SOLUTION
Unless stated otherwise, use π= $22/7 $1. Find the area of the shaded region in Fig, if PQ = 24 $cm$, PR = 7 $cm$ and O is the centre of the circle. Area of shaded region = Area of semicircle - area of triangle QPR In ∆QPR, ∠QPR=$90^0$, therefore ∆QPR is a right triangle. Therefore $QR^2=PR^2+PQ^2$ $QR^2=24^2+7^2$ $QR=25$ $cm$ is the diameter of the circle area of ∆QPR = $1/2× base× height$ = $1/2× PR× PQ$ = $1/2× 7×24$ = $84$ $cm^2$ Area of semicircle = ${πr^2}/2$ = $1/2× 22/7× 25/2×25/2 $ = $245.5 $ $cm^2$ Area of shaded region = Area of semicircle - area of triangle QPR = (245.5 - 84) $cm^2$ = 161.5 $cm^2$2. Find the area of the shaded region in Fig, if radii of the two concentric circles with centre O are 7 $cm$ and 14 $cm$ respectively and ∠AOC = 40°. Area of shaded region= area of sector(OAC) - area of sector(OBD) Area of sector (OAC) = $40/360$ ×π×$14^2$ Area of sector (OBD) = $40/360$ ×π×$7^2$ Area of shaded region= $40/360$ ×π×$14^2$ - $40/360$ ×π×$7^2$ = $(1/9 ×π)(14^2-7^2)$ = $(1/9 ×22/7)×(21)×(7)$ =$1/3× 22× 7$ =$154/3$ $cm^2$3. Find the area of the shaded region in Fig., if ABCD is a square of side 14 $cm$ and APD and BPC are semicircles. Area of shaded region= Area of square - 2 × area of semi circle Area of square = side × side = $14 × 14$ = 196 $cm^2$ Area of semi-circle = ${πr^2}/2$ = $1/2× 22/7 × 7×7 $ = $11×7$ = $77$ $cm^2$ Area of shaded region= Area of square - 2 × area of semi circle = 196 - 2 × 77 = 42 $cm^2$4. Find the area of the shaded region, 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. Area of shaded region= area of equilateral triangle OAB + Area of major sector Area of equilateral triangle OAB =$√3/4a^2$ =$√3/4 × 12 × 12 $ =$36√3 $ $cm^2$ Area of major sector= area of circle-area of minor sector = $πr^2$ -$(θ/360)$ ×π $r^2$ = $πr^2 ×({360-θ}/360)$ = $πr^2 ×({360-60}/360)$ = $πr^2 ×({300}/360)$ = $5/6 × πr^2$ = $5/6 × 22/7 ×6×6 $ = $660/7 $ $cm^2$ Area of shaded region = $36√3 $ + $660/7 $ $cm^2$ = ${252√3+660}/7$ $cm^2$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 Fig. 12.23. Find the area of the remaining portion of the square. Area of shaded area= area of square - 4 × area of quadrant - area of circle Side of square= 4 $cm$ area of square= side × side. = $4×4$= $16$ $cm^2$ Radius of quadrant= 1 $cm$ Area of quadrant = $1/4 × πr^2 $ = $1/4 ×22/7× 1 ×1$ = $11/14$ $cm^2$ Radius of circle= 1 $cm$ Area of circle = $πr^2$ = $22/7 ×1×1 $ = $22/7$ $cm^2$ Area of shaded area= area of square - 4 × area of quadrant - area of circle = $(16 - 4×11/14 - 22/7)$ $cm^2$ = $(16 - 22/7 - 22/7)$ $cm^2$ = $(16 - 44/7)$ $cm^2$ = $68/7$ $cm^2$6. In a circular table cover of radius 32 $cm$, a design is formed leaving an equilateral triangle ABC in the middle . Find the area of the design. 7. In Fig.,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 the shaded region. Area of shaded region= area of square - 4× area of quadrant side of square = 14 $cm$ Area of square = side ×side = $14 × 14$ = $196$ $cm^2$ Since each circle touches 2 circles, therefore radius 0f each circle= 7 $cm$ Area of quadrant = $1/4 × πr^2 $ = $1/4 × 22/7 × 7 × 7$ = ${22 × 7}/{4}$ $cm^2$ Therefore, Area of shaded region= area of square - 4 ×area of quadrant = $196 - 4×{22 × 7}/{4} $ = $196 - 154 $ = $42 $ $cm^2$8. Fig. 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. Since the distance between 2 inner parallel line of the jogging track is 60 $m$ and end portion is a semicircle, therefore the diameter of the inner semicircle = 60 $m$ Also, given the width of the track is 10 $m$, therefore the diameter of outer semicircle = $60 + 2 ×10$ =$80$ $m$ (i) the distance around the track along its inner edge = =2× length of parallel track + 2× perimeter of inner semicircle = $2 × l+ 2×(πr)$ = $2 ×106+ 2×(22/7× 30)$ = $212+ 2×(660/7)$ = ${1484+1320}/7$ = $2804/7$ $m$ (ii) the area of the track.= area of outer part - area of inner part Area of outer part(A1) = area of rectangle of dimension 106 × 80 + 2 × area of semicircle of radius 40 meter. area of rectangle = 106 × 80 = 8480 $m^2$ area of semicircle= $1/2× πr^2 $ = $1/2 × 22/7 × 40×40$= $17600/7$ $m^2$ Area of outer part($A_1$) = 8480+$2 ×17600/7$ Area of outer part($A_1$) = ${59360+35200}/7$ = $94560/7$ $m^2$ Area of inner part (A2) = area of rectangle of dimension 106 × 60 + 2 × area of semicircle of radius 30 meter. area of rectangle = 106 × 60 = 6360 $m^2$ area of semicircle= $1/2× πr^2 $ = $1/2 × 22/7 × 30×30$= $9900/7$ $m^2$ Area of inner part (A2)= 6360+$2 ×9900/7$ $m^2$ = $64320/7$ $m^2$ Therefore the area of the track = $94560/7$-$64320/7$ Therefore the area of the track = $30240/7$ = $4320$ $m^2$9. In Fig. , 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. 10. The area of an equilateral triangle ABC is 17320.5 $cm^2$. 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 Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and (√3 = 1.73205) 11. On a square handkerchief, nine circular designs each of radius 7 $cm$ are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief. 12. In Fig. 12.30, 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. (i) Area of quadrant = $(90/360)$ ×π $r^2$ = $1/4 ×22/7× 3.5 ^2$ = $1/4 ×22× 5/10× 35/10$ = $1/4 ×22× 1/2× 35/10$ = $1/4 ×22× 1/2× 35/10$ = $77/8$ $cm^2$ (ii) Area of shaded region= area of quadrant - area of triangle BOD Area of triangle BOD = $1/2×OB ×OD $ = $1/2×3.5 ×2 $ = $7/2 $ $cm^2$ Therefore Area of shaded region = $77/8-7/2$ = $(77-28)/8$ = $49/8$ $cm^2$13. In Fig., a square OABC is inscribed in a quadrant OPBQ. If OA = 20 $cm$, find the area of the shaded region. (Use π= 3.14) Square OABC is inscribed in the quadrant. Join OA, so that OAB is a right angle triangle. Therefore $OB^2=OA^2+AB^2$ $OB=√{OA^2+AB^2}$ $OB=√{20^2+20^2}$ $OB=√{800}$ $OB=20√{2}$ $cm$ Also OB is the radius of the quadrant OPBQ Area of the quadrant = $(90/360)$ ×π $r^2$ = $(1/4) ×22/7 × 800$ = $22/7 × 200$ = $4400/7$ $cm^2$ Area of square = $20 ×20$ = $400$ $cm^2$ Therefore the area of required shaded region = area of quadrant -area of square = $4400/7 -400 $ = $(4400-2800)/7 $ = $(1600)/7 $ $cm^2$14. AB and CD are respectively arcs of two concentric circles of radii 21 $cm$ and 7 $cm$ and centre O (see Fig. 12.32). If ∠AOB = 30°, find the area of the shaded region. Area of shaded region= Area of sector (AOB) - Area of sector (COD) = $(θ/360)$ ×π $r_1^2$ -$(θ/360)$ ×π $r_2^2$ =$θ/360×π$ $(r_1^2 - r_2^2)$ =$30/360×π$ $(21^2 - 7^2)$ =$1/12×22/7$× $(14)×(28)$ =$308/3$ Ans: Area of shaded region = $308/3$ $cm^2$15. In Fig., 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. Area of shaded area= area of semi circle - area of segment (COBR) Area of segment (COBR) = area of sector (ACRB) - area of ∆ABC Area of sector = $(θ/360)$ ×π $r^2$ = $(90/360)$ ×π $r^2$ = $1/4 ×22/7 × 14 ×14 $ = $1078/7 $ $cm^2$ = $154 $ $cm^2$ Area of ∆ABC = $1/2 × AC × AB$ = $1/2 × 14 × 14$ = $98$ $cm^2$ Therefore area of segment = $154-98$ $cm^2$ = $56$ $cm^2$ Since ∆ABC is a right angle triangle, therefore $BC^2=AB^2+AC^2$ $BC=√{AB^2+AC^2}$ $BC=√{14^2+14^2}$ $BC=14√2$ $cm$ Since BC is the diameter of the circle, so radius $r$= $7√2$ $cm$ Area of semicircle= $1/2×πr^2 $ = $1/2×22/7× (7√2)× (7√2)$ = $22×7$ = $154$ $cm^2$ Area of shaded area= area of semi circle - area of segment (COBR) = $154- 56$ $cm^2$ = $98$ $cm^2$16. Calculate the area of the region in Fig.common between the two quadrants of circles of radius 8 $cm$ each. Area of unshaded region= Area of square - area of minor sector side of square= radius of circle= 8 $cm$ Area of square= side× side = 8× 8 =64 $cm^2$ Area of sector =$90/360$ ×π×$r^2$ = $90/360 ×22/7×8 ×8$ = $22/7×2 ×8$ = $50.28$ $cm^2$ Area of unshaded area= $64-50.28$=$13.72$ $cm^2$ So area of shaded area= area of square - 2× area of unshaded area = $64 - 2 × 13.72 $ = $(64 - 27.44)$ $cm^2$ = $36.56$ $cm^2$