Question

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.​

Answer 1

Step-by-step explanation:

Answer

The distance around the track along its inner edge

=EF+HG + circumference of semicircle with diameter EH+circumference of semicircle with diameter FG

circumference of semicircle with diameter EH + Circumference of semicircle with diameter FG

=2×π×30=60πm

∴ The distance around the track along its inner edge

=60π+106+106

=

7

2804

m

∴ Area of track = Area of rectangle ABFE + Area of rectangle HGCD + (Area of semicircle with diameter AD − Area of semicircle with diameter EH) + (Area of semicircle with diameter BC − Area of semicircle with diameter FG)

=(106×10)+(106×10)+(

2

π

×40×40−

2

π

×30×30)+(

2

π

×40×40−

2

π

×30×30)

=2120+2200=4320m

2

Answer 2

$\\ \\☯ \color{red}{\large \rm \underline{ \: Question :- }}$

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.

$\\ \\☯ \color{red}{\large \rm \underline{ \: Solution :- }}$

$\\ \\ \sf \: At \: first, \\ \sf We \: have \: to \: find \: the \: radius \: of \: the \: inner \: track :$

$\\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: Radius = \frac{ Diameter}{2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sf\frac{60}{2} \\ \\ \: \: \: \: \: \: \: \ \sf \: \boxed{ \sf \: Radius = 30 \: m}$

$\\ \\ \sf We \: have \: to \: find \: the \: distance \: of \: the \: inner \: track :$

$\\ \\ \sf \: Distance \: = \: 2\pi r + length + length$

$\\ \\ \sf \: Distance \: = \: 2\pi (30) + 106 + 106 \\ \\ \sf \: \: \: \: \: \: = 2\pi(30) + 212 \\ \\ \sf \boxed{ \sf Distance = 400.57 \: m }$

Now,

$\\ \\ \sf We \: have \: to \: find \: the \: area \: of \: the \: inner \: track :$

$\\ \\ \: \: \sf Area \: = \pi {r}^{2} + (length \: \times \: breadth) \\ \\ \sf \: \: = \pi(30) ^{2} + (106 \times 60) \\ \\ \boxed{ \sf \: Area = 9188.57 \: {m}^{2} }$

Now,

$\\ \\ \sf We \: have \: to \: find \: the \: area \: of \: the \: whole\: track :$

$\\ \\ \: \: \sf Area \: = \pi {r}^{2} + (length \: \times \: breadth)$

$\\ \\ \sf \star \: \: Radius = 30+10 = 40 m \\ \star \sf \: Width = 60 +10+10 = 60+20 = 80m$

$\\ \sf\star \: Area = π(40)^2+(160 \times 80 )$

$\\ \\ \sf\star \: Area = π(40)^2+(160 \times 80 ) \\ \\ \underline{ {\boxed{ \sf Area = \: 13508.57}}}$

$\\ \\ \sf \: The \: area \: of \: track \: is$

$\\ \\ \sf \: \: \: \: \: \: \: \: \: \: Area = 13508.57 - 9188.57$

$\\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{\boxed{\boxed{ \pink{ \sf Area = 4320m ^{2} }} }}$

$\\ \\ \sf \bold{ Answer :-} \:\underline{ \: \: (1) 400.57 m \: \: (2)4320 m^2 \: \: }$