Areas Related to Circles Ex 12.3 Solutions
Question - 1 : - Find the area of theshaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre ofthe circle.
Answer - 1 : - 
Solution:
Here, P is in the semi-circle and so,
P = 90°
So, it can be concluded that QR is hypotenuseof the circle and is equal to the diameter of the circle.
∴ QR = D
Using Pythagorean theorem,
QR2 = PR2+PQ2
Or, QR2 = 72+242
QR= 25 cm = Diameter
Hence, the radius of the circle = 25/2 cm
Now, the area of the semicircle = (πR2)/2
= (22/7)×(25/2)×(25/2)/2 cm2
= 13750/56 cm2 = 245.54 cm2
Also, area of the ΔPQR = ½×PR×PQ
=(½)×7×24 cm2
= 84 cm2
Hence, the area of the shaded region = 245.54cm2-84 cm2
= 161.54 cm2
Find the area of theshaded region in Fig. 12.20, if radii of the two concentric circles with centreO are 7 cm and 14 cm respectively and AOC = 40°.
Answer - 2 : - 
Given,
Angle made by sector = 40°,
Radius the inner circle = r = 7 cm, and
Radius of the outer circle = R = 14 cm
We know,
Area of the sector = (θ/360°)×πr2
So, Area of OAC = (40°/360°)×πr2 cm2
= 68.44 cm2
Area of the sector OBD = (40°/360°)×πr2 cm2
= (1/9)×(22/7)×72 = 17.11 cm2
Now, area of the shaded region ABDC = Area ofOAC – Area of the OBD
= 68.44 cm2 – 17.11 cm2 =51.33 cm2
Find the area of theshaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPCare semicircles.
Answer - 3 : - 
Question - 4 : - Find the area of theshaded region in Fig. 12.22, where a circular arc of radius 6 cm has been drawnwith vertex O of an equilateral triangle OAB of side 12 cm as centre.
Answer - 4 : - 
Question - 5 : - From each corner of a square of side 4 cm aquadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cmis cut as shown in Fig. 12.23. Find the area of the remaining portion of thesquare.
Answer - 5 : -
Side of the square = 4 cm
Radius of the circle = 1 cm
Four quadrant of a circle are cut from cornerand one circle of radius are cut from middle.
Area of square = (side)2= 42 =16 cm2
Area of the quadrant = (πR2)/4 cm2 =(22/7)×(12)/4 = 11/14 cm2
∴ Total area of the 4 quadrants = 4 ×(11/14) cm2 =22/7 cm2
Area of the circle = πR2 cm2 =(22/7×12) = 22/7 cm2
Area of the shaded region = Area of square –(Area of the 4 quadrants + Area of the circle)
= 16 cm2-(22/7) cm2+(22/7)cm2
= 68/7 cm2
Question - 6 : - In a circular table cover of radius 32 cm, adesign is formed leaving an equilateral triangle ABC in the middle as shown inFig. 12.24.
Answer - 6 : -
Find the area of the design.
Question - 7 : - In Fig. 12.25, ABCD is a square of side 14 cm.With centres A, B, C and D, four circles are drawn such that each circle touchexternally two of the remaining three circles.
Answer - 7 : - Find the area of the shaded region
Question - 8 : - Fig. 12.26 depicts aracing track whose left and right ends are semicircular.
The distance betweenthe two inner parallel line segments is 60 m and they are each 106 m long. Ifthe track is 10 m wide,
Answer - 8 : -
find:
(i) the distance around the track along its inner edge
(ii) the area of the track.
Question - 9 : - In Fig. 12.27, ABand CD are two diameters of a circle (with centre O) perpendicular to eachother and OD is the diameter of the smaller circle. If OA = 7 cm, find the areaof the shaded region.
Answer - 9 : - 
Question - 10 : - The area of anequilateral triangle ABC is 17320.5 cm2. With each vertex of thetriangle as centre, a circle is drawn with radius equal to half the length ofthe side of the triangle (see Fig. 12.28).
Answer - 10 : -
Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205)
Solution:
ABC is an equilateral triangle.
∴ ∠ A = ∠ B = ∠ C = 60°
There are three sectors each making 60°.
Area of ΔABC = 17320.5 cm2
⇒ √3/4 ×(side)2 = 17320.5
⇒ (side)2 =17320.5×4/1.73205
⇒ (side)2 = 4×104
⇒ side = 200 cm
Radius of the circles = 200/2 cm = 100 cm
Area of the sector = (60°/360°)×π r2 cm2
= 1/6×3.14×(100)2 cm2
= 15700/3cm2
Area of 3 sectors = 3×15700/3 = 15700 cm2
Thus, area of the shaded region = Area ofequilateral triangle ABC – Area of 3 sectors
= 17320.5-15700 cm2 = 1620.5cm2