krit.club logo

Measuring Space: Perimeter and Area - Perimeter of a Shape

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Perimeter is the continuous line forming the boundary of a closed geometrical figure. It represents the total distance around the shape.

•

For any polygon, the perimeter is calculated by finding the sum of the lengths of all its sides. For example, if a triangle has sides aa, bb, and cc, the perimeter is a+b+ca + b + c.

•

In regular polygons (where all sides and angles are equal), the perimeter is given by n×sn \times s, where nn is the number of sides and ss is the length of one side.

•

The perimeter of a circle is specifically called the circumference. It is proportional to the radius rr or diameter dd.

•

In Grade 9 geometry, the semi-perimeter (ss) is frequently used, especially in Heron's formula for the area of a triangle, defined as s=a+b+c2s = \frac{a + b + c}{2}.

•

Units of perimeter are linear, such as cmcm, mm, or kmkm.

📐Formulae

P=2(l+b)P = 2(l + b) (Perimeter of a Rectangle)

P=4sP = 4s (Perimeter of a Square)

C=2πrC = 2\pi r (Circumference of a Circle)

P=πr+2rP = \pi r + 2r (Perimeter of a closed Semi-circle)

s=a+b+c2s = \frac{a + b + c}{2} (Semi-perimeter of a Triangle)

P=n×sP = n \times s (Perimeter of a Regular Polygon with nn sides)

💡Examples

Problem 1:

Calculate the perimeter of a rectangular park whose length is 45 m45\text{ m} and breadth is 25 m25\text{ m}. Also, find the cost of fencing it at the rate of Rs 15 per meter.

Solution:

Given: l=45 ml = 45\text{ m}, b=25 mb = 25\text{ m}. Using the formula for the perimeter of a rectangle: P=2(l+b)P = 2(l + b) P=2(45+25)P = 2(45 + 25) P=2(70)P = 2(70) P=140 mP = 140\text{ m} Now, calculating the cost of fencing: Cost = Perimeter×RatePerimeter \times Rate Cost = 140×15140 \times 15 Cost = Rs 2100

Explanation:

We first find the total boundary length (perimeter) using the length and breadth, then multiply that total length by the unit cost to find the total expenditure.

Problem 2:

A triangle has sides of 12 cm12\text{ cm}, 15 cm15\text{ cm}, and 9 cm9\text{ cm}. Find its semi-perimeter ss.

Solution:

Given sides: a=12 cma = 12\text{ cm}, b=15 cmb = 15\text{ cm}, c=9 cmc = 9\text{ cm}. To find the perimeter PP, we sum the sides: 1215+936\begin{array}{r} 12 \\ 15 \\ + 9 \\ \hline 36 \end{array} So, P=36 cmP = 36\text{ cm}. The semi-perimeter ss is half of the perimeter: s=a+b+c2s = \frac{a + b + c}{2} s=362s = \frac{36}{2} s=18 cms = 18\text{ cm}

Explanation:

The semi-perimeter is calculated by adding all three sides of the triangle and dividing the resulting sum by 22. This value is essential for applying Heron's Formula.

Problem 3:

Find the circumference of a circular wire with a radius of 7 cm7\text{ cm} (Take π=227\pi = \frac{22}{7}).

Solution:

Given: r=7 cmr = 7\text{ cm}. Using the formula for circumference: C=2πrC = 2\pi r C=2×227×7C = 2 \times \frac{22}{7} \times 7 Cancelling the 77 in the numerator and denominator: C=2×22C = 2 \times 22 C=44 cmC = 44\text{ cm}

Explanation:

The circumference is the perimeter of the circle. By substituting the given radius into the formula 2πr2\pi r, we find the total length of the wire.