Question

13. A race track is in the form of a ring whose inner circumference is 220 m and outer circumference is
550 m. Find the width of the track.​

Answer 1

Step-by-step explanation:

Answer done mark me as brainlisest

Answer 2

It is said that a race track is in the form of a ring whose inner circumference is 220 m and outer circumference is 550 m respectively. For finding the width firstly we have to find out radius of the both inner and outer face and thereafter we will subtract the inner radius from the outer one and then we will got our answer.

Let's do it now :-

Formula need to know :

${\boxed{\bf{Circumference \: of \: circle \: = 2 \pi r}}}$

• Outer Circumference → 550 m.

• Inner Circumference → 220 m.

Let us we consider the outer and inner radii of the ring be r1 metres and r2 metres respectively.

Now,

Substituting the respective values, we get :

$\dashrightarrow\:\:\sf Circumference \: of \: circle = 2 \pi r = 550 \: m$

$\dashrightarrow\:\:\sf 550 = 2 \times \dfrac{22}{7} \times r \: m$

$\dashrightarrow\:\:\sf r_1 = \dfrac{550 \times 7}{22 \times 2} \: m$

$\dashrightarrow\:\:\sf r_1 = \dfrac{3850}{44}\: m$

$\dashrightarrow\:\:\sf r_1 = 87.5 \: m$

Similarly ,

Substituting the respective values, we get :

$\dashrightarrow\:\:\sf Circumference \: of \: circle = 2 \pi r = 220 \: m$

$\dashrightarrow\:\:\sf 220 = 2 \times \dfrac{22}{7} \times r_2 \: m$

$\dashrightarrow\:\:\sf r_2 = \dfrac{220 \times 7}{22 \times 2} \: m$

$\dashrightarrow\:\:\sf r_2 = \dfrac{10 \times 7}{2}\: m$

$\dashrightarrow\:\:\sf r_2 = \dfrac{70}{2}\: m$

$\dashrightarrow\:\:\sf r_2 = 35 \: m$

For Width :-

→ Outer Radius - Inner Radius

= 87.5 - 35

Therefore, Width of the track = 52.5 m.

$\dashrightarrow\:\: \underline{ \boxed{\sf Width \: of \: the \: track \: = 52 \: m }}$

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