Area of Polygons | IT

Question 3

Context: Consider two regions, Region 1 and Region 2, such that Perimeter of Region 1 > Perimeter of Region 2, but Area of Region 1 < Area of Region 2.

Q. Find two rectangles that are examples of such regions. If needed, use a grid paper (given at the end of the book) for this.

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Solution

We need to find two rectangles that satisfy the given conditions.

Step 1 — Define Region 1

Let us choose a long and thin rectangle for Region 1. This shape will have a relatively large perimeter for its area. Let the length of Region 1 be L1L_1 and its width be W1W_1. We choose L1=10 unitsL_1 = 10 \text{ units} and W1=1 unitW_1 = 1 \text{ unit}. The perimeter of a rectangle is 2×(length+width)2 \times (\text{length} + \text{width}). The area of a rectangle is length×width\text{length} \times \text{width}.

P1=2(L1+W1)P_1 = 2(L_1 + W_1)

P1=2(10+1)P_1 = 2(10 + 1)

P1=2(11)P_1 = 2(11)

P1=22 units\boxed{P_1 = 22 \text{ units}}

A1=L1×W1A_1 = L_1 \times W_1

A1=10×1A_1 = 10 \times 1

A1=10 square units\boxed{A_1 = 10 \text{ square units}}

Diagram 1

Step 2 — Define Region 2

Now we need to find a second rectangle, Region 2. It must have a smaller perimeter than Region 1. It must have a larger area than Region 1. Let the length of Region 2 be L2L_2 and its width be W2W_2. We choose L2=4 unitsL_2 = 4 \text{ units} and W2=3 unitsW_2 = 3 \text{ units}.

P2=2(L2+W2)P_2 = 2(L_2 + W_2)

P2=2(4+3)P_2 = 2(4 + 3)

P2=2(7)P_2 = 2(7)

P2=14 units\boxed{P_2 = 14 \text{ units}}

A2=L2×W2A_2 = L_2 \times W_2

A2=4×3A_2 = 4 \times 3

A2=12 square units\boxed{A_2 = 12 \text{ square units}}

Diagram 2

Step 3 — Verify the conditions

Let us check if our chosen rectangles satisfy the given conditions. Condition 1: Perimeter of Region 1 > Perimeter of Region 2. We found P1=22 unitsP_1 = 22 \text{ units} and P2=14 unitsP_2 = 14 \text{ units}.

P1>P2P_1 > P_2

22>1422 > 14 This condition is satisfied.

Condition 2: Area of Region 1 < Area of Region 2. We found A1=10 square unitsA_1 = 10 \text{ square units} and A2=12 square unitsA_2 = 12 \text{ square units}.

A1<A2A_1 < A_2

10<1210 < 12 This condition is satisfied.

Answer

(i) Region 1 can be a rectangle with length 10 units and width 1 unit. (ii) Region 2 can be a rectangle with length 4 units and width 3 units.

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Q3

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