Perimeter and Area - Area and perimeter of triangles, parallelograms, and circles
Hello, bright young mathematicians! Welcome to another exciting learning session.
Have you ever wondered how much ribbon you need to put around a circular birthday card? Or how much grass seed is needed to cover a triangular park? To solve these real-world puzzles, we use two very special mathematical concepts: Perimeter and Area.
Before we dive in, let's quickly refresh our basic definitions:
In Class 6, you learned about rectangles and squares. Today, in Class 7, we are expanding our toolset to master three incredible shapes: Parallelograms, Triangles, and Circles!
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1. The Parallelogram: A Tilted Rectangle
What is a Parallelogram?
A parallelogram is a four-sided flat shape (quadrilateral) where opposite sides are parallel and equal in length. Think of it as a rectangle that has been pushed slightly from the top corner!
Base and Height (Altitude)
To calculate the area of a parallelogram, we need two key measurements:
β οΈ Teacher's Warning: Never confuse the slanted side with the height! Height is always a straight, vertical line perpendicular to the base.
```
A _______________ B
/| /
/ | /
/ | h /
/___|__________/
D E C
<--- base --->
```
Area of a Parallelogram
Imagine taking a pair of scissors and cutting off the right-angled triangle ($\triangle ADE$) from the left side of a parallelogram and gluing it to the right side. What shape do you get? A Rectangle!
Because a parallelogram transforms into a rectangle:
$$\text{Area of Rectangle} = \text{Length} \times \text{Breadth}$$
Replacing "Length" with Base ($b$) and "Breadth" with Height ($h$):
$$\text{Area of a Parallelogram} = \text{Base} \times \text{Height} = b \times h$$
Perimeter of a Parallelogram
$$\text{Perimeter} = \text{Sum of all 4 sides} = 2 \times (\text{Side 1} + \text{Side 2})$$
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2. The Triangle: Half a Parallelogram
Perimeter of a Triangle
The perimeter of any triangle is simply the sum of its three side lengths.
$$\text{Perimeter} = a + b + c$$
Area of a Triangle
Let's do a quick visual experiment! Take any parallelogram and draw a diagonal line from one corner to the opposite corner. What do you see?
You get two identical (congruent) triangles!
Since two identical triangles make up one parallelogram:
$$\text{Area of 1 Triangle} = \frac{1}{2} \times \text{Area of Parallelogram}$$
Therefore:
$$\text{Area of a Triangle} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times b \times h$$
```
/\
/ |\
/ | \ h
/___|__\
<-- base -->
```
π‘ Pro-Tip: The height must always correspond to the base you choose! If you pick side $BC$ as the base, the height must be the line perpendicular to $BC$ from the opposite vertex $A$.
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3. The Circle: Curves, Radii, and Pi ($\pi$)
Unlike triangles or parallelograms, circles don't have straight sides! So, how do we measure them?
Key Parts of a Circle
$$\text{Diameter} = 2 \times \text{Radius} \quad (d = 2r)$$
```
. - ~ - .
/ \
/ r \
| O---------> |
|---|
\ /
\ /
' - _ - '
```
Circumference (Perimeter) of a Circle
The distance around a circular edge is called its Circumference ($C$).
If you measure the circumference of *any* circle (a coin, a plate, or a bicycle wheel) and divide it by its diameter, you will always get the same special number: approximately $3.14$ or $\frac{22}{7}$!
We call this constant value $\pi$ (Pi).
$$\frac{\text{Circumference}}{\text{Diameter}} = \pi$$
Rearranging this gives us our formulas:
$$\text{Circumference } (C) = \pi \times d = 2\pi r$$
Area of a Circle
Imagine cutting a pizza into 16 thin slices and arranging them alternately facing up and down. They form a shape that looks almost like a rectangle!
$$\text{Area of Circle} = \text{Length} \times \text{Breadth} = (\pi r) \times r = \pi r^2$$
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π Quick Formula Cheat Sheet
| Shape | Perimeter / Circumference | Area |
|---|---|---|
| Parallelogram | $2 \times (\text{Side}_1 + \text{Side}_2)$ | $\text{Base} \times \text{Height}$ ($b \times h$) |
| Triangle | $\text{Side}_1 + \text{Side}_2 + \text{Side}_3$ | $\frac{1}{2} \times \text{Base} \times \text{Height}$ ($\frac{1}{2} \times b \times h$) |
| Circle | $2 \pi r$ or $\pi d$ | $\pi r^2$ |
*(Use $\pi = \frac{22}{7}$ unless $3.14$ is specified in the question!)*
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π Practice Questions with Detailed Step-by-Step Solutions
Now, let's test our understanding with 3 practice problems, ranging from straightforward to real-world applications!
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Question 1: Parallelogram & Triangle
One side of a parallelogram is $14\text{ cm}$ and its corresponding height is $8\text{ cm}$. A triangle has a base of $16\text{ cm}$. If the area of the triangle is equal to the area of the parallelogram, find the height of the triangle.
Solution:
Step 1: Calculate the area of the parallelogram.
$$\text{Area of Parallelogram} = b_1 \times h_1 = 14 \times 8 = 112\text{ cm}^2$$
Step 2: Use the area equality to find the triangle's height.
$$\text{Area of Triangle} = \frac{1}{2} \times b_2 \times h_2$$
$$112 = \frac{1}{2} \times 16 \times h_2$$
$$112 = 8 \times h_2$$
$$h_2 = \frac{112}{8} = 14\text{ cm}$$
Answer: The height of the triangle is $14\text{ cm}$.
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Question 2: Bending Wire into Shapes (Circle)
A wire is in the shape of a square of side $11\text{ cm}$. It is rebent into the shape of a circle. Find the radius of the circle and calculate its area. (Take $\pi = \frac{22}{7}$)
Solution:
Step 1: Find the length of the wire (Perimeter of the square).
$$\text{Perimeter of square} = 4 \times s = 4 \times 11 = 44\text{ cm}$$
Since the same wire is bent to form a circle, the Circumference of the circle = Perimeter of the square = $44\text{ cm}$.
Step 2: Find the radius ($r$) of the circle.
$$\text{Circumference} = 2 \pi r$$
$$44 = 2 \times \frac{22}{7} \times r$$
$$44 = \frac{44}{7} \times r$$
$$r = \frac{44 \times 7}{44} = 7\text{ cm}$$
Step 3: Calculate the area of the circle.
$$\text{Area} = \pi r^2 = \frac{22}{7} \times 7 \times 7 = 22 \times 7 = 154\text{ cm}^2$$
Answer: The radius of the circle is $7\text{ cm}$, and its area is $154\text{ cm}^2$.
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Question 3: Real-World Park Problem (Combined Shapes)
A circular park has a radius of $21\text{ m}$. Inside the park, there is a triangular play area with a base of $20\text{ m}$ and an altitude of $15\text{ m}$. The remaining part of the park is covered with grass. Find the area covered with grass. (Take $\pi = \frac{22}{7}$)
Solution:
```
Area of Grass = Area of Circular Park - Area of Triangular Play Area
```
Step 1: Find the total area of the circular park.
$$\text{Area of Park} = \pi r^2 = \frac{22}{7} \times 21 \times 21$$
$$\text{Area of Park} = 22 \times 3 \times 21 = 1386\text{ m}^2$$
Step 2: Find the area of the triangular play area.
$$\text{Area of Triangle} = \frac{1}{2} \times b \times h = \frac{1}{2} \times 20 \times 15 = 10 \times 15 = 150\text{ m}^2$$
Step 3: Subtract the triangular area from the total circular area.
$$\text{Area covered with grass} = 1386 - 150 = 1236\text{ m}^2$$
Answer: The area covered with grass is $1236\text{ m}^2$.
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π Teacher's Final Tip for Success!
Whenever you solve Mensuration problems:
Keep practicing, stay curious, and enjoy math! You've got this! Fantastic job learning today! π