CBSE Class 10 Mathematics Areas Related to Circles Assignment Set 10

Excluded from NCERT: Exercise 1 and Exercise 2 (Questions 1 to 6)

Question 1. The radii of the two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.
Answer:
Circumference of the 1st circle \( = 2\pi \times 19 = 38\pi \) cm and of the 2nd circle \( = 2\pi \times 9 = 18\pi \) cm. Their sum \( = 56\pi \) cm.
Let the radius of the 3rd circle be R. Then \( 2\pi R = 56\pi \Rightarrow R = 28 \) cm.
Hence, the required radius is 28 cm.
In simple words: Add the two circumferences and divide by 2 pi.

Teacher's Note:
1. Circumference is 2 pi r and area is pi r squared; compare them carefully.
2. State the unit of the answer.

Question 2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
Answer:
Area of the 1st circle \( = \pi(8)^2 = 64\pi \) cm² and of the 2nd circle \( = \pi(6)^2 = 36\pi \) cm². Their sum \( = 100\pi \) cm².
Let the radius of the 3rd circle be R. Then \( \pi R^2 = 100\pi \Rightarrow R^2 = 100 \Rightarrow R = 10 \) cm (a radius cannot be negative).
Hence, the required radius is 10 cm.
In simple words: Add the areas, not the radii.

Teacher's Note:
1. Radii do not add; the areas do.
2. Reject the negative root of R squared.

Question 3. The given figure depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing the Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.
Answer:

Diameter of the Gold region \( = 21 \) cm, so \( r_1 = \frac{21}{2} = 10.5 \) cm. Each other band is 10.5 cm wide, so \( r_2 = 21 \), \( r_3 = 31.5 \), \( r_4 = 42 \) and \( r_5 = 52.5 \) cm.
Area of the Gold region \( = \pi r_1^2 = \frac{22}{7} \times 10.5^2 = 346.5 \) cm².
Area of the Red region \( = \pi(21^2 - 10.5^2) = \frac{22}{7} \times 31.5 \times 10.5 = 1039.5 \) cm².
Area of the Blue region \( = \pi(31.5^2 - 21^2) = \frac{22}{7} \times 52.5 \times 10.5 = 1732.5 \) cm².
Area of the Black region \( = \pi(42^2 - 31.5^2) = \frac{22}{7} \times 73.5 \times 10.5 = 2425.5 \) cm².
Area of the White region \( = \pi(52.5^2 - 42^2) = \frac{22}{7} \times 94.5 \times 10.5 = 3118.5 \) cm².
Hence, the areas are 346.5, 1039.5, 1732.5, 2425.5 and 3118.5 cm² for Gold, Red, Blue, Black and White respectively.
In simple words: Each ring's area is the area of the bigger circle minus the area of the smaller circle inside it.

Teacher's Note:
1. Use a squared minus b squared equals (a + b)(a minus b) to simplify.
2. The radii grow in steps of 10.5 cm.

Question 4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
Answer:
Radius of the wheel \( = 40 \) cm, so its circumference \( = 2\pi r = 2 \times \frac{22}{7} \times 40 = \frac{1760}{7} \) cm.
Speed \( = 66 \) km/h \( = 66 \times 10^5 \) cm per hour. Distance covered in 10 minutes \( = \frac{66 \times 10^5 \times 10}{60} = 1100000 \) cm.
Number of revolutions \( = \frac{1100000}{1760/7} = \frac{1100000 \times 7}{1760} = 4375 \).
Hence, each wheel makes 4375 revolutions.
In simple words: Distance in 10 minutes divided by one wheel circumference gives the revolutions.

Teacher's Note:
1. Convert km/h into cm per minute carefully.
2. In one revolution the wheel covers its circumference.

Question 5. If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(a) 2 units
(b) \( \pi \) units
(c) 4 units
(d) 7 units
Answer:
Perimeter \( = \) area \( \Rightarrow 2\pi r = \pi r^2 \Rightarrow r^2 - 2r = 0 \Rightarrow r(r - 2) = 0 \Rightarrow r = 2 \) (as \( r \neq 0 \)).
Hence, the radius is 2 units: option (a).
In simple words: Equate 2 pi r and pi r squared and cancel pi r.

Teacher's Note:
1. Cancel the common factor pi r only after noting that r is not zero.
2. Check: perimeter = 4 pi and area = 4 pi when r = 2.

Question 6. Find the area of the shaded region in the given figure, if \( PQ = 24 \) cm, \( PR = 7 \) cm and O is the centre of the circle.
Answer:

RQ passes through the centre O, so RQ is a diameter and \( \angle RPQ = 90^\circ \) (angle in a semicircle).
By the Pythagoras theorem in \( \triangle PQR \): \( QR^2 = PR^2 + PQ^2 = 7^2 + 24^2 = 625 \Rightarrow QR = 25 \) cm. So the radius is \( \frac{25}{2} \) cm.
Area of the semicircle \( RQS = \frac{\pi R^2}{2} = \frac{1}{2} \times \frac{22}{7} \times \frac{25}{2} \times \frac{25}{2} = \frac{6875}{28} \) cm². Area of \( \triangle PQR = \frac{1}{2} \times 7 \times 24 = 84 \) cm².
Area of the shaded region \( = \) area of the semicircle \( RQS - \) area of \( \triangle PQR = \frac{6875}{28} - 84 = \frac{6875 - 2352}{28} = \frac{4523}{28} \) cm².
Hence, the area of the shaded region is \( \frac{4523}{28} \) cm² (about 161.54 cm²).
In simple words: The angle in a semicircle is a right angle, so the Pythagoras theorem gives the diameter.

Teacher's Note:
1. A chord through the centre is a diameter.
2. Shaded area = semicircle minus triangle.

Question 7. Find the area of the shaded region in the given figure. If the radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and \( \angle AOC = 40^\circ \).
Answer:

Angle of the sector \( = 40^\circ \), inner radius \( r = 7 \) cm and outer radius \( R = 14 \) cm.
Area of the sector \( OAC = \frac{40^\circ}{360^\circ} \times \frac{22}{7} \times 14 \times 14 = \frac{616}{9} \) cm² and area of the sector \( OBD = \frac{40^\circ}{360^\circ} \times \frac{22}{7} \times 7^2 = \frac{154}{9} \) cm².
Area of the shaded region \( ABDC = \frac{616}{9} - \frac{154}{9} = \frac{462}{9} = \frac{154}{3} \) cm².
Hence, the area of the shaded region is \( \frac{154}{3} \) cm².
In simple words: Subtract the small sector from the big sector with the same angle.

Teacher's Note:
1. Both sectors share the same angle at O.
2. Shaded part = outer sector minus inner sector.

Question 8. Find the area of the shaded region in the given figure if ABCD is a square of side 14 cm and APD and BPC are semicircles.
Answer:

Side of the square \( = 14 \) cm, so the area of square \( ABCD = 14 \times 14 = 196 \) cm².
The sides AD and BC are diameters of the semicircles, so the radius of each semicircle is 7 cm. Area of semicircle \( APD = \frac{1}{2} \times \frac{22}{7} \times 7 \times 7 = 77 \) cm², and area of semicircle \( BPC = 77 \) cm².
Area of the shaded region \( = \) area of square \( - \) area of semicircle APD \( - \) area of semicircle BPC \( = 196 - 77 - 77 = 42 \) cm².
Hence, the area of the shaded region is 42 cm².
In simple words: Remove the two equal semicircles from the square.

Teacher's Note:
1. The side of the square equals the diameter of each semicircle.
2. Shaded area = square minus two semicircles.

Question 9. Find the area of the shaded region in the figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.
Answer:

\( \triangle OAB \) is equilateral, so each angle is \( 60^\circ \). The radius of the circle is 6 cm and the side of the triangle is 12 cm.
Area of the equilateral triangle \( OAB = \frac{\sqrt{3}}{4} \times 12^2 = 36\sqrt{3} \) cm².
Area of the circle \( = \pi r^2 = \frac{22}{7} \times 6^2 = \frac{792}{7} \) cm². Area of the sector \( OPQ = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 6^2 = \frac{132}{7} \) cm².
The sector OPQ lies inside the triangle, so it is counted once. Shaded area \( = \) area of \( \triangle OAB + \) area of the circle \( - \) area of the sector \( OPQ = 36\sqrt{3} + \frac{792}{7} - \frac{132}{7} = 36\sqrt{3} + \frac{660}{7} \) cm².
Hence, the area of the shaded region is \( \left(36\sqrt{3} + \frac{660}{7}\right) \) cm² (about 156.6 cm²).
In simple words: The shaded part is the triangle plus the circle, with the overlapping 60 degree sector subtracted once.

Teacher's Note:
1. Find the angle of the overlapping sector from the triangle: 60 degrees.
2. Add the two areas and subtract the overlap.

Question 10. From each corner of a square of side 4 cm, a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the figure. Find the area of the remaining portion of the square.
Answer:

Side of the square \( = 4 \) cm, so its area \( = 4^2 = 16 \) cm².
The four quadrants of radius 1 cm together make one full circle: area \( = 4 \times \frac{\pi \times 1^2}{4} = \frac{22}{7} \) cm².
The circle cut from the middle has diameter 2 cm, so its radius is 1 cm and its area \( = \frac{22}{7} \) cm².
Area of the remaining portion \( = 16 - \frac{22}{7} - \frac{22}{7} = \frac{112 - 44}{7} = \frac{68}{7} \) cm².
Hence, the remaining area is \( \frac{68}{7} \) cm² (about 9.71 cm²).
In simple words: Four quarter circles make one whole circle, so two circles of radius 1 cm are removed in all.

Teacher's Note:
1. Four quadrants at the corners add up to one full circle.
2. Remaining area = square minus the cut parts.

Question 11. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. Find the area of the design.
Answer:

Let AD be the median of the equilateral \( \triangle ABC \), so \( BD = \frac{BC}{2} \). The centroid O divides the median in the ratio \( 2 : 1 \), so \( OD = \frac{OA}{2} = 16 \) cm (as \( OA = 32 \) cm is the radius).
\( AD = OA + OD = 32 + 16 = 48 \) cm. In \( \triangle ADB \), \( \sin 60^\circ = \frac{AD}{AB} \Rightarrow \frac{\sqrt{3}}{2} = \frac{48}{AB} \Rightarrow AB = \frac{96}{\sqrt{3}} = 32\sqrt{3} \) cm.
Area of \( \triangle ABC = \frac{\sqrt{3}}{4} \times (32\sqrt{3})^2 = 768\sqrt{3} \) cm². Area of the circle \( = \frac{22}{7} \times 32 \times 32 = \frac{22528}{7} \) cm².
Area of the design \( = \) area of the circle \( - \) area of \( \triangle ABC = \left(\frac{22528}{7} - 768\sqrt{3}\right) \) cm².
Hence, the area of the design is \( \left(\frac{22528}{7} - 768\sqrt{3}\right) \) cm² (about 1888.1 cm²).
In simple words: The triangle is inscribed in the circle, so its centroid is the centre and the circle's radius is two-thirds of the median.

Teacher's Note:
1. The circumradius of an equilateral triangle is two-thirds of its median.
2. Design = circle minus triangle.