Question

Unless stated otherwise, use t = 22/7

1. The radii of two circles are 19 cm and 9 cm respectively, Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

3. depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.​

Answer 1

steps of 1)

1)let's take radius of big circle=R

and small circle =r

given circumference of 3rd circle =sum circumference of the other 2 circles

so 2πr3=2πR+2πr

taking 2π common and cancelling

we get r3=R+r

so radius of wanted circle=19+9=28cm

steps of 2)

Like in 1) let's take radius of 1st circle R

and the other circle r

third circle radius be =r'

so given πr'²=πr²+πR²

taking π common and cancelling

we get r'²=r²+R²

so our wanted radius =64+36=100=r'²

so r'=10cm :)

Answer 2

Answer:

1. 28 cm

2. 10 cm

3. Area of Gold region: 346.5$cm^{2}$

Area of Red region: 1039.5$cm^{2}$

Area of Blue region: 2079$cm^{2}$

Area of Black region: 3465$cm^{2}$

Area of White region: 5197.5$cm^{2}$

Step-by-step explanation:

So, it is stated that $\pi$=$\frac{22}{7}$ and the value is the same throughout the questions. This should not be forgotten.

1)The circumference for the first circle:

2$\pi$r

=38$\pi$

The circumference for the second circle:

2$\pi$r

=18$\pi$

Sum of the circumferences of the two circles:

18$\pi$+38$\pi$

=56$\pi$

So, the circumference of the other circle must be 56$\pi$.

Radius: 56$\pi$÷2$\pi$=28 (cm)

2)The area of the first circle:

$\pi r^{2}$

=64$\pi$

The area of the second circle:

$\pi r^{2}$

=36$\pi$

Sum of the areas of the two circles:

64$\pi$+36$\pi$

=100$\pi$

So, the area of the other circle must be 100$\pi$.

Radius: $\sqrt{100}$=10 (cm)

3)The radius of the region representing Gold:

21cm÷2

=10.5cm

Area of the region representing Gold:

$\pi r^{2}$

=110.25$\pi$

=346.5$cm^{2}$

Radius of the region representing Red and Gold:

Radius of the region representing Gold+ Wideness of Red band

=10.5cm+10.5cm

=21cm

We must do this because the band itself is not a full circle. We must take the total area of Gold and Red region (because it will form a circle), and minus the Gold region after that.

Area of the region representing Red:

$\pi r^{2}$-110.25$\pi$

=441$\pi$-110.25$\pi$

=330.75$\pi$

=1039.5$cm^{2}$

Notice that after this, the radius will only keep increasing by 10.5cm.

So, radius of the region representing Blue (and previous):

21cm+10.5cm

=31.5cm

Area of the region representing Blue:

$\pi r^{2}$-330.75$\pi$

=992.25$\pi$-330.75$\pi$

=661.50$\pi$

=2079$cm^{2}$

Radius of the region representing Black (and previous):

31.5cm+10.5cm

=42cm

Area of the region representing Black (using same area formula):

1764$\pi$-661.50$\pi$

=1102.5$\pi$

=3465$cm^{2}$

Radius of the region representing White (and previous):

42cm+10.5cm

=52.5cm

Area of the region representing White (using same area formula):

2756.25$\pi$-1102.5$\pi$

=1653.75$\pi$

=5197.5$cm^{2}$

The last question was relatively the hardest. Thank you for asking such a meaningful question.