Question

✌✌Any Maths Aryabhatta Here. Solve the Maths Question in Attachement. ✌✌​

Answer 1

Gɪᴠᴇɴ :-

• Length of Rectangular part = 106m.
• Breadth = 60m.
• width of Track = 10m.

Tᴏ Fɪɴᴅ :-

• The distance Around the track along its inner edge..
• Area of Track.

Fᴏʀᴍᴜʟᴀ ᴜsᴇᴅ :-

• Perimeter of circle = 2πr
• Area of rectangle = Length * Breadth
• Area of circle = π * r²

Sᴏʟᴜᴛɪᴏɴ :-

→ The distance Around the track along its inner edge = Length + Breadth + Perimeter of 2 half circle with diameter equal to bredth .

Or,

→ The distance Around the track along its inner edge = Length + Breadth + Perimeter of 1 full circle with diameter equal to bredth .

→ The distance Around the track along its inner edge = 106 + 60 + (2 * π * 30) . { As semicircles are on breadth, and radius is half of diameter.}

So,

→ The distance Around the track along its inner edge = 166 + 60*3.14 = 166 + 188.4 = 354.4m (Ans.)

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Now,

to find the area of track we have to find area of outer full figure and subtract area of inner part.

Also,

Notice that, track is made of one rectangle and 2 semi-circles or we can say that, one rectangle and one full circle..

Lets calculate now,

→ Area of inner figure = Rectangle area + circle area with radius equal to half of breath .

→ Area of inner figure = (106*60) + {π * (30)²} = 6360 + (3.14*900) = 6360 + 2826 = 9186m².

Now,

→ Length of outer figure = same = 106m.

→ Breath of outer figure = 60 + 2(width of track) = 60 + 20 = 80m.

→ radius of circle = (80/2) = 40m.

So,

→ Area of outer figure = Rectangle area + circle area with radius equal to half of breath .

→ Area of inner figure = (106*80) + {π * (40)²} = 8480 + (3.14*1600) = 6360 + 5024 = 11384m².

Therefore,

→ Area of Track = 11384 - 9186 = 2198m². (Ans.)

Hence, Area of Track will be 2198m².

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

• Distance between the two inner parallel line segments is 60 m
• they are each 106m long.
• the track is 10m wide

${\bf{\blue{\underline{To\:Find:}}}}$

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

${\bf{\blue{\underline{Now:}}}}$

Let r be the radius of inner semi-circle.

$: \implies{\sf{ 2r = 60m}}$

$: \implies{\sf{ r = \frac{60}{2} m}}$

$: \implies{\sf{ r = 30m}}$

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$\dagger \boxed{\sf{ distance \: around \: the \: edge \: along \: its \: inner \: edge = AB+ BC + CD + DA}}$

$: \implies{\sf{ 106m + \pi \: \times 30m + 106m + \pi \times 30m}}$

$: \implies{\sf{ 212m + 2 \times \frac{22}{7} \times 30m }}$

$: \implies{\sf{ 212m + \frac{1320}{7} m}}$

$: \implies{\sf{ \frac{1484 + 1320}{7} m}}$

$: \implies{\sf{ \frac{2804}{7} m}}$

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✷Area of track = Area of rectangular ABFE+Area of semi-circle BFGC+Area of rectangular DCGH+

Area of semi circle AEHD.

$: \implies{\sf{ 106m \times 10m + \frac{1}{2} \pi( {40}^{2} m - {30}^{2}m) + 106m \times 10m + \frac{1}{2}( {40}^{2} m - {30}^{2}m) }}$

$: \implies{\sf{ 2 \times 106 \times 10{m}^{2} + \frac{22}{7} (40 - 30)(40 + 30) {m}^{2} }}$

$: \implies{\sf{ 2120 {m}^{2} + 220 {m}^{2} }}$

$: \implies{\sf{ 4320 {m}^{2} }}$

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Hence,

• The distance around the track along its inner edge = 2804/7m
• The area of track = 4320m²