QuestionFind $\frac{a}{b} + \frac{c}{d}$ where $a$ is the circumscribed circle radius, $b$ is the inscribed circle radius, $c$ is the larger square side, and $d$ is the smaller square side.

Studdy Solution

STEP 1

Assumptions1. $a$ is the radius of the circumscribed circle for a square. . $b$ is the radius of the inscribed circle for the same square.
3. $c$ and $d$ are the side lengths of the bigger and the smaller square respectively.
4. All sides of a square are equal.
5. The diameter of the circumscribed circle is equal to the diagonal of the square.
6. The diameter of the inscribed circle is equal to the side length of the square.

STEP 2

First, let's express the side length of the bigger square $c$ in terms of the radius of the circumscribed circle $a$ . Since the diameter of the circumscribed circle is equal to the diagonal of the square, and the diagonal of a square is $\sqrt{2}$ times the side length, we have $c = \frac{2a}{\sqrt{2}}$

STEP 3

implify the expression to get $c$ in terms of $a$ .
$c = \sqrt{2}a$

STEP 4

Next, let's express the side length of the smaller square $d$ in terms of the radius of the inscribed circle $b$ . Since the diameter of the inscribed circle is equal to the side length of the square, we have $d =2b$

STEP 5

Now, we can substitute $c$ and $d$ into the expression $\frac{a}{b}+\frac{c}{d}$ .
$\frac{a}{b}+\frac{c}{d} = \frac{a}{b}+\frac{\sqrt{2}a}{2b}$

STEP 6

Since $a$ and $b$ are common factors in both terms, we can factor them out.
$\frac{a}{b}+\frac{\sqrt{2}a}{2b} = a\left(\frac{1}{b}+\frac{\sqrt{2}}{2b}\right)$

STEP 7

Combine the terms in the parentheses.
$a\left(\frac{1}{b}+\frac{\sqrt{2}}{2b}\right) = a\left(\frac{2+\sqrt{2}}{2b}\right)$

STEP 8

implify the expression.
$a\left(\frac{2+\sqrt{2}}{2b}\right) = \frac{a(2+\sqrt{2})}{2b}$

STEP 9

Finally, we can see that the value of $\frac{a}{b}+\frac{c}{d}$ is $\frac{a(2+\sqrt{2})}{2b}$ .
The value of $\frac{a}{b}+\frac{c}{d}$ is $\frac{a(2+\sqrt{2})}{2b}$ .

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QuestionFind $\frac{a}{b} + \frac{c}{d}$ where $a$ is the circumscribed circle radius, $b$ is the inscribed circle radius, $c$ is the larger square side, and $d$ is the smaller square side.

Studdy Solution

STEP 1

STEP 2

STEP 3

implify the expression to get $c$ in terms of $a$ .
$c = \sqrt{2}a$

STEP 4

Next, let's express the side length of the smaller square $d$ in terms of the radius of the inscribed circle $b$ . Since the diameter of the inscribed circle is equal to the side length of the square, we have $d =2b$

STEP 5

Now, we can substitute $c$ and $d$ into the expression $\frac{a}{b}+\frac{c}{d}$ .
$\frac{a}{b}+\frac{c}{d} = \frac{a}{b}+\frac{\sqrt{2}a}{2b}$

STEP 6

Since $a$ and $b$ are common factors in both terms, we can factor them out.
$\frac{a}{b}+\frac{\sqrt{2}a}{2b} = a\left(\frac{1}{b}+\frac{\sqrt{2}}{2b}\right)$

STEP 7

Combine the terms in the parentheses.
$a\left(\frac{1}{b}+\frac{\sqrt{2}}{2b}\right) = a\left(\frac{2+\sqrt{2}}{2b}\right)$

STEP 8

implify the expression.
$a\left(\frac{2+\sqrt{2}}{2b}\right) = \frac{a(2+\sqrt{2})}{2b}$

STEP 9

Finally, we can see that the value of $\frac{a}{b}+\frac{c}{d}$ is $\frac{a(2+\sqrt{2})}{2b}$ .
The value of $\frac{a}{b}+\frac{c}{d}$ is $\frac{a(2+\sqrt{2})}{2b}$ .

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QuestionFind ab+cd\frac{a}{b} + \frac{c}{d}ba​+dc​ where aaa is the circumscribed circle radius, bbb is the inscribed circle radius, ccc is the larger square side, and ddd is the smaller square side.

Studdy Solution

STEP 1

STEP 2

STEP 3

implify the expression to get ccc in terms of aaa.c=2ac = \sqrt{2}ac=2​a

STEP 4

Next, let's express the side length of the smaller square ddd in terms of the radius of the inscribed circle bbb. Since the diameter of the inscribed circle is equal to the side length of the square, we haved=2bd =2bd=2b

STEP 5

Now, we can substitute ccc and ddd into the expression ab+cd\frac{a}{b}+\frac{c}{d}ba​+dc​.ab+cd=ab+2a2b\frac{a}{b}+\frac{c}{d} = \frac{a}{b}+\frac{\sqrt{2}a}{2b}ba​+dc​=ba​+2b2​a​

STEP 6

Since aaa and bbb are common factors in both terms, we can factor them out.ab+2a2b=a(1b+22b)\frac{a}{b}+\frac{\sqrt{2}a}{2b} = a\left(\frac{1}{b}+\frac{\sqrt{2}}{2b}\right)ba​+2b2​a​=a(b1​+2b2​​)

STEP 7

Combine the terms in the parentheses.a(1b+22b)=a(2+22b)a\left(\frac{1}{b}+\frac{\sqrt{2}}{2b}\right) = a\left(\frac{2+\sqrt{2}}{2b}\right)a(b1​+2b2​​)=a(2b2+2​​)

STEP 8

implify the expression.a(2+22b)=a(2+2)2ba\left(\frac{2+\sqrt{2}}{2b}\right) = \frac{a(2+\sqrt{2})}{2b}a(2b2+2​​)=2ba(2+2​)​

STEP 9

Finally, we can see that the value of ab+cd\frac{a}{b}+\frac{c}{d}ba​+dc​ is a(2+2)2b\frac{a(2+\sqrt{2})}{2b}2ba(2+2​)​.The value of ab+cd\frac{a}{b}+\frac{c}{d}ba​+dc​ is a(2+2)2b\frac{a(2+\sqrt{2})}{2b}2ba(2+2​)​.

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QuestionFind $\frac{a}{b} + \frac{c}{d}$ where $a$ is the circumscribed circle radius, $b$ is the inscribed circle radius, $c$ is the larger square side, and $d$ is the smaller square side.

implify the expression to get $c$ in terms of $a$ .
$c = \sqrt{2}a$

Next, let's express the side length of the smaller square $d$ in terms of the radius of the inscribed circle $b$ . Since the diameter of the inscribed circle is equal to the side length of the square, we have $d =2b$

Now, we can substitute $c$ and $d$ into the expression $\frac{a}{b}+\frac{c}{d}$ .
$\frac{a}{b}+\frac{c}{d} = \frac{a}{b}+\frac{\sqrt{2}a}{2b}$

Since $a$ and $b$ are common factors in both terms, we can factor them out.
$\frac{a}{b}+\frac{\sqrt{2}a}{2b} = a\left(\frac{1}{b}+\frac{\sqrt{2}}{2b}\right)$

Combine the terms in the parentheses.
$a\left(\frac{1}{b}+\frac{\sqrt{2}}{2b}\right) = a\left(\frac{2+\sqrt{2}}{2b}\right)$

implify the expression.
$a\left(\frac{2+\sqrt{2}}{2b}\right) = \frac{a(2+\sqrt{2})}{2b}$

Finally, we can see that the value of $\frac{a}{b}+\frac{c}{d}$ is $\frac{a(2+\sqrt{2})}{2b}$ .
The value of $\frac{a}{b}+\frac{c}{d}$ is $\frac{a(2+\sqrt{2})}{2b}$ .