CBSE Class 10 Mathematics Areas Related to Circles Assignment Set 11

Excluded from NCERT: Exercise 2 (Questions 7 to 16)

Question 1. In the given figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region.
Answer:

Side of the square \( = 14 \) cm, so its area \( = 14^2 = 196 \) cm². Circles with centres at neighbouring corners touch each other externally, so the radius of each circle is \( \frac{14}{2} = 7 \) cm.
Area of a quadrant \( = \frac{\pi r^2}{4} = \frac{22}{7} \times \frac{7^2}{4} = \frac{77}{2} \) cm². The square contains one quadrant of each circle, so the total area of the four quadrants \( = 4 \times \frac{77}{2} = 154 \) cm².
Area of the shaded region \( = \) area of the square \( - \) area of the four quadrants \( = 196 - 154 = 42 \) cm².
Hence, the area of the shaded region is 42 cm².
In simple words: Four quarter circles of radius 7 cm are cut from the corners of the square.

Teacher's Note:
1. Break the shaded region into simple figures (circles, sectors, triangles, squares) and add or subtract their areas.
2. Write every area with its unit.

Question 2. The figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find: (i) the distance around the track along its inner edge (ii) the area of the track.
Answer:

The inner semicircles have diameter 60 m, so the inner radius is \( r = 30 \) m. The outer radius is \( R = 30 + 10 = 40 \) m.
(i) Distance around the track along its inner edge \( = 106 + 106 + 2 \times \) (inner semicircle) \( = 212 + 2\pi r = 212 + 2 \times \frac{22}{7} \times 30 = 212 + \frac{1320}{7} = \frac{2804}{7} \) m.
(ii) The track consists of two rectangular strips, each \( 106 \times 10 \), and two semicircular rings that together make one full ring.
Area of the track \( = 2 \times (106 \times 10) + \pi(R^2 - r^2) = 2120 + \frac{22}{7}(40^2 - 30^2) = 2120 + \frac{22}{7} \times 70 \times 10 = 2120 + 2200 = 4320 \) m².
Hence, the distance is \( \frac{2804}{7} \) m (about 400.57 m) and the area of the track is 4320 m².
In simple words: The two curved ends together make one ring; the two straight parts are rectangles.

Teacher's Note:
1. Add the two straight 106 m edges to the circumference of a full circle of radius 30 m.
2. Area of the ring uses pi (R squared minus r squared).

Question 3. In the figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If \( OA = 7 \) cm, find the area of the shaded region.
Answer:

Radius of the larger circle \( OA = OC = OD = 7 \) cm and \( AB = 2 \times 7 = 14 \) cm. The smaller circle has diameter \( OD = 7 \) cm, so its radius is \( \frac{7}{2} \) cm.
Area of \( \triangle BCA = \frac{1}{2} \times AB \times OC = \frac{1}{2} \times 14 \times 7 = 49 \) cm².
Area of the larger semicircle \( ACB = \frac{1}{2} \times \frac{22}{7} \times 7^2 = 77 \) cm². Area of the smaller circle \( = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{2} \) cm².
Area of the shaded region \( = \) area of the larger semicircle \( - \) area of the triangle \( + \) area of the smaller circle \( = 77 - 49 + \frac{77}{2} = \frac{133}{2} = 66.5 \) cm².
Hence, the area of the shaded region is 66.5 cm².
In simple words: Two shaded segments make the semicircle minus the triangle; add the small circle.

Teacher's Note:
1. Break the shaded region into simple figures (circles, sectors, triangles, squares) and add or subtract their areas.
2. Write every area with its unit.

Question 4. The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region. (Use \( \pi = 3.14 \) and \( \sqrt{3} = 1.73205 \))
Answer:

Let the side of the triangle be a. \( \frac{\sqrt{3}}{4}a^2 = 17320.5 \Rightarrow a^2 = 17320.5 \times \frac{4}{1.73205} = 40000 \Rightarrow a = 200 \) cm.
Radius of each circle \( = \frac{200}{2} = 100 \) cm. Each angle of the triangle is \( 60^\circ \), so each circle contributes a sector of \( 60^\circ \) inside the triangle.
Area of one sector \( = \frac{60^\circ}{360^\circ} \times 3.14 \times 100^2 = \frac{15700}{3} \) cm², so the area of the 3 sectors \( = 3 \times \frac{15700}{3} = 15700 \) cm².
Area of the shaded region \( = \) area of \( \triangle ABC - \) area of the 3 sectors \( = 17320.5 - 15700 = 1620.5 \) cm².
Hence, the area of the shaded region is 1620.5 cm².
In simple words: The three 60 degree sectors cover everything in the triangle except the small curved triangle in the middle.

Teacher's Note:
1. Each circle touches the next because the radius is half the side.
2. Three 60 degree sectors make half a circle in all: 15700 cm².

Question 5. On a square handkerchief, nine circular designs each of radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.
Answer:

Radius of each design \( = 7 \) cm, so its diameter is 14 cm. There are three circles along each side of the square, so the side of the square \( = 3 \times 14 = 42 \) cm.
Area of the square \( = 42 \times 42 = 1764 \) cm². Area of one circle \( = \frac{22}{7} \times 7 \times 7 = 154 \) cm², so the area of the nine designs \( = 9 \times 154 = 1386 \) cm².
Area of the remaining portion \( = 1764 - 1386 = 378 \) cm².
Hence, the area of the remaining portion is 378 cm².
In simple words: Three circles side by side give the side of the square.

Teacher's Note:
1. Count the circles along one side to find the side of the square.
2. Remaining area = square minus nine circles.

Question 6. In the given figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If \( OD = 2 \) cm, find the area of the (i) quadrant OACB, (ii) shaded region.
Answer:

Radius of the quadrant \( = 3.5 = \frac{7}{2} \) cm.
(i) Area of the quadrant \( OACB = \frac{\pi r^2}{4} = \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{8} \) cm².
(ii) Area of \( \triangle BOD = \frac{1}{2} \times OD \times OB = \frac{1}{2} \times 2 \times \frac{7}{2} = \frac{7}{2} \) cm².
Area of the shaded region \( = \) area of the quadrant \( - \) area of \( \triangle BOD = \frac{77}{8} - \frac{7}{2} = \frac{49}{8} \) cm².
Hence, the quadrant has area \( \frac{77}{8} \) cm² and the shaded region has area \( \frac{49}{8} \) cm².
In simple words: Subtract the right triangle BOD from the quadrant.

Teacher's Note:
1. OB is a radius, so it is 3.5 cm.
2. The triangle has its right angle at O.

Question 7. In the figure, a square OABC is inscribed in a quadrant OPBQ. If \( OA = 20 \) cm, find the area of the shaded region. (Use \( \pi = 3.14 \))
Answer:

Join OB. In the right \( \triangle OAB \), \( OA = AB = 20 \) cm (sides of the square), so \( OB^2 = 20^2 + 20^2 = 800 \Rightarrow OB = 20\sqrt{2} \) cm. OB is the radius of the quadrant.
Area of the quadrant \( OPBQ = \frac{1}{4} \times 3.14 \times (20\sqrt{2})^2 = \frac{1}{4} \times 3.14 \times 800 = 628 \) cm². Area of the square \( OABC = 20^2 = 400 \) cm².
Area of the shaded region \( = 628 - 400 = 228 \) cm².
Hence, the area of the shaded region is 228 cm².
In simple words: The diagonal of the square is the radius of the quadrant.

Teacher's Note:
1. Diagonal of a square of side a is a root 2.
2. Shaded area = quadrant minus square.

Question 8. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If \( \angle AOB = 30^\circ \), find the area of the shaded region.
Answer:

Area of the sector \( OAEB = \frac{30^\circ}{360^\circ} \times \frac{22}{7} \times 21^2 = \frac{1}{12} \times \frac{22}{7} \times 441 = \frac{231}{2} \) cm².
Area of the sector \( OCFD = \frac{30^\circ}{360^\circ} \times \frac{22}{7} \times 7^2 = \frac{1}{12} \times \frac{22}{7} \times 49 = \frac{77}{6} \) cm².
Area of the shaded region \( = \frac{231}{2} - \frac{77}{6} = \frac{693 - 77}{6} = \frac{616}{6} = \frac{308}{3} \) cm².
Hence, the area of the shaded region is \( \frac{308}{3} \) cm² (about 102.67 cm²).
In simple words: Big sector minus small sector, both with the same angle 30 degrees.

Teacher's Note:
1. Both sectors have the same central angle.
2. Use (R squared minus r squared) to cross-check: pi/12 x 392.

Question 9. In the given figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.
Answer:

Radius of the quadrant \( ABC = 14 \) cm, so \( AB = AC = 14 \) cm. Area of the quadrant \( = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 154 \) cm². Area of the right \( \triangle ABC = \frac{1}{2} \times 14 \times 14 = 98 \) cm².
Area of the segment (between chord BC and the arc of the quadrant) \( = 154 - 98 = 56 \) cm².
By the Pythagoras theorem \( BC^2 = 14^2 + 14^2 = 392 \Rightarrow BC = 14\sqrt{2} \) cm, so the radius of the semicircle is \( 7\sqrt{2} \) cm. Area of the semicircle \( = \frac{1}{2} \times \frac{22}{7} \times (7\sqrt{2})^2 = 154 \) cm².
Area of the shaded region \( = \) area of the semicircle \( - \) area of the segment \( = 154 - 56 = 98 \) cm².
Hence, the area of the shaded region is 98 cm².
In simple words: The shaded crescent is the semicircle on BC minus the segment cut by the quadrant's arc.

Teacher's Note:
1. Find BC with the Pythagoras theorem to get the radius of the semicircle.
2. Segment = sector minus triangle.

Question 10. Calculate the area of the designed region in the given figure, common between the two quadrants of circles of radius 8 cm each.
Answer:

The designed region is the part common to the sector \( BAEC \) (centre B) and the sector \( DAFC \) (centre D); it is made of two equal segments with the common chord AC.
Area of the sector \( BAEC = \frac{90^\circ}{360^\circ} \times \frac{22}{7} \times 8^2 = \frac{352}{7} \) cm². Area of \( \triangle BAC = \frac{1}{2} \times BA \times BC = \frac{1}{2} \times 8 \times 8 = 32 \) cm².
Area of one segment \( AEC = \frac{352}{7} - 32 = \frac{128}{7} \) cm².
Area of the designed region \( = 2 \times \) (area of segment AEC) \( = 2 \times \frac{128}{7} = \frac{256}{7} \) cm².
Hence, the area of the designed region is \( \frac{256}{7} \) cm² (about 36.57 cm²).
In simple words: The common part of two quadrants is two equal segments joined along the diagonal.

Teacher's Note:
1. Each segment is a quadrant minus a right triangle.
2. Double the segment area.