\(\triangle\)  PQR is an isosceles right triangle
⇒ Q = 90°, PQ = RQ = x (say)
⇒ 2x2 = (7\(\sqrt2\))2
⇒(\(\sqrt2\)x)² = (7\(\sqrt2\))²
⇒ x = 7 cm

Now, semi-perimeter(s) = \({7+7+7\sqrt2 \over2} = {14+7\sqrt2 \over2}\) 
Also, we know that radius of inscribed circle,
r = \(\triangle\over s\)
⇒ r = \(\frac { \frac { 1 }{ 2 } \times 7\times 7 }{ \frac { 14+7\sqrt { 2 } }{ 2 } } =\frac { 7\times 7 }{ 7(2+\sqrt { 2 } ) } \)
= \(\frac { 7 }{ 2+\sqrt { 2 } } \)cm
Area of the circle inscribed in \(\triangle\) PQR
= \(\frac { 22 }{ 7 } \times \frac { 7 }{ 2+\sqrt { 2 } } \times \frac { 7 }{ 2+\sqrt { 2 } } \)
= \(\frac { 152 }{ { \left( 2+\sqrt { 2 } \right) }^{ 2 } } { cm }^{ 2 }\)

Here,  OA = OB = r = 12cm, \(\angle \)AOB = 120°
In \(\triangle\)AOB, AO = BO = 12 cm, draw OD ⊥ AB
Since \(\triangle\)AOB is an isosceles and OD ⊥ AB
ஃ OD is the angle bisector as well as median

Now, in rt.  \(\triangle\)ADO, \(\angle \)D = 90° , \(\angle \)AOD = 60°
∴ \(OD\over AO\) = cos 60°
= 12 x 1/2 =6 cm
and  \(AD\over AO\) = sin 60°

⇒ AD = AO, sin 60° = 12 x \(\sqrt3 \over2\) = 6\(\sqrt3\) cm
∴ AB = 2 AD = 2\(\sqrt3\) x = 12\(\sqrt3\) cm
Area of minor segment 
= Area of sector AOB - Area of \(\triangle\)AOB
= \(\theta\over360°\) x \(\pi\)r² - \(1\over2\) x AB x OD
= \(120° \over360°\) x 3.14 x 12 x 12 - \(1\over2\) x 12\(\sqrt3\) x 6
= 150.72 - 62.28 = 88.44 cm².

The area of rectangular sheet ABCD = 40 x 28 
= 1120 cm² 
The diameter of semicircle = 28 cm 
ஃ Radius of semicircle = 14 cm 
Area of semicircle = \(\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times 14\times 14\quad { cm }^{ 2 }\)
= 308 cm² 
Thus, area of remaining paper = Area of paper - Area of semicircle 
= 1120 - 308 = 812 
∴ Required area = 812 cm² 

Let r be the radius of the circular track.
Now, area of the circular track = 5544 m²  [given]
⇒ \(\pi\)r² = 5544
⇒ r² = \(5544\over \pi\) ⇒  r² = \(5544 \times7 \over22\)
⇒ r² = 1764 ⇒ r = \(\sqrt{1764}\) = 42 m
Circumference of the circular track = 2\(\pi\)r
= 2 x \(22\over7\)x42
= 264 m
ஃ Distance covered in 12\(1\over2\) rounds = 264 x \(25\over2\)
= 3300 m
= 3.3 km
Time taken to cover 3.3 km = 1 hour [given] Hence, time taken to cover 3300 m (12\(1\over2\) rounds) around the circular track is 1 hour.

Area of bigger circle
= \(\pi { R }^{ 2 }=\frac { 22 }{ 7 } \times 14\times 14=616\quad { cm }^{ 2 }\)
Area of smaller circle
= \(\pi { R }^{ 2 }=\frac { 22 }{ 7 } \times 7\times 7=154\quad { cm }^{ 2 }\)
Area of sector AOC
= \(\frac { 40 }{ 360 } \times \pi \times { 14 }^{ 2 }=\frac { 1 }{ 9 } \times \frac { 22 }{ 7 } \times 14\times 14\) = \(\frac { 616 }{ 9 } \) cm²
Area of sector BOD
= \(\frac { 40 }{ 360 } \times \pi \times { 7 }^{ 2 }=\frac { 1 }{ 9 } \times \frac { 22 }{ 7 } \times 7\times 7\) = \(154\over9\) cm²
Area of shaded part
= area of bigger circle - area of smaller circle - area of ABCD
= \(\frac { 40 }{ 360 } \times \pi \times { 7 }^{ 2 }=\frac { 1 }{ 9 } \times \frac { 22 }{ 7 } \times 7\times 7\) cm²
= \(1232\over3\) cm² 

Area of shaded part (I + II)]
= Area of square ABCD - Area of two semicircles
= a² - 2 x \(\frac { \pi { r }^{ 2 } }{ 2 } \) = 14 x 14 - \(22\over7\) x 7 x 7
= 196 - 154 = 42 cm²
Similarly,  Area of shaded part (III + IV) = 42 cm²
∴ Area of shaded part = 42 + 42 = 84 cm²

Area of the segment 
= area of the shaded portion 
= area of the sector - area of the triangle 
= \(\frac { 90° }{ 360° } \times \frac { 22 }{ 7 } \times { (10) }^{ 2 }-\frac { 1 }{ 2 } \times 10\times 10\)
= \(\left( \frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times 100-50 \right) \) cm²
= (78.57 - 50) cm²
= 28.57 cm²

Let AO = OB = r
Then perimeter of semicircle APB
= \(2\pi r \over 2\) = \(\pi\)r
Perimeter of semicircle AQO
= \(\frac { 2\pi \frac { r }{ 2 } }{ 2 } =\frac { \pi r }{ 2 } \)
(∵ radius of semicircle AQO = \(r\over2\))
∴ Perimeter of shaded region
\(=\ \pi r+\frac { \pi r }{ 2 } +r=\frac { 2\pi r+\pi r+2r }{ 2 } \)
But perimeter of shaded region
ஃ \(\frac { 2\pi r+\pi r+2r }{ 2 } =40\)
⇒ r(2\(\pi\) + \(\pi\) + 2) = 80
⇒  r(3\(\pi\) + 2) = 80
⇒ \(r\left( 3\times \frac { 22 }{ 7 } +2 \right) =80\)
⇒ \(r\left( \frac { 66 }{ 7 } +2 \right) =80\)
⇒  \(r\left( \frac { 80 }{ 7 } \right) =80\)
ஃ  \(r=\frac { 80\times 7 }{ 80 } =7cm\)
Now, Area of APB = \(\frac { \pi { r }^{ 2 } }{ 2 } =\frac { 22\times 7\times 7 }{ 7\times 2 } \)
= 77 cm2
Area of AQO = \(\frac { \pi { r }^{ 2 } }{ 2 } \)
 \(=\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \times \frac { 1 }{ 2 } \)
= \(77\over4\) cm²
ஃ Area of shaded region
= 77 + \(\frac { 77 }{ 4 } =\frac { 308+77 }{ 4 } \)
= \(\frac { 385 }{ 4 } \) = 96.25 cm²

Triangle AED is right-angled at E.
ஃ AD² = AE² + ED²
⇒ AD = \(\sqrt { { 9 }^{ 2 }+{ 12 }^{ 2 } } \) = \(\sqrt { 81+144 } \)
= \(\sqrt{225}\) = 15 cm
ஃ  BC = AD = 15 cm = Diameter of semicircle
Now area of shaded portion
= Area of rectangle + Area of semicircle - Area of triangle AED
= \(\left[ 20\times 15+\frac { 1 }{ 2 } \pi { \left( \frac { 15 }{ 2 } \right) }^{ 2 }-\frac { 1 }{ 2 } \times 9\times 12 \right] { cm }^{ 2 }\)
= \(\left[ 300+\frac { 1 }{ 2 } \times 3.14\times { \left( \frac { 15 }{ 2 } \right) }^{ 2 }-54 \right] { cm }^{ 2 }\)
= [246 + 88.31] cm² = 334.31 cm² 

Diameter of Semicircle PSR, PAQ and QTR are 10 cm, 7 cm and 3 cm respectively.

Perimeter of shaded region
= length of arc PSR + length of arc PAQ + length of arc QTR
= [\(\pi\)(5) + \(\pi\)(3.5) + \(\pi\)(1.5)] cm
= x 10 cm = 3.14 x 10 cm = 31.4 cm

Perimeter of 2 inner semicircle = (400 - 2 x 90) m = (400 - 180) m = 220 m
Radius of each semicircle = \(\frac { 220 }{ 2\pi } \) = \(\frac { 220\times 7 }{ 2\times 22 } \) = 35 m
Perimeter of 2 outer semicircle = (488 - 180) m = 308 m
Radius of each outer semicircle = \(\frac { 308 }{ 2\pi } =\frac { 308\times 7 }{ 2\times 22 } =\quad 49m\)
width of the track = outer radius - inner radius = 49 - 35 = 14 m
Area of rectangular tracks = 2 x area of rectangle
= 2 x i x b = 2 x 90 x 14       (i = 90, b = 14 m)
= 28 x 90 = 2520 m²
Area of two semicircle rings = area of one circle ring
\(=\quad \pi \left( { R }^{ 2 }-{ r }^{ 2 } \right) =\frac { 22 }{ 7 } \left( { 49 }^{ 2 }-{ 35 }^{ 2 } \right) { m }^{ 2 }\)
= \(22\over7\) x (49 - 35) (49 - 35) m²
= \(22\over7\) x 14 x 84 m² = 44 x 84 = 3696 m²
Total area of track = (2520 + 3696) m² = 6216 m²
Cost of developing the track at the rate of Rs.12.50/ m² = Rs. 6216 x 12.50 = Rs.77,700

In OBD
cos 60°=\(OD \over OB\)and sin 60°=\(BD\over OB\)
⇒ \(1\over2\)= \(OD\over6\) and \(\frac { \sqrt { 3 } }{ 2 } \) = \(BD\over6\)
⇒ \(6\over2\)=OD and \(\frac { \sqrt { 3 } }{ 2 } \)x6 = BD
⇒ OD = 3 and BD = 3\(\sqrt3\)
BC = 2BD = 2 x 3\(\sqrt3\)=6\(\sqrt3\)
Area of the shaded region = area of circle - area of \(\triangle\)ABC
= \(2\pi { (6) }^{ 2 }-\frac { \sqrt { 3 } }{ 2 } { \left( 6\sqrt { 3 } \right) }^{ 2 }\)
= 3.14 x 6 x - \(\frac { \sqrt { 3 } }{ 4 } \) x 6\(\sqrt3\) x 6 \(\sqrt3\)
=  113.04 - 27 x 1.732 = 113.04 - 46.76 = 66.28 cm²

Radius of circle = 7cm
Side of equilateral triangle = 12cm

∴ Area of shaded portion = (Area of a circle — area of sector of central angle 60⁰) + area of equilateral triangle OAB
\(=\left[\left\{\pi(7)^2-{60^0\over 360^0}\times\pi(7)^2\right\}+{\sqrt3\over 4}(12)^2\right]cm^2\)
\(=\left[{5\over6}\times{22\over 7}\times49+{\sqrt3\over4}\times144\right]cm^2\)
=(128.33+62.35)cm²=190.68cm²

Let the side of equilateral triangle be a cm
∴ Area \(={\sqrt3\over 2}\times a^2\)
\(⇒\ 49\sqrt3={\sqrt3\over 4}\times a^2\)
⇒ a²=49 x 4 ⇒ a=14
∴ Radius of circle=1/2 (side of triangle)
∴ Radius of circle=7cm
Area of 1 sector of circle with central angle 60⁰
\(={60^0\over 360^0}\times \pi\times7\times7={49\pi\over 6}cm^2\)
Area of 3 sector=\({3\times49\pi\over 6}cm^2\)
\(={49\over 2}\times{22\over 7}cm^2=77cm^2\)
Area of triangle not included in the circle
= area of triangle — area of 3 sectors
=49√3  - 77 = 49 x 1.73-77
 = 84.77 -77 = 7.77 cm²

In right-angled ΔABC,
AB²+AC²=BC²
(3)²+(4)²=BC²
9+16=BC²
BC=5units
Now,
Area of shaded region = area of semicircle on side AB + area of semicircle on side AC - area of semicircle on side BC + area of  ΔABC
Area of semicircle on side \(AB={\pi r^2\over 2}={22\over 7}\times{9\over 2}={9\over2}sq.units\)......(i)
Area of semicircle on side \(AC={\pi r^2\over 2}={22\over 7}\times4\times{1\over 2}={44\over 7}sq.units\) .....(ii)
Area of semicircle on side \(BC={\pi r^2\over 2}={22\over 7}\times{25\over 4}\times{1\over 2}={275\over 28}sq.units\)....(iii)
Area of ΔABC=\({1\over 2}\times AB \times AC={1\over 2}\times3\times4=6 sq.units\) ....(iv)
Now, Area of shaded region = (i)+(ii)-(iii)+(iv)
\(={99\over 28}+{44\over 7}-{275\over 28}+6={99+176-275+168\over 28}={168\over 28}\)
∴ Area of shaded region=6 square units.