Sol.         We have,

Area of ring = (R² – r²)

= × (14² – 7²)

= 462 cm²

Figure 

Area of the region ABDC = Area of sector AOC – Area of sector BOD

= \left( {\frac{{40}}{{360}} \times \frac{{22}}{7} \times 14 \times 14 - \frac{{40}}{{360}} \times \frac{{22}}{7} \times 7 \times 7} \right)cm²

= \left( {\frac{1}{9} \times 22 \times 14 \times 2 - \frac{1}{9} \times 22 \times 7 \times 1} \right) cm²

=\frac{{22}}{9} \times \left( {28 - 7} \right)  cm² = \frac{{154}}{3}cm²

Hence,  Required shaded area =\left( {462 - \frac{{154}}{3}} \right) cm² =  \frac{{1232}}{3}cm² = 410.67 cm²

Sol.         PS = Diameter of a circle of radius 6 cm = 12 cm

Figure 

∴  PQ = QR = RS = \frac{{12}}{3} = 4 cm

QS = QR + RS = (4 + 4) cm = 8 cm

Hence, required perimeter = Arc of semi-circle of radius 6 cm + Arc of semi-circle of radius 4 cm

+ Arc of semi-circle of radius 2 cm

= (× 6 + × 4 + × 2) cm

= 12 cm

Required area   = Area of semi-circle with PS as diameter + Area of semi-circle with PQ as

diameter – Area of semi-circle with QS as diameter.

= \frac{1}{2} \times \frac{{22}}{7} (6)² +\frac{1}{2} \times \frac{{22}}{7}  × 2² – \frac{1}{2} \times \frac{{22}}{7} × (4)²

= \frac{1}{2} \times \frac{{22}}{7} (6² + 2² – 4²)

=\frac{1}{2} \times \frac{{22}}{7}× 24 = \frac{{264}}{7}cm² = 37.71 cm²

Sol.         We have,

Length of the rectangle ABCD = AB = 20 cm

Breadth of the rectangle ABCD = BC = 14 cm

∴     Area of rectangle ABCD = (20 × 14) cm² = 280 cm²

Diameter of the semi-circle = BC = 14 cm

∴     Radius of the semi-circle = 7 cm

Figure 

Area of the semi-circular portion cut off from the rectangle ABCD

=\frac{1}{2} (r²) = \left( {\frac{1}{2} \times \frac{{22}}{7} \times {7^2}} \right) cm² = 77 cm²

∴   Area of the remaining part = Area of rectangle ABCD – Area of semi-circle

= (280 – 77)cm² = 203 cm²

Sol.         Let ABCD be the given square each side of which is 14 cm long. Clearly, the radius of each circle is 7 cm.

We have,

Area of the square of side 14 cm long = (14 × 14) cm² = 196 cm²

Area of each quadrant of a circle of radius 7 cm

Figure 

= \frac{1}{4} (r²) = \left\{ {\frac{1}{4} \times \frac{{22}}{7} \times {{(7)}^2}} \right\} cm² = 38.5 cm²

∴  Area of 4 quadrants = 4 × 38.5 cm² = 154 cm²

Hence,  Area of the shaded region = Area of the square ABCD – Area of 4 quadrants

⇒     Area of the shaded region = (196 – 154) cm² = 42 cm²

Sol.         Let ABC be an equilateral triangle of side 24 cm, and let AD be perpendicular from A on BC. Since the triangle is equilateral, so bisects BC.

Figure 

∴   BD= CD = 12 cm

The centre of the inscribed circle will coincide with the centroid of Δ ABC.

∴   OD = \frac{{AD}}{3}

In Δ ABD, we have

AB² = AD² + BD²                 [Using Phythagoras Theorem]

⇒   24² = AD² + 12²

⇒   AD = \sqrt {{{24}^2} - {{12}^2}} =\sqrt {\left( {24 - 12} \right)\left( {24 + 12} \right)}   =  \sqrt {36 \times 12} = 12 \sqrt 3 = cm

∴   OD = \frac{1}{3}AD =\left( {\frac{1}{3} \times 12\sqrt 3 } \right)  cm = 4\sqrt 3 cm

Area of the incircle = (OD)² = \left\{ {\frac{{22}}{7} \times {{\left( {4\sqrt 3 } \right)}^2}} \right\}cm² = \left\{ {\frac{{22}}{7} \times 48} \right\}cm2 = 150.85 cm²

Area of the triangle ABC = \frac{{\sqrt 3 }}{4} (Side)² =\frac{{\sqrt 3 }}{4} (24)⁴ = 249.4 cm²

∴  Area of the remaining portion of the triangle = (249.4 – 150.85)cm² = 98.55 cm²

Sol.

AC² = AB² + BC²

⇒   AC² = 14² + 14²

⇒   AC = \sqrt {2 \times {{14}^2}} = 14\sqrt 2 cm

⇒   \frac{{{\rm{AC}}}}{{\rm{2}}}= \frac{{14\sqrt 2 }}{2} cm = 7\sqrt 2

Now,

Figure 

Required area  = Area APCQA

= Area ACQA – Area ACPA

= Area ACQA – (Area ABCPA – Area of ΔABC)

= (Area of semi-circle with AC as diameter)

– [Area of a quadrant of a circle with AB as radius – Area of ΔABC]

=  \left[ {\frac{1}{2}\left\{ {\frac{{22}}{7} \times {{\left( {7\sqrt 2 } \right)}^2}} \right\} - \left\{ {\frac{1}{4} \times \frac{{22}}{7} \times {{14}^2} - \frac{1}{2} \times 14 \times 14} \right\}} \right]

=\left\{ {\frac{1}{2} \times \frac{{22}}{7} \times 45 \times 2 - \frac{1}{4} \times \frac{{22}}{7} \times 14 \times 14 + \frac{1}{2} \times 14 \times 14} \right\} cm² = 85.4 cm²

Sol.      Area of ΔABC = 49 \sqrt 3 cm²

⇒   \frac{{\sqrt 3 }}{4}a² = 49\sqrt 3     [ Area =\frac{{\sqrt 3 }}{4} (Side)²]

⇒   a² = 49 × 4

⇒   a = 14 cm

Thus,

Radius of each circle is 7 cm

Now,

Figure 

Required area = Area of ΔABC –  3 × (Area of a sector of angle 60° in a circle of radius 7 cm)

⇒   Required area =\left\{ {49\sqrt 3 -3\left( {\frac{{60}}{{360}} \times \frac{{22}}{7} \times {7^2}} \right)} \right\}  cm²,

⇒   Required area = [/katex](49\sqrt 3 -77)[/katex]cm²

= (49 × 1.73 – 77) cm² = 7.77 cm².

Sol.         Since AB || CD and ∠ABC = 90°. Therefore∠BCD = 90°. Also,

∠BAD = 60°

∴   ∠CDA   = 180° – 60° = 120°

(i)    We have,

Total area of the four sectors = Area of sector at A + Area of sector at B + Area of sector at C + Area of sector at D.

=  \left\{ {\left( {\frac{1}{6} + \frac{1}{4} + \frac{1}{4} + \frac{1}{3}} \right) \times \pi  \times {{(17.5)}^2}} \right\} m²

=  \pi  \times {\left( {\frac{{35}}{2}} \right)^2}m2 =\frac{{22}}{7} \times \frac{{35}}{2} \times \frac{{35}}{2} m²

=  962.5 m²

Figure 

(ii)   Let DL be perpendicular drawn from D on AB. Then,

AL = AB – BL = AB – CD = (75 – 50)m

= 25 m

In ΔALD, we have tan 60° = \frac{{{\rm{DL}}}}{{{\rm{AL}}}}

⇒    \sqrt {\rm{3}} =\frac{{{\rm{DL}}}}{{{\rm{25}}}}

⇒    DL =  {\rm{25}}\sqrt {\rm{3}} m

∴  Area of trapezium ABCD =  \frac{{\rm{1}}}{{\rm{2}}} (AB + CD) × DL

=  \frac{{\rm{1}}}{{\rm{2}}} (75 + 50) × {\rm{25}}\sqrt {\rm{3}} m²

=  1562.5 × 1.732 m²

=  2706.25 m²

Hence,

Area of the remaining portion = Area of trapezium ABCD – Area of 4 sector

= 2706.25 m² – 962.5 m²

= 1743.75 m².

Sol.         Area of a circle with centre O and radius (6 cm)

= r² = \frac{{{\rm{22}}}}{{\rm{7}}} × 6 × 6 = \frac{{{\rm{792}}}}{{\rm{7}}} cm²

Area of triangle OAB = \frac{{\sqrt {\rm{3}} }}{{\rm{4}}} a2 =\frac{{\sqrt {\rm{3}} }}{{\rm{4}}}  × 12 × 12 cm²

= \sqrt {\rm{3}} × 3 × 12 = 36 \sqrt {\rm{3}} cm²

Area of sector OCD with angle

= \frac{{\pi  \times {{\rm{r}}^{\rm{2}}}\theta }}{{360^\circ }} = \frac{{{\rm{22}}}}{{\rm{7}}} ×\frac{{6 \times 6 \times 60^\circ }}{{360^\circ }}  =  \frac{{132}}{7}cm²

∴   Required area (shaded portion)

Figure

=  Area of a circle + Area of ΔOAD – Area of sector OCD

= \left[ {\frac{{792}}{7} + \frac{{36\sqrt 6 }}{1} - \frac{{132}}{7}} \right] cm² = \left[ {\frac{{660}}{7} + 36\sqrt 3 } \right] cm²

Figure 

Sol.    ΔABC is an equilateral Δ

Let AD ⊥ BC. M is the point of intersecton of three medians.

∴    AM : MD = 2 : 1

32 : MD  = 2 : 1       [ Radius = 32 cm]

⇒   MD =  \frac{{32 \times 1}}{2}= 16 cm

∴      BM = AM = Radius = 32 cm

In right angled ΔMDB, BD² = BM² – DM² = (32)² – (16)²

BD²   = (32 + 16) (32 – 16) = 48 × 16

Figure 

∴ BD = \sqrt {16 \times 3 \times 16} =  16\sqrt 3 cm

⇒   BC = 2BD = 2 × 16\sqrt 3 cm = 32\sqrt 3 cm.

∴  Area of e  qualateral ΔABC =\frac{{\rm{1}}}{{\rm{2}}}  × BC × AD

=  \frac{1}{2} \times 32\sqrt 3  \times 48  cm²

=  32 \times 24\sqrt 3 cm²                                                

Also area of the circle with radius

32 cm = r²

=  \frac{{{\rm{22}}}}{{\rm{7}}}× 32 × 32 cm²

Hence, required area (shaded portion with design)

= Area of circle – area of ΔABC

=  \left[ {\frac{{22}}{7} \times 1024-32 \times 24\sqrt 3 } \right]cm².

=  \left( {\frac{{22528}}{7}-768\sqrt 3 } \right)cm².

Figure  (i)                           Figure  (ii)

Sol.         Area of the quadrant ABPD

=  \frac{{\pi {r^2}\theta }}{{360^\circ }} = \frac{{22}}{7} \times \frac{{{\rm{A}}{{\rm{B}}^2} \times 90^\circ }}{{360^\circ }}cm²

=  \frac{{22}}{7} \times \frac{{8 \times 8 \times 1}}{4}  = \frac{{352}}{7}cm²

Area of the triangle ABD = \frac{{\rm{1}}}{{\rm{2}}} (AB × AD)

=  \frac{{\rm{1}}}{{\rm{2}}} × 8 × 8 = 32 cm²

⇒   Area of the segment BPD =  Area of quadrant – Area of ΔABD

=  \left[ {\frac{{352}}{7}-32} \right]cm² = \left[ {\frac{{352-224}}{7}} \right] cm² =  \frac{{128}}{7}cm².

Hence, area of the design BPDQB = 2 × Area of segment BPD

= 2 × \frac{{128}}{7} cm² = \frac{{256}}{7} cm².