Ch 11: Areas of Combinations of Plane Figures
Master the "Area of Shaded Region" word problems from the Old NCERT Syllabus (Exercise 12.3). These are high-weightage board exam questions.
Exercise 12.3 (Questions 1 to 8)
💡 Strategy: Area of Shaded Region = Area of Outer Figure - Area of Unshaded Inner Figure.
Q1. Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
Solution:
Since RQ is a diameter passing through O, the angle in a semicircle is a right angle. So, $\angle RPQ = 90^\circ$.
By Pythagoras Theorem in $\Delta RPQ$:
$RQ^2 = RP^2 + PQ^2 = 7^2 + 24^2 = 49 + 576 = 625$
$RQ = 25$ cm (Diameter). Radius $r = 25/2$ cm.
Area of Semicircle = $\frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times \frac{25}{2} \times \frac{25}{2} = \frac{6875}{28} \text{ cm}^2$.
Area of $\Delta RPQ$ = $\frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 7 \times 24 = 84 \text{ cm}^2$.
Area of shaded region = Area of Semicircle - Area of Triangle
= $\frac{6875}{28} - 84 = \frac{6875 - 2352}{28} =$ $\frac{4523}{28} \text{ cm}^2$.
Q3. Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
Solution:
Area of Square ABCD = $\text{side}^2 = 14 \times 14 = 196 \text{ cm}^2$.
The two semicircles APD and BPC combine to form one full circle.
Diameter of semicircle = side of square = 14 cm. Radius $r = 7$ cm.
Area of two semicircles (one full circle) = $\pi r^2 = \frac{22}{7} \times 7 \times 7 = 154 \text{ cm}^2$.
Area of shaded region = Area of Square - Area of 2 Semicircles
= $196 - 154 =$ $42 \text{ cm}^2$.
Q8. Fig. 12.26 depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60m and they are each 106m long. If the track is 10m wide, find: (i) the distance around the track along its inner edge (ii) the area of the track.
Solution:
Inner diameter = 60m $\implies$ Inner radius $r = 30$m.
Outer radius $R = 30 + 10 (\text{width}) = 40$m.
Length of straight parallel sections = 106m.
(i) Distance along inner edge:
= $2 \times (\text{straight segments}) + 2 \times (\text{inner semicircles})$
= $(2 \times 106) + (2 \times \frac{1}{2} \times 2\pi r) = 212 + 2\pi(30)$
= $212 + 2 \times \frac{22}{7} \times 30 = 212 + \frac{1320}{7} =$ $\frac{2804}{7} \text{ m}$.
(ii) Area of the track:
= Area of 2 straight rectangular strips + Area of 2 semicircular ends.
Area of straight strips = $2 \times (\text{Length} \times \text{Width}) = 2 \times (106 \times 10) = 2120 \text{ m}^2$.
Area of curved ends (Forms a ring) = $\pi(R^2 - r^2) = \frac{22}{7} \times (40^2 - 30^2) = \frac{22}{7} \times (1600 - 900) = \frac{22}{7} \times 700 = 2200 \text{ m}^2$.
Total Area = $2120 + 2200 =$ $4320 \text{ m}^2$.
Exercise 12.3 (Questions 9 to 16)
Q9. In Fig. 12.27, AB and CD are two diameters of a circle perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.
Solution:
Radius of large circle $R = OA = 7$ cm.
Radius of small circle $r = OD/2 = 7/2$ cm.
Area of small circle = $\pi r^2 = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{2} = 38.5 \text{ cm}^2$.
The remaining shaded region is in the semicircle ABC.
Area of Semicircle ABC = $\frac{1}{2} \pi R^2 = \frac{1}{2} \times \frac{22}{7} \times 7^2 = 77 \text{ cm}^2$.
Area of $\Delta ABC$ (Base AB = 14, Height OC = 7) = $\frac{1}{2} \times 14 \times 7 = 49 \text{ cm}^2$.
Shaded region in semicircle = $77 - 49 = 28 \text{ cm}^2$.
Total Shaded Area = Area of small circle + Shaded region in semicircle
= $38.5 + 28 =$ $66.5 \text{ cm}^2$.
Q16. Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each.
Solution:
The design is formed by the intersection of two quadrants drawn from opposite corners of a square (side 8 cm).
Area of 2 Quadrants = $2 \times (\frac{90}{360} \times \pi r^2) = 2 \times \frac{1}{4} \times \frac{22}{7} \times 8^2 = \frac{1}{2} \times \frac{22}{7} \times 64 = \frac{704}{7} \text{ cm}^2$.
When you add the areas of the two quadrants, the common designed region is added twice. So, to find the design area, subtract the area of the square from the sum of the two quadrants.
Area of Square = $8 \times 8 = 64 \text{ cm}^2$.
Area of Design = Area of 2 Quadrants - Area of Square
= $\frac{704}{7} - 64 = \frac{704 - 448}{7} =$ $\frac{256}{7} \text{ cm}^2$.