Question

pls answer the questions​

Answer 1

Answer 5 :

Given :

• Horse is tied to a pole with 28m long string.

To Find :

• Area where horse can graze

Solution :

The area covered :

Radius = Length of the string

Area horse can graze is area of circle with radius.

★ Area of circle = πr²

→ $\sf\dfrac{22}{7}$ × 28 × 28

→ $\sf\cancel\dfrac{17248}{7}$

→ 2464m²

$\therefore$ Area where horse can graze is 2464m²

____________________________

Answer 6 :

Given :

• Race track is in the form of a ring

• Inner circumference is 352 m

• Outer circumference is 396 m

To Find :

• Area of the track

Solution :

Let inner radius be r

★ Circumference = 2πr

→ 352 = 2 ×$\sf\dfrac{22}{7}$ × r

→ 352 = $\sf\dfrac{44}{7}$ × r

→ r = 56 m

Let outer radius be R

★ Circumference = 2πr

→ 396 = 2 ×$\sf\dfrac{22}{7}$ × r

→ 396 = $\sf\dfrac{44}{7}$ × r

→ r = 63 m

Area of inner circle :

★ Area of circle = πr²

→ $\sf\dfrac{22}{7}$ × 56 × 56

→ $\sf\cancel\dfrac{68992}{7}$

→ 9856m²

Area of outer circle :

★ Area of circle = πr²

→ $\sf\dfrac{22}{7}$ × 63 × 63

→ $\sf\cancel\dfrac{87318}{7}$

→ 12474m²

★ Area of Track = Outer Area - Inner Area

→ 12474m² - 9856m²

→ 2618m²

$\therefore$ Area of the track is 2618m²

Answer 2

Question 5

A horse is tied to a pole with 28 m long string. Find the area where the horse can graze

Solution

• Length of string = 28m

Area where the horse can graze = Area of circle

➠ πr²

➠ 22/7 × 28 × 28

➠ 22 × 4 × 28

➠ 2464m²

Hence, area where the horse can graze is 2464m²

Question 6

A race track in the form of ring whose inner circumference is 352m, and the outer circumference is 396m. Find the area of track .

Solution

• Inner circumference = 352m

★ Let the inner radius be " r "

➠ inner circumference = 352m

➠ 2πr = 352

➠ πr = 352/2

➠ πr = 176

➠ r = 176×7/22

➠ r = 56m

So, inner radius is 56m

★ Let the outer radius be " R "

➠ outer circumference = 396m

➠ 2πR = 396

➠ πR = 396/2

➠ πR = 198

➠ R = 198×7/22

➠ R = 63m

So, outer radius is 63m

Now, Area of track

Area of outer circle - area of inner circle

➠ πR² - πr²

➠ π ( R² - r² )

➠ 22/7 [(63)² - (56)²]

➠ 22/7 [(63+56)(63-56)]

➠ 22/7 [ 119 × 7 ]

➠ 22/7 × 833

➠ 2618 m²

Hence, area of track is 2618m²