a side length of \( \sqrt{(\frac{s}{2})^2 + (\frac{s}{2})^2} = \sqrt{\frac{s^2}{4} + \frac{s^2}{4}} = \sqrt{\frac{s^2}{2}} = \frac{s}{\sqrt{2}} \). Its perimeter is \( 4 \times \frac{s}{\sqrt{2}} = 2\sqrt{2}s \). Since \( \pi \approx 3.14 \) and \( 2\sqrt{2} \approx 2 \times 1.414 = 2.828 \), \( \pi s > 2\sqrt{2}s \), so the circumference is larger.
In simple words: The bigger square on the outside has a longer edge (perimeter) than the smaller square inside. Also, the round edge of the circle (circumference) is longer than the edge of that smaller square.

Exam Tip: When comparing perimeters and circumferences of nested shapes, assign a variable (like 's' for side or 'r' for radius) to one dimension and express all other dimensions in terms of it. Then calculate and compare the values using formulas.

 

Try These (Page 222)

 

Question 1. Draw circles of different radii on a graph paper. Find the area by counting the number of squares. Also find the area by using the formula. Compare the two answers.
Answer: Students should perform this activity themselves. Draw multiple circles of varying radii on graph paper. For each circle, estimate its area by counting the full squares enclosed and approximating for partial squares. Then, calculate the area of each circle using the formula \( A = \pi r^2 \). Compare the estimated area from counting squares with the calculated area. You will find that counting squares provides a good approximation, but the formula gives a more precise value, especially as the number of squares inside the circle increases. The accuracy of the square-counting method improves with smaller grid sizes.
In simple words: Draw circles of different sizes on graph paper. Count the squares inside each circle to guess its size (area). Then, use the math rule (formula) to find the exact area. See how close your guess was to the real answer.

Exam Tip: This exercise highlights the difference between estimation and precise calculation. When counting squares, be consistent in how you handle partially filled squares (e.g., count if more than half, ignore if less, or average).

 

Try These (Page 225)

 

Question 1. Convert the following:
(i) 50 cm\(^2\) in mm\(^2\)
(ii) 2 ha in m\(^2\)
(iii) 10 m\(^2\) in cm\(^2\)
(iv) 1000 cm\(^2\) in m\(^2\)
Answer:
(i) 50 cm\(^2\) in mm\(^2\)
We know that 1 cm = 10 mm.
Therefore, 1 cm\(^2\) = \( (10 \, \text{mm}) \times (10 \, \text{mm}) = 100 \, \text{mm}^2 \)
So, 50 cm\(^2\) = \( 50 \times 100 \, \text{mm}^2 = 5000 \, \text{mm}^2 \)
(ii) 2 ha in m\(^2\)
We know that 1 ha = 10000 m\(^2\).
Therefore, 2 ha = \( 2 \times 10000 \, \text{m}^2 = 20000 \, \text{m}^2 \)
(iii) 10 m\(^2\) in cm\(^2\)
We know that 1 m = 100 cm.
Therefore, 1 m\(^2\) = \( (100 \, \text{cm}) \times (100 \, \text{cm}) = 10000 \, \text{cm}^2 \)
So, 10 m\(^2\) = \( 10 \times 10000 \, \text{cm}^2 = 100000 \, \text{cm}^2 \)
(iv) 1000 cm\(^2\) in m\(^2\)
We know that 10000 cm\(^2\) = 1 m\(^2\).
Therefore, 1 cm\(^2\) = \( \frac{1}{10000} \, \text{m}^2 \)
So, 1000 cm\(^2\) = \( 1000 \times \frac{1}{10000} \, \text{m}^2 = \frac{1}{10} \, \text{m}^2 = 0.1 \, \text{m}^2 \)
In simple words: To change square units, remember how many smaller units are in one bigger unit (e.g., 10 mm in 1 cm). Then, square that number to find how many square smaller units are in one square bigger unit. For hectares to square meters, just use the direct conversion factor.

Exam Tip: Be careful with unit conversions for area; if 1 unit of length is 10 times another, then 1 unit of area will be \( 10 \times 10 = 100 \) times. Always square the conversion factor for linear units to get the conversion factor for square units.