Question

solve this question in given attachment ​

Answer 1

$\green{\bold{\underline{answer}}}$

Given :-

• length of parallel line = 106

• width of the track = 10 m

To Find:-

1. distance around the track along its inner edge.
2. the area of track

Solution : - 1

• distance around the track along its inner edge = PQ + RS +2 × circumference of semi circle POR

• $106 + 106 + 2( \frac{2\pi( \frac{60}{2} }{2} )$
• 22 + 60 π

$↪212 + 60 \times \frac{22}{7}$

$↪212 + \frac{1320}{7}$

$↪ \frac{1484 + 1320}{7}$

$↪ \frac{2804}{7} m$

Solution :- 2

• The area of track

• The area of track = area of outer track - area if inner track

• [Area of rectangle ABCD + 2× area of semi circle ATD.] - [Area of rectangle PQRS +2 × Area of semi circle POS ]

• [AB × AD + 2× π/ 2 × ( AD / 2) ² - [PQ ×PR +2 ×π/ 2(PR/2)²]

• [106×(60+20)+22/7(60+20/2)²] - [ 106×60+ 22/7 (60/2)²

• ( 106×80 +22/7 ×40×40 ) - ( 106 ×60 + 22/7 30×30 )

• (8480+35200/7) - (6360 + 19800/7 )

• 2120+15400/7 = 2120 + 2200 = 4320 M²

▶We have 2 method to find out area of Track

• Alternate method

• Area of the track = 2× area of rectangle APQB + 2 × Area of semi circle ring ATDROPA

• 2×106 ×10+2×π/2 [(40)²-(30)²]

• 2× 106 × 10 + 22/7 ×70×10

• 2120 - 2200 = 4320 m ²

$\large\mathtt \green {Important formula related to circle }$

• circumference of circle= 2π r

• circumference of semi - circle = (πr+2r)

• circumference of quadrant of circle = (πr/2+2r)

• Area of semi -circle = 1/2 πr²

• Area of Quadrant of circle = πr²/ 4

• Area of ring = π (R²- r²)

Answer 2

Given:

• length of parallel line = 106
• width of the track = 10 m

To Find:-

• distance around the track along its inner edge.
• the area of track

Solution : - 1

• distance around the track along its

• $106+106+2\frac{2\pi(\frac{60}{2}) }{2}$

• 22 + 60 π

Solution :- 2

• The area of track
• The area of track = area of outer track - area if inner track
• [Area of rectangle ABCD + 2× area of semi circle ATD.] - [Area of rectangle PQRS +2 × Area of semi circle POS ]
• [AB × AD + 2× π/ 2 × ( AD / 2) ² - [PQ ×PR +2 ×π/ 2(PR/2)²]
• [106×(60+20)+22/7(60+20/2)²] - [ 106×60+ 22/7 (60/2)²
• ( 106×80 +22/7 ×40×40 ) - ( 106 ×60 + 22/7 30×30 )
• (8480+35200/7) - (6360 + 19800/7 )
• 2120+15400/7 = 2120 + 2200 = 4320 M²

▶We have 2 method to find out area of Track

• Alternate method

• Area of the track = 2× area of rectangle APQB + 2 × Area of semi circle ring ATDROPA

• 2×106 ×10+2×π/2 [(40)²-(30)²]

• 2× 106 × 10 + 22/7 ×70×10

• 2120 - 2200 = 4320 m ²

$\huge\underlined{\underline{\underline{\sf{\boxed{Important\: formula\: related\: to\: circle}}}}$

• circumference of circle= 2π r

• circumference of semi - circle = (πr+2r)

• circumference of quadrant of circle = (πr/2+2r)

• Area of semi -circle = 1/2 πr²

• Area of Quadrant of circle = πr²/ 4

• Area of ring = π (R²- r²)