Maharashtra Board Class 8 Maths Chapter 15 Area Set 15.6 Solutions

Question 1. Radii of the circles are given below, find their areas.
i. 28 cm
ii. 10.5 cm
iii. 17.5 cm
Answer:
i. Radius of the circle (r) = 28 cm ... [Given]
Area of the circle = \( \pi r^2 \)
= \( \frac{22}{7} \) x \( (28)^2 \)
= \( \frac{22}{7} \) x 28 x 28
= 22 x 4 x 28
= 2464 sq. cm
ii. Radius of the circle (r) = 10.5 cm ... [Given]
Area of the circle = \( \pi r^2 \)
= \( \frac{22}{7} \) x \( (10.5)^2 \)
= \( \frac{22}{7} \) x 10.5 x 10.5
= 22 x 1.5 x 10.5
= 346.5 sq. cm
iii. Radius of the circle (r) = 17.5 cm ... [Given]
Area of the circle = \( \pi r^2 \)
= \( \frac{22}{7} \) x \( (17.5)^2 \)
= \( \frac{22}{7} \) x 17.5 x 17.5
= 22 x 2.5 x 17.5
= 962.5 sq. cm
In simple words: The area of a circle is calculated using the formula \( \pi r^2 \), where 'r' is the radius. We substitute the given radius values into this formula to find the area for each circle.

🎯 Exam Tip: Remember the formula for the area of a circle and be careful with calculations involving decimals and fractions for full marks.

Question 2. Areas of some circles are given below, find their diameters.
i. 176 sq.cm
ii. 394.24 sq. cm
iii. 12474 sq. cm
Answer:
i. Area of the circle = 176 sq. cm ... [Given]
Area of the circle = \( \pi r^2 \)

\( \therefore \) 176 = \( \frac{22}{7} \) x \( r^2 \)

\( \therefore r^2 = 176 \) x \( \frac{7}{22} \)

\( \therefore r^2 = 56 \)

\( \therefore r = \sqrt{56} \) ... [Taking square root of both sides]
Diameter = 2r = \( 2\sqrt{56} \) CM
ii. Area of the circle = 394.24 sq. cm ... [Given]
Area of the circle = \( \pi r^2 \)

\( \therefore \) 394.24 = \( \frac{22}{7} \) x \( r^2 \)

\( \therefore r^2 = 394.24 \) x \( \frac{7}{22} \)

\( \therefore r^2 = \frac{394.24 \times 100}{1 \times 100} \times \frac{7}{22} \)

\( \therefore r^2 = \frac{39424}{100} \times \frac{7}{22} \)

\( \therefore r^2 = \frac{1792}{100} \times 7 \)

\( \therefore r^2 = \frac{12544}{100} \)

\( \therefore r^2 = \frac{112^2}{10^2} \)

\( \therefore r = \frac{112}{10} \) ... [Taking square root of both sides]

\( \therefore \) r = 11.2 cm

\( \therefore \) Diameter = 2r = 2 x 11.2 = 22.4 cm
iii. Area of the circle = 12474 sq. cm ... [Given]
Area of the circle = \( \pi r^2 \)

\( \therefore \) 12474 = \( \frac{22}{7} \) x \( r^2 \)

\( \therefore r^2 = 12474 \) x \( \frac{7}{22} \)

\( \therefore r^2 = 567 \times 7 \)

\( \therefore r^2 = 3969 \)

\( \therefore r = 63 \) ... [Taking square root of both sides]

\( \therefore \) Diameter = 2r = 2 x 63 = 126cm
In simple words: To find the diameter from the area, we first use the area formula \( A = \pi r^2 \) to calculate the radius 'r', and then double the radius (2r) to get the diameter.

🎯 Exam Tip: Remember to calculate the radius first, and then double it to find the diameter. Ensure correct square root calculations.

Question 3. Diameter of the circular garden is 42 m. There is a 3.5 m wide road around the garden. Find the area of the road.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्ताकार बगीचे और उसके चारों ओर एक सड़क को दर्शाता है। इसमें दो संकेंद्रित वृत्त हैं, एक आंतरिक वृत्त बगीचे को और बाहरी वृत्त सड़क सहित कुल क्षेत्र को दर्शाता है। चित्र में बगीचे की त्रिज्या 21 मीटर और सड़क की चौड़ाई 3.5 मीटर दिखाई गई है।
Diameter of the circular garden is 42 m. ... [Given]

\( \therefore \) Radius of the circular garden (r) = \( \frac{42}{2} \) = 21 m
Width of the road = 3.5 m ... [Given]
Radius of the outer circle (R)
= radius (r) + width of the road
= 21 + 3.5
= 24.5 m
Area of the road = area of outer circle - area of circular garden
= \( \pi R^2 - \pi r^2 \)
= \( \pi (R^2 - r^2) \)
= \( \frac{22}{7} \) [\( (24.5)^2 - (21)^2 \)]
= \( \frac{22}{7} \) (24.5 + 21) (24.5 - 21)
...[ \( \therefore a^2-b^2 = (a+b)(a-b) \)]
= \( \frac{22}{7} \) x 45.5 x 3.5
= 22 x 45.5 x 0.5
= 500.50 sq. m

\( \therefore \) The area of the road is 500.50 sq. m.
In simple words: To find the area of the road around the garden, we calculate the area of the larger circle (garden + road) and subtract the area of the smaller circle (garden only). This uses the difference of squares formula for simplification.

🎯 Exam Tip: Clearly identify the inner and outer radii. Using the \( a^2-b^2 = (a+b)(a-b) \) identity simplifies calculations significantly.

Question 4. Find the area of the circle if its circumference is 88 cm.
Answer:
Circumference of the circle = 88 cm ...[Given]
Circumference of the circle = \( 2\pi r \)

\( \therefore \) 88 = 2 x \( \frac{22}{7} \) x r

\( \therefore r = \frac{88 \times 7}{2 \times 22} \)

\( \therefore \) r = 14cm
Area of the circle = \( \pi r^2 = \frac{22}{7} \) x \( (14)^2 \)
= \( \frac{22}{7} \) x 14 x 14 = 22 x 2 x 14 = 616 sq. cm

\( \therefore \) The area of circle is 616 Sq cm
In simple words: Given the circumference, we first find the radius using the formula \( C = 2\pi r \). Once the radius is known, we then calculate the area using the formula \( A = \pi r^2 \).

🎯 Exam Tip: This question requires two steps: first finding the radius from the circumference, and then using that radius to find the area. Ensure both formulas are applied correctly.

Maharashtra Board Class 8 Maths Chapter 15 Area Practice Set 15.6 Intext Questions And Activities

Question 1. Draw a circle of radius 28mm. Draw any one triangle and draw a trapezium on the graph paper. Find the area of these figures by counting the number of small squares on the graph paper. Verify your answers using formula for area of these figures. Observe that smaller the squares of graph paper, better is the approximation of area. (Textbook pg. no. 105)
Answer:
(Students should do this activity on their own.)
In simple words: This activity involves drawing shapes on graph paper, estimating their areas by counting squares, and then verifying those estimates using the standard area formulas for circles, triangles, and trapeziums.

🎯 Exam Tip: While direct calculation is often preferred, understanding area approximation through counting squares on graph paper helps build foundational geometric intuition.