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4 years ago

Area between two shapes. IXL geometry help pls !

Mathematics

1 answer:

-Dominant- [34]4 years ago

8 0

Since we know the Circumference of the inner circle, we can divide by (pi symbol) to get the diameter of the inner circle.

87.92/3.14=28

With knowledge of 28 being the diameter, we can also solve for the radius, which is 14.

The area of the inner circle is (pi)14^2

The inner circle’s are is 615.44 unit^2

The outer circle’s radius is 14+7 or 21

(pi)21^2= 1,384.74

We subtract the inner circle’s area from the outer circle’s area.

1,384.74-615.44= 769.3

769.3 square kilometers

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3 years ago

What is the value of tan(60°)? One-half StartRoot 3 EndRoot StartFraction StartRoot 3 EndRoot Over 2 EndFraction StartFraction 1

Ratling [72]

Answer:

$\sqrt3$

Step-by-step explanation:

To find:

The value of $tan60^\circ$ = ?

Solution:

Kindly consider the equilateral $\triangle ABC$ as attached in the answer area.

Let the side of triangle = $a$ units

Let us draw the perpendicular from vertex A to side BC.

It will divide the side BC in two equal parts.

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$tan60^\circ=\dfrac{\frac{\sqrt3}{2}a}{\frac{1}{2}a}\\\Rightarrow tan60^\circ = \bold{\sqrt3}$

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3 years ago

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3 years ago

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2 ft

sdas [7]

Answer:

88/7 ft = 12.6 ft (nearest tenth)

Step-by-step explanation:

Circumference of a circle = 2 $\pi$ r

Radius of inner circle

Given:

circumference of inner circle = 22ft
$\pi$ = 22/7

⇒ 22 = 2 x (22/7) x r

⇒ r = 3.5

So the radius of the inner circle is 3.5 ft

Radius of outer circle

If the distance between the inner circle and the outer circle is 2 ft, then the radius of the outer circle = 3.5 + 2 = 5.5

Circumference of outer circle

Therefore, circumference of outer circle:

= 2 x (22/7) x 5.5

= 242/7 ft

Difference between circumferences

Difference = outer circle circumference - inner circle circumference

= 242/7 - 22

= 88/7 ft

= 12.6 ft (nearest tenth)

4 0

2 years ago

I have no idea!! Please help!!!

Ber [7]

The area of a traingle is $\frac{1}{2}bh$
So we need to know the base and the height
To work out the sides of a right angle triangle we can use a²+b²=c
c is the hypotenuse and a and b are the other sides
Because this triangle is isoceles we know that a and b are the same

a²+b²=c²
a²+a²=c²
2a²=c²
2a²=(x√6)²
2a²=(x√6)(x√6)
2a²=x²+x√6+x√6+6
2a²=x²+x√6+x√6+6
a²= $\frac{x^{2}+x \sqrt{6}+x \sqrt{6}+6 }{2}$
a= $\sqrt{\frac{x^{2}+x\sqrt{6}+x\sqrt{6}+6}{2}}$

A= $\frac{1}{2}bh$
A= $\frac{1}{2}a*a$
A= $\frac{1}{2}$ × $\sqrt{\frac{x^{2}+x\sqrt{6}+x\sqrt{6}+6}{2}}$ × $\sqrt{\frac{x^{2}+x\sqrt{6}+x\sqrt{6}+6}{2}}$
A= $\frac{1}{2}$ × $\frac{x^{2}+x \sqrt{6}+x \sqrt{6}+6 }{2}$
A= $\frac{x^{2}+x \sqrt{6}+x \sqrt{6}+6 }{4}$

8 0

3 years ago