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Q7:
Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Solution :
Given Variables & Initial Setup
The problem requires us to find the total perimeter of a closed geometric figure consisting of a semicircular arc and its straight-line diameter. While the specific numerical value is derived from the adjoining figure in standard textbook contexts (typically $d = 10 \text{ cm}$), we will first establish the universal algebraic formula before applying the standard dimension.
- Diameter of the semicircle = $d$
- Radius of the semicircle = $r = \frac{d}{2}$
- Standard Applied Dimension = $d = 10 \text{ cm}$ (Standard for this specific problem)
Step 1: Formulating the Perimeter Equation
The perimeter ($P$) of any two-dimensional figure is the total continuous length of its boundary. [Per the axioms of Euclidean geometry], the boundary of a closed semicircle is composed of two distinct parts:
- The curved semicircular arc.
- The straight-line diameter that closes the shape.
Therefore, the total perimeter is expressed as:
$P = \text{Length of Semicircular Arc} + \text{Length of Diameter}$
Step 2: Calculating the Length of the Semicircular Arc
The circumference of a full circle is given by the formula $C = \pi d$ or $C = 2\pi r$. Because a semicircle represents exactly one-half of a circle, the length of its curved arc is half of the full circumference:
$\text{Arc Length} = \frac{1}{2} \times (2\pi r) = \pi r$
Given the standard diameter $d = 10 \text{ cm}$, the radius is $r = 5 \text{ cm}$. Substituting this into our arc length formula (using $\pi \approx \frac{22}{7}$):
$\text{Arc Length} = \frac{22}{7} \times 5 \text{ cm} = \frac{110}{7} \text{ cm} \approx 15.71 \text{ cm}$
(Note: If using $\pi \approx 3.14$, the arc length is exactly $15.7 \text{ cm}$.)
Step 3: Computing the Total Perimeter
To find the total perimeter, we must add the straight-line diameter back to the arc length. Failing to add the diameter is a common error that only yields the length of the open curve, not the closed figure.
$P = \pi r + d$
Substituting the calculated values:
$P = 15.71 \text{ cm} + 10 \text{ cm}$
$P = 25.71 \text{ cm}$
General Algebraic Proof
For any semicircle of diameter $d$, the perimeter can be factored as follows:
$P = \frac{\pi d}{2} + d$
$P = d \left( \frac{\pi}{2} + 1 \right)$
Substituting $d = 10 \text{ cm}$ into the factored form yields $10 \left( \frac{3.14}{2} + 1 \right) = 10(1.57 + 1) = 10(2.57) = 25.7 \text{ cm}$, confirming our step-by-step arithmetic.
Final Solution: The total perimeter of the adjoining figure (the semicircle including its diameter) is 25.71 cm (or exactly 25.7 cm if using π = 3.14).
More Questions from Class 9 Mathematics Coordinate Geometry EXERCISE 9.2
- Q1(a): Find the circumference of the circles with the following radius: (Take $\pi = \frac{22}{7}$) (a) $14$ cm
- Q1(b): Find the circumference of the circles with the following radius: (Take $\pi = \frac{22}{7}$) (b) $28$ mm
- Q1(c): Find the circumference of the circles with the following radius: (Take $\pi = \frac{22}{7}$) (c) $21$ cm
- Q10: From a circular card sheet of radius $14$ cm, two circles of radius $3.5$ cm and a rectangle of length $3$ cm and breadth $1$cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. (Take $\pi = \frac{22}{7}$)
- Q11: A circle of radius $2$ cm is cut out from a square piece of an aluminium sheet of side $6$ cm. What is the area of the left over aluminium sheet? (Take $\pi = 3.14$)
- Q12: The circumference of a circle is $31.4$ cm. Find the radius and the area of the circle? (Take $\pi = 3.14$)
- Q13: A circular flower bed is surrounded by a path $4$ m wide. The diameter of the flower bed is $66$ m. What is the area of this path? ($\pi = 3.14$)
- Q14: A circular flower garden has an area of $314$ m$^2$. A sprinkler at the centre of the garden can cover an area that has a radius of $12$ m. Will the sprinkler water the entire garden? (Take $\pi = 3.14$)
- Q15: Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take $\pi = 3.14$)
- Q16: How many times a wheel of radius $28$ cm must rotate to go $352$ m? (Take $\pi = \frac{22}{7}$)
- Q17: The minute hand of a circular clock is $15$ cm long. How far does the tip of the minute hand move in $1$ hour. (Take $\pi = 3.14$)
- Q2(a): Find the area of the following circles, given that: (a) radius = $14$ mm (Take $\pi = \frac{22}{7}$)
- Q2(b): Find the area of the following circles, given that: (b) diameter = $49$ m
- Q2(c): Find the area of the following circles, given that: (c) radius = $5$ cm
- Q3: If the circumference of a circular sheet is $154$ m, find its radius. Also find the area of the sheet. (Take $\pi = \frac{22}{7}$)
- Q4: A gardener wants to fence a circular garden of diameter $21$ m. Find the length of the rope he needs to purchase, if he makes $2$ rounds of fence. Also find the cost of the rope, if it costs ₹ $4$ per meter. (Take $\pi = \frac{22}{7}$)
- Q5: From a circular sheet of radius $4$ cm, a circle of radius $3$ cm is removed. Find the area of the remaining sheet. (Take $\pi = 3.14$)
- Q6: Saima wants to put a lace on the edge of a circular table cover of diameter $1.5$ m. Find the length of the lace required and also find its cost if one meter of the lace costs ₹ $15$. (Take $\pi = 3.14$)
- Q8: Find the cost of polishing a circular table-top of diameter $1.6$ m, if the rate of polishing is ₹ $15/m^2$. (Take $\pi = 3.14$)
- Q9: Shazli took a wire of length $44$ cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take $\pi = \frac{22}{7}$)
CBSE Solutions for Class 9 Mathematics Coordinate Geometry
Chapters in CBSE - Class 9 Mathematics
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