GSEB Class 10 Maths Solutions Chapter 12 Areas Related to Circles Exercise 12.3

Question 1. Find the area of shaded region in the given figure. If PQ = 24 cm, PR = 7 cm, and O is the centre of the circle. (CBSE 2012)
Answer: \( \angle QPR \) is a right angle because an angle subtended in a semicircle is a right angle.
Therefore,
\( QR^2 = PR^2 + PQ^2 = 24^2 + 7^2 = 576 + 49 \)
\( QR^2 = 625 \)
\( QR = 25 \) cm
Radius of circle \( r = \frac{QR}{2} = \frac{25}{2} \) cm
Area of semicircle \( = \frac{1}{2}\pi r^2 \)
Area of \( \triangle PQR = \frac{1}{2} \times PQ \times PR \)
Area of shaded region = Area of semicircle – Area of \( \triangle PQR \)
\( = \frac{1}{2} \times \pi \times (\frac{25}{2})^2 - \frac{1}{2} \times 24 \times 7 \)
\( = \frac{1}{2} \times \frac{22}{7} \times \frac{25}{2} \times \frac{25}{2} - \frac{1}{2} \times 24 \times 7 \)
\( = \frac{6875}{28} - 84 \)
\( = \frac{6875 - 84 \times 28}{28} \)
\( = \frac{6875 - 2352}{28} \)
\( = \frac{4523}{28} \text{ cm}^2 \)
Hence, the area of the shaded region is \( \frac{4523}{28} \text{ cm}^2 \).
In simple words: First, we find the diameter of the circle using the Pythagorean theorem. Then, we calculate the area of the semicircle and subtract the area of the triangle from it to get the shaded area.

Exam Tip: Remember that an angle formed inside a semicircle with its vertex on the circumference is always a right angle, a key property for many geometry problems.

Question 2. Find the area of the shaded region in given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and \( \angle AOC = 40^\circ \). (CBSE 2012)
Answer: Let \( r_1 \) be the radius of the inner circle and \( r_2 \) be the radius of the outer circle.
Here, \( r_1 = 7 \) cm and \( r_2 = 14 \) cm. The angle \( \theta = 40^\circ \).
Area of shaded region = Area of sector OAC – Area of sector OBD
\( = \frac{\pi r_2^2 \theta}{360^\circ} - \frac{\pi r_1^2 \theta}{360^\circ} \)
\( = \frac{\pi \theta}{360^\circ} (r_2^2 - r_1^2) \)
\( = \frac{22}{7} \times \frac{40}{360} (14^2 - 7^2) \)
\( = \frac{22}{7} \times \frac{1}{9} (196 - 49) \)
\( = \frac{22}{63} \times 147 \)
\( = \frac{22}{9} \times 21 \)
\( = \frac{22 \times 7}{3} \)
\( = \frac{154}{3} \text{ cm}^2 \)
In simple words: To find the shaded area between two circles, we subtract the area of the smaller sector from the area of the larger sector. We use the formula for a sector's area and the given radii and angle.

Exam Tip: For concentric circles, the area of the region between two sectors with the same central angle can be calculated by factoring out common terms, simplifying the calculation.

Question 3. Find the area of shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
Answer: Side of square = 14 cm
Radius of semicircle = \( \frac{14}{2} = 7 \) cm
Area of shaded region = Area of square – 2 \( \times \) Area of semicircle
Area of square = Side \( \times \) Side = \( 14 \times 14 = 196 \text{ cm}^2 \)
Area of a semicircle = \( \frac{1}{2} \pi r^2 \)
Area of 2 semicircles = \( 2 \times \frac{1}{2} \pi r^2 = \pi r^2 \)
\( = \frac{22}{7} \times 7 \times 7 = 22 \times 7 = 154 \text{ cm}^2 \)
Area of shaded region = \( 196 - 154 = 42 \text{ cm}^2 \)
In simple words: We find the area of the square first. Then, we calculate the area of the two semicircles combined. Finally, we take the square's area and subtract the area of the two semicircles to get the shaded part.

Exam Tip: Recognize that two semicircles with the same diameter form a full circle. This simplifies calculations, allowing you to use the area of a full circle rather than two separate semicircles.

Question 4. Find the area of the shaded region in the given 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. (CBSE 2016)
Answer: Here, radius of the arc \( r = 6 \) cm.
\( \triangle AOB \) is an equilateral triangle.
Hence, the angle \( \theta = 60^\circ \).
Side of the equilateral triangle \( a = 12 \) cm.
Area of shaded region = Area of equilateral triangle + Area of circle – Area of sector
Area of equilateral triangle \( = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \times 12^2 = \frac{\sqrt{3}}{4} \times 144 = 36\sqrt{3} \text{ cm}^2 \)
Area of circle \( = \pi r^2 = \pi \times 6^2 = 36\pi \text{ cm}^2 \)
Area of sector \( = \frac{\pi r^2 \theta}{360^\circ} = \frac{\pi \times 6^2 \times 60}{360} = \frac{36\pi \times 60}{360} = 6\pi \text{ cm}^2 \)
Area of shaded region \( = 36\sqrt{3} + 36\pi - 6\pi \)
\( = 36\sqrt{3} + 30\pi \)
\( = [36\sqrt{3} + 30 \times \frac{22}{7}] \)
\( = [\frac{660}{7} + 36\sqrt{3}] \text{ cm}^2 \)
In simple words: We calculate the area of the equilateral triangle, the full circle, and the sector separately. Then, we add the triangle's area to the circle's area and subtract the sector's area to find the total shaded region.

Exam Tip: Remember the formulas for the area of an equilateral triangle \( \left(\frac{\sqrt{3}}{4} a^2\right) \), a circle \( (\pi r^2) \), and a sector \( \left(\frac{\theta}{360^\circ} \pi r^2\right) \) for similar problems.

Question 5. 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 figure. Find the area of the remaining portion of the square.
Answer: Side of the square = 4 cm
Radius of the circle cut from the center = \( \frac{\text{diameter}}{2} = \frac{2}{2} = 1 \) cm
Radius of each quadrant cut from the corners = 1 cm
Area of shaded region = Area of square – (Area of 4 quadrants + Area of central circle)
Area of square = \( 4 \times 4 = 16 \text{ cm}^2 \)
Area of 4 quadrants = \( 4 \times \frac{1}{4} \pi r^2 = \pi r^2 \)
Area of central circle = \( \pi r^2 \)
Total area of cut portions = \( \pi r^2 + \pi r^2 = 2\pi r^2 \)
Given \( r = 1 \) cm, so total cut area = \( 2 \times \pi \times 1^2 = 2\pi \text{ cm}^2 \)
Using \( \pi = \frac{22}{7} \)
Total cut area = \( 2 \times \frac{22}{7} = \frac{44}{7} \text{ cm}^2 \)
Area of remaining portion = \( 16 - \frac{44}{7} \)
\( = \frac{16 \times 7 - 44}{7} \)
\( = \frac{112 - 44}{7} = \frac{68}{7} \text{ cm}^2 \)
In simple words: First, we find the area of the entire square. Then, we calculate the combined area of the four corner quadrants and the single central circle. Finally, we subtract this combined cut-out area from the square's total area to find the remaining part.

Exam Tip: Remember that four quadrants of the same radius effectively form one full circle, simplifying the calculation of their combined area.

Question 6. In a round table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in figure. Find the area of the design (Shaded region).
Answer: The radius of the circular table cover \( r = 32 \) cm.
Since \( \triangle ABC \) is an equilateral triangle, its centroid coincides with the center of the circle (O).
The ratio of OD to OA is \( \frac{OD}{OA} = \frac{1}{2} \). So, \( \frac{OD+OA}{OA} = \frac{3}{2} \).
This means the height of the triangle (h) is related to the radius (r) as \( h = \frac{3}{2}r \).
\( h = \frac{3}{2} \times 32 = 48 \) cm.
In the right-angled triangle ADB (where D is the midpoint of BC):
\( AB^2 = AD^2 + BD^2 \)
Let the side of the equilateral triangle be \( a \). So \( BD = \frac{a}{2} \) and \( AD = h = 48 \) cm.
\( a^2 = h^2 + (\frac{a}{2})^2 \)
\( a^2 = 48^2 + \frac{a^2}{4} \)
\( a^2 - \frac{a^2}{4} = 48^2 \)
\( \frac{3a^2}{4} = 48^2 \)
\( a^2 = \frac{48^2 \times 4}{3} = \frac{2304 \times 4}{3} = 768 \times 4 = 3072 \)
Area of design or shaded region = Area of circle – Area of equilateral triangle
Area of circle \( = \pi r^2 = \pi \times 32^2 = 1024\pi \text{ cm}^2 \)
Area of equilateral triangle \( = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \times 3072 = 768\sqrt{3} \text{ cm}^2 \)
Area of shaded region \( = (1024\pi - 768\sqrt{3}) \text{ cm}^2 \)
Using \( \pi = \frac{22}{7} \) and \( \sqrt{3} = 1.73205 \):
\( = (1024 \times \frac{22}{7} - 768 \times 1.73205) \)
\( = (\frac{22528}{7} - 1330.1376) \)
\( \approx (3218.28 - 1330.14) \approx 1888.14 \text{ cm}^2 \)
*Note: The OCR solution for Question 6 seems incomplete and then skips to a different method and numbers. I will generate the full answer based on standard geometric principles using the given radius and the fact that it's an equilateral triangle.*
*Revisiting the OCR for solution of Q6:*
*The OCR solution starts with \( \frac{OD}{OA} = \frac{1}{2} \). Radius \( r = 32 \) cm. So OA = r = 32 cm. Then \( h = 32 \times \frac{3}{2} = 48 \) cm. This calculation of h is incorrect based on the typical properties of centroid for circumcircle. In a circumcircle, the radius R is \( \frac{2}{3} \) of the median (height) h. So \( r = \frac{2}{3} h \implies h = \frac{3}{2} r \). This is correct.*
*The side of the equilateral triangle \( a \) is related to the circumradius \( R \) by \( R = \frac{a}{\sqrt{3}} \).
So \( a = R\sqrt{3} = 32\sqrt{3} \) cm.
Area of equilateral triangle \( = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (32\sqrt{3})^2 = \frac{\sqrt{3}}{4} \times 1024 \times 3 = 768\sqrt{3} \text{ cm}^2 \).
Area of shaded region \( = \pi r^2 - \frac{\sqrt{3}}{4} a^2 = \pi (32)^2 - 768\sqrt{3} = (1024\pi - 768\sqrt{3}) \text{ cm}^2 \).
The calculation for \( a^2 = 64 \times 48 \) is not clear how it arises from the preceding steps. Let's follow the OCR's steps as closely as possible, correcting where it's demonstrably wrong or using the final given values.
Let's follow OCR provided steps even if it has some error in derivation. The OCR states:
\( \frac{OD}{OA} = \frac{1}{2} \)
\( \frac{OD + OA}{OA} = \frac{3}{2} \)
Given \( OA \) is radius \( r \). So \( h \) (height of equilateral triangle) is \( AD \). \( OA = r \). OD is altitude from O to side. In an equilateral triangle, centroid divides median in ratio 2:1. So \( OA:OD = 2:1 \). \( OA = r = 32 \). So \( OD = \frac{1}{2} OA = 16 \). Total height \( h = AD = AO + OD = 32 + 16 = 48 \). This part of OCR seems correct.
Then \( AB^2 = AD^2 + BD^2 \). \( a^2 = h^2 + (\frac{a}{2})^2 \).
\( a^2 = 48^2 + \frac{a^2}{4} \)
\( \frac{3a^2}{4} = 48^2 \implies a^2 = \frac{48 \times 48 \times 4}{3} = 48 \times 16 \times 4 = 3072 \). This is also correct. The OCR then has \( a^2 = 64 \times 48 \) which is \( 3072 \). This is okay, just a different multiplication order.
So \( a^2 = 3072 \).
Area of design or shaded region \( = \pi r^2 – \frac{\sqrt{3}}{4} a^2 \)
\( = \frac{22}{7} \times 32 \times 32 – \frac{\sqrt{3}}{4} \times 3072 \)
\( = \frac{22528}{7} – 768\sqrt{3} \)
\( = (\frac{22528}{7} – 768\sqrt{3}) \text{ cm}^2 \). This final expression matches the OCR. I will stick to this one.

In simple words: We find the total area of the circular table cover. Then, we determine the area of the equilateral triangle in the middle. The design's area is the circle's area minus the triangle's area.

Exam Tip: For problems involving equilateral triangles inscribed in circles, remember the relationship between the circle's radius and the triangle's height or side length for accurate calculations.

Question 7. In figure, ABCD is a square of side 14 cm with centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of shaded region.
Answer: Side of square = 14 cm.
From the figure, the radius of each circle is half the side of the square.
Radius of each quadrant \( r = \frac{14}{2} = 7 \) cm.
The four circles are drawn with centers A, B, C, D. Each circle touches two others externally. This means the side of the square is equal to the diameter of two circles (2r), or rather, from the image, each corner circle has radius \( r = \frac{14}{2} = 7 \) cm.
Area of shaded region = Area of square – 4 \( \times \) Area of quadrants
Area of square = Side \( \times \) Side = \( 14 \times 14 = 196 \text{ cm}^2 \)
Area of 4 quadrants = \( 4 \times \frac{1}{4} \pi r^2 = \pi r^2 \)
\( = \frac{22}{7} \times 7 \times 7 = 22 \times 7 = 154 \text{ cm}^2 \)
Area of shaded region = \( 196 - 154 = 42 \text{ cm}^2 \)
In simple words: We determine the square's area. Then, we calculate the combined area of the four quadrants, which essentially form a full circle. Subtracting this combined area from the square's area gives us the shaded region.

Exam Tip: When four quadrants are at the corners of a square and their radii sum up to the square's side, they often combine to form a single full circle for simpler area calculations.

Question 8. 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:
(i) Distance between two inner parallel lines = 60 m
Radius of inner circular ends \( r_1 = \frac{60}{2} = 30 \) m
Length of inner parallel segments = 106 m
Distance around the track along its inner edge = Length of 2 straight sections + Circumference of 2 semicircles
\( = 106 + 106 + 2 \times \frac{1}{2} \times 2\pi r_1 \)
\( = 212 + 2\pi r_1 \)
\( = 212 + 2 \times \frac{22}{7} \times 30 \)
\( = 212 + \frac{1320}{7} \)
\( = \frac{212 \times 7 + 1320}{7} \)
\( = \frac{1484 + 1320}{7} = \frac{2804}{7} \) m
(ii) The track is 10 m wide.
Radius of outer circular ends \( r_2 = r_1 + \text{width} = 30 + 10 = 40 \) m
Area of the track = Area of two rectangular tracks + Area of (Outer semicircles – Inner semicircles)
Area of two rectangular tracks = \( 2 \times \text{length} \times \text{width} = 2 \times 106 \times 10 = 2120 \text{ m}^2 \)
Area of the region between semicircles = \( 2 \times \frac{1}{2} \pi (r_2^2 - r_1^2) = \pi (r_2^2 - r_1^2) \)
\( = \pi (40^2 - 30^2) \)
\( = \frac{22}{7} (1600 - 900) \)
\( = \frac{22}{7} \times 700 = 22 \times 100 = 2200 \text{ m}^2 \)
Total area of the track = \( 2120 + 2200 = 4320 \text{ m}^2 \)
In simple words: For part (i), we add the lengths of the two straight sections to the circumference of a full circle formed by the two semicircular ends. For part (ii), we calculate the area of the two straight rectangular parts and then add the area of the annular region (the ring shape) formed by the semicircular ends.

Exam Tip: When dealing with composite shapes like a racing track, break the figure down into simpler geometric shapes (rectangles and semicircles) to calculate distances and areas systematically.

Question 9. In 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. (CBSE 2000, 2010, 2013)
Answer: Here, OA is the radius of the larger circle, \( r = 7 \) cm.
Since AB is a diameter, \( AB = 2 \times OA = 2 \times 7 = 14 \) cm.
CD is also a diameter, so \( CD = 14 \) cm.
OD is the radius of the larger circle, which is 7 cm.
OD is also the diameter of the smaller circle.
Radius of smaller circle \( r_2 = \frac{OD}{2} = \frac{7}{2} \) cm.
Area of shaded region = Area of smaller circle + Area of semicircle (ACB) – Area of \( \triangle ABC \)
Area of smaller circle \( = \pi r_2^2 = \pi \times (\frac{7}{2})^2 = \frac{49}{4}\pi \text{ cm}^2 \)
Area of semicircle ACB \( = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (7)^2 = \frac{49}{2}\pi \text{ cm}^2 \)
For \( \triangle ABC \), base \( AB = 14 \) cm, and height \( OC = r = 7 \) cm.
Area of \( \triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times OC = \frac{1}{2} \times 14 \times 7 = 49 \text{ cm}^2 \)
Area of shaded region \( = \frac{49}{4}\pi + \frac{49}{2}\pi - 49 \)
\( = \frac{49\pi + 98\pi}{4} - 49 \)
\( = \frac{147\pi}{4} - 49 \)
Using \( \pi = \frac{22}{7} \):
\( = \frac{147}{4} \times \frac{22}{7} - 49 \)
\( = \frac{21}{4} \times 22 - 49 \)
\( = \frac{21 \times 11}{2} - 49 \)
\( = \frac{231}{2} - 49 \)
\( = 115.5 - 49 = 66.5 \text{ cm}^2 \)
In simple words: To find the shaded area, we add the area of the small circle to the area of the large semicircle, then subtract the area of the triangle formed by the large circle's diameter and radius.

Exam Tip: Remember to identify the different geometric shapes that make up the shaded region and apply their respective area formulas. Pay attention to how diameters and radii are defined in relation to each other.

Question 10. 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 side of the triangle (see figure). Find the area of shaded region. (Use \( \pi = 3.14 \) and \( \sqrt{3} = 1.73205 \))
Answer: Area of an equilateral triangle ABC \( = 17320.5 \text{ cm}^2 \).
Formula for area of equilateral triangle \( = \frac{\sqrt{3}}{4} a^2 \)
So, \( \frac{\sqrt{3}}{4} a^2 = 17320.5 \)
Using \( \sqrt{3} = 1.73205 \):
\( \frac{1.73205}{4} a^2 = 17320.5 \)
\( a^2 = \frac{17320.5 \times 4}{1.73205} \)
\( a^2 = 10000 \times 4 = 40000 \)
\( a = \sqrt{40000} = 200 \) cm
Radius of each circle \( r = \frac{a}{2} = \frac{200}{2} = 100 \) cm.
The central angle of each sector in an equilateral triangle is \( \theta = 60^\circ \).
Area of shaded region = Area of equilateral triangle – 3 \( \times \) Area of sector
Area of one sector \( = \frac{\pi r^2 \theta}{360^\circ} = \frac{3.14 \times 100^2 \times 60}{360} \)
\( = \frac{3.14 \times 10000 \times 1}{6} \)
\( = \frac{31400}{6} = 5233.33 \text{ cm}^2 \) (approx.)
Area of 3 sectors \( = 3 \times \frac{31400}{6} = \frac{31400}{2} = 15700 \text{ cm}^2 \)
Area of shaded region \( = 17320.5 - 15700 = 1620.5 \text{ cm}^2 \)
In simple words: First, we use the given area of the equilateral triangle to find its side length, and then the radius of the circles. Next, we calculate the area of three sectors formed within the triangle. Finally, we subtract the total area of these three sectors from the triangle's area to find the shaded region.

Exam Tip: For equilateral triangles, remember that all angles are 60 degrees. This is crucial when calculating the area of sectors with vertices at the triangle's corners.

Question 11. In 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 circular design \( r = 7 \) cm.
Diameter of each circular design \( d = 2 \times r = 2 \times 7 = 14 \) cm.
From the figure, the side of the square handkerchief ABCD is equal to 3 times the diameter of one circular design.
Side of square ABCD = \( 3 \times d = 3 \times 14 = 42 \) cm.
Area of square ABCD = Side \( \times \) Side = \( 42 \times 42 = 1764 \text{ cm}^2 \).
Area of one circular design \( = \pi r^2 = \frac{22}{7} \times 7 \times 7 = 22 \times 7 = 154 \text{ cm}^2 \).
Area of 9 circular designs = \( 9 \times 154 = 1386 \text{ cm}^2 \).
Area of remaining portion of the handkerchief = Area of square ABCD – Area of 9 circular designs
\( = 1764 - 1386 = 378 \text{ cm}^2 \).
In simple words: We find the side of the square handkerchief by using the diameter of the circular designs. Then, we calculate the area of the square and the total area of the nine circular designs. The remaining area is found by subtracting the designs' area from the square's area.

Exam Tip: For arrangements of circles within a square, carefully determine the relationship between the circle's radius/diameter and the square's side length to correctly find areas.

Question 12. In figure OACB is a quadrant of circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
(i) quadrant OACB
(ii) shaded region. (CBSE 2017)
Answer: Radius of quadrant OACB \( r = 3.5 \) cm.
(i) Area of quadrant OACB \( = \frac{1}{4} \pi r^2 \)
\( = \frac{1}{4} \times \frac{22}{7} \times (3.5)^2 \)
\( = \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \)
\( = \frac{1}{4} \times 11 \times \frac{7}{2} \)
\( = \frac{77}{8} \text{ cm}^2 \)
(ii) For the shaded region, we need to find the area of \( \triangle BOD \).
Base of \( \triangle BOD = OB = r = 3.5 \) cm.
Height of \( \triangle BOD = OD = 2 \) cm.
Area of \( \triangle BOD = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OB \times OD \)
\( = \frac{1}{2} \times 3.5 \times 2 = 3.5 \text{ cm}^2 \)
Area of shaded region = Area of quadrant OACB – Area of \( \triangle BOD \)
\( = \frac{77}{8} - 3.5 \)
\( = \frac{77}{8} - \frac{7}{2} \)
\( = \frac{77 - 28}{8} = \frac{49}{8} \text{ cm}^2 \)
In simple words: For part (i), we use the formula for a quadrant's area with the given radius. For part (ii), we find the area of the small triangle within the quadrant and subtract it from the quadrant's area to get the shaded part.

Exam Tip: Clearly distinguish between the area of the quadrant and any triangular or other shapes within it. Break down complex shaded regions into simpler, calculable parts.

Question 13. In figure, a square OABC is inscribed in a quadrant OPBQ If OA = 20 cm, find the area of the shaded region. (Use \( \pi = 3.14 \)) (CBSE 2012, 2014)
Answer: OABC is a square inscribed in a quadrant OPBQ.
Side of the square \( OA = 20 \) cm.
In a square, \( OA = AB = BC = OC = 20 \) cm.
To find the radius of the quadrant (OB), we use the Pythagorean theorem in right-angled \( \triangle OAB \).
\( OB^2 = OA^2 + AB^2 \)
\( OB^2 = 20^2 + 20^2 \)
\( OB^2 = 400 + 400 = 800 \)
\( OB = \sqrt{800} = \sqrt{400 \times 2} = 20\sqrt{2} \) cm.
The radius of the quadrant \( r = OB = 20\sqrt{2} \) cm.
Area of shaded region = Area of quadrant OPBQ – Area of square OABC
Area of quadrant OPBQ \( = \frac{1}{4} \pi r^2 \)
\( = \frac{1}{4} \times 3.14 \times (20\sqrt{2})^2 \)
\( = \frac{1}{4} \times 3.14 \times (400 \times 2) \)
\( = \frac{1}{4} \times 3.14 \times 800 \)
\( = 3.14 \times 200 = 628 \text{ cm}^2 \)
Area of square OABC = \( OA \times OA = 20 \times 20 = 400 \text{ cm}^2 \).
Area of shaded region = \( 628 - 400 = 228 \text{ cm}^2 \).
In simple words: First, we find the radius of the quadrant using the diagonal of the inscribed square. Then, we calculate the area of the entire quadrant and the area of the square. Subtracting the square's area from the quadrant's area gives us the shaded region.

Exam Tip: When a square is inscribed in a quadrant, the diagonal of the square is equal to the radius of the quadrant. Use the Pythagorean theorem to find this relationship.

Question 14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and with common center O as shown in the figure. If \( \angle AOB = 30^\circ \), find the area of shaded region.
Answer: Radius of the inner circle \( r_1 = 7 \) cm.
Radius of the outer circle \( r_2 = 21 \) cm.
Central angle \( \theta = 30^\circ \).
Area of shaded region = Area of sector OAB – Area of sector OCD
\( = \frac{\pi r_2^2 \theta}{360^\circ} - \frac{\pi r_1^2 \theta}{360^\circ} \)
\( = \frac{\pi \theta}{360^\circ} (r_2^2 - r_1^2) \)
\( = \frac{22}{7} \times \frac{30}{360} (21^2 - 7^2) \)
\( = \frac{22}{7} \times \frac{1}{12} (441 - 49) \)
\( = \frac{22}{7} \times \frac{1}{12} (392) \)
\( = \frac{22}{7} \times \frac{392}{12} \)
\( = \frac{22}{7} \times \frac{98}{3} \)
\( = \frac{22 \times 14}{3} \) (since \( \frac{98}{7} = 14 \))
\( = \frac{308}{3} \text{ cm}^2 \)
In simple words: To find the shaded area, we calculate the area of the larger sector and then subtract the area of the smaller sector, both sharing the same central angle.

Exam Tip: For regions between concentric sectors, always use the difference of the squares of the radii \( (R^2 - r^2) \) in the sector area formula to simplify computations.

Question 15. In 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 shaded region. (CBSE 2008,2012)
Answer: Radius of quadrant ABC \( r = 14 \) cm.
In right-angled \( \triangle BAC \):
\( BC^2 = AB^2 + AC^2 \)
\( BC^2 = 14^2 + 14^2 \)
\( BC^2 = 196 + 196 = 2 \times 196 \)
\( BC = \sqrt{2 \times 196} = 14\sqrt{2} \) cm.
BC is the diameter of the semicircle.
Radius of semicircle \( r_1 = \frac{BC}{2} = \frac{14\sqrt{2}}{2} = 7\sqrt{2} \) cm.
Area of shaded region = Area of semicircle – Area of segment BPC
Area of segment BPC = Area of quadrant ABC – Area of \( \triangle ABC \)
Area of quadrant ABC \( = \frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 14^2 \)
\( = \frac{1}{4} \times \frac{22}{7} \times 196 = \frac{22 \times 28}{4} = 22 \times 7 = 154 \text{ cm}^2 \).
Area of \( \triangle ABC = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 14 \times 14 = 7 \times 14 = 98 \text{ cm}^2 \).
Area of segment BPC \( = 154 - 98 = 56 \text{ cm}^2 \).
Area of semicircle (with diameter BC) \( = \frac{1}{2} \pi r_1^2 = \frac{1}{2} \times \frac{22}{7} \times (7\sqrt{2})^2 \)
\( = \frac{1}{2} \times \frac{22}{7} \times (49 \times 2) \)
\( = \frac{1}{2} \times \frac{22}{7} \times 98 \)
\( = 11 \times 14 = 154 \text{ cm}^2 \).
Area of shaded region = Area of semicircle – Area of segment BPC
\( = 154 - 56 = 98 \text{ cm}^2 \).
In simple words: First, we find the length of BC, which becomes the diameter of the semicircle. Then, we calculate the area of the semicircle and the area of the segment BPC (which is the quadrant's area minus the triangle's area). Finally, we subtract the segment's area from the semicircle's area to get the shaded region.

Exam Tip: In figures with multiple overlapping shapes, calculate the area of individual components and then add or subtract them carefully to find the desired shaded area.

Question 16. Calculate the area of the diagonal region in figure common between the two quadrants of circles of radius 8 cm each.
Answer: Radius of each quadrant \( r = 8 \) cm.
The common diagonal region is formed by the overlap of two quadrants. This region can be found by summing the areas of the two quadrants and then subtracting the area of the square that encloses them. However, a simpler approach is to consider the segments.
Area of one segment = Area of quadrant – Area of corresponding triangle
Area of quadrant \( = \frac{1}{4} \pi r^2 = \frac{1}{4} \times \pi \times 8^2 = \frac{1}{4} \times 64\pi = 16\pi \text{ cm}^2 \).
Area of triangle (e.g., \( \triangle ADC \)) \( = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 8 = 32 \text{ cm}^2 \).
Area of one segment \( = 16\pi - 32 \text{ cm}^2 \).
The shaded region is formed by two such segments. So, the area of the shaded region is twice the area of one segment.
Area of shaded region \( = 2 \times (16\pi - 32) \)
\( = (32\pi - 64) \text{ cm}^2 \).
Using \( \pi = \frac{22}{7} \):
\( = (32 \times \frac{22}{7} - 64) \)
\( = (\frac{704}{7} - 64) \)
\( = \frac{704 - 64 \times 7}{7} \)
\( = \frac{704 - 448}{7} = \frac{256}{7} \text{ cm}^2 \).
In simple words: We find the area of one circular segment by subtracting the area of the square's corner triangle from the quadrant's area. Since the shaded region is made of two identical segments, we double the area of one segment to get the total shaded area.

Exam Tip: For overlapping circular shapes, it's often easiest to find the area of a segment (area of sector minus area of triangle) and then combine segments as needed for the final shaded region.