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mathematics chaleng ( 2 ) - How does the length BFBF relate to the original rectangle?

- août 15, 2017 0

mathematics chaleng ( 2 ) - How does the length BFBF relate to the original rectangle?

Given rectangle ABCDABCD, ABAB is extended to EE such that BE=BCBE=BC. AEAE forms the diameter of a circle and CBCB is extended to FF which lies on the circle.

How does the length

B F

relate to the original rectangle?

solution :

Consider the following diagram.

As triangle

A E F

is in a semi-circle,

∠ A F E = 90 \Rightarrow ∠ F A B + ∠ F E B = 90

But

∠ F A B + ∠ A F B = 90 \Rightarrow ∠ A F B = ∠ F E B

. In the same way,

∠ B F E = ∠ F A B

.

Thus the right angled triangles

F A B

and

F E B

are similar.

∴ \frac{A B}{F B} = \frac{F B}{B E} \Rightarrow A B \cdot B E = (F B)^{2}

But

B E = B C

, so

A B \cdot B C = (F B)^{2}

Hence the square on

F B

is equal to the area of the rectangle

A B C D

and it can be seen that this construction process has "squared" the rectangle.

Show how this method can be adapted to construct the square root of a given length

A B = x

.

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Calculate the values of x and y

- août 14, 2017 0

Calculate the values of $x$ and $y$ , if $z^{3} = \sqrt{36^{5}}$ , $y = x^{- 5}$ and $x = \log_{6} (z) - \frac{2}{3}$

x = \log_{6} (z) - \frac{2}{3}

x = \log_{6} (z) - \frac{2}{3}

x = \log_{6} (z) - \frac{2}{3}

Three mathematical statements are given to find the values of

x

and

y

in this logarithm problem.

(1)

z^{3} = \sqrt{36^{5}}

(2)

x = \log_{6} (z) - \frac{2}{3}

(2)

y = x^{- 5}

The first statement expresses the value of

z

in terms of a radical having an exponential term as its radicand. The second statement is useful to find the value of

x

by substituting the value of

z

in it. Finally, the third statement is useful to find the value

y

by substituting the value of

x

in it.

Step: 1

Solve the first statement and find the value of

z

.

z^{3} = \sqrt{36^{5}}

⟹ z^{3} = \sqrt{{(6^{2})}^{5}}

Apply the power rule of an exponential term to simplify this expression.

⟹ z^{3} = \sqrt{6^{2 \times 5}}

⟹ z^{3} = \sqrt{6^{5 \times 2}}

⟹ z^{3} = \sqrt{{(6^{5})}^{2}}

⟹ z^{3} = 6^{5}

Take cube root both sides to find the value of

z

.

⟹ \sqrt[3]{z^{3}} = \sqrt[3]{6^{5}}

∴ z = {(6)}^{\frac{5}{3}}

Step: 2

Now, substitute the value of

z

in the second statement to obtain the value of

x

.

x = \log_{6} (z) - \frac{2}{3}

⟹ x = \log_{6} {(6)}^{\frac{5}{3}} - \frac{2}{3}

Use the power law of logarithm of an exponential term to simplify the equation.

⟹ x = \frac{5}{3} \log_{6} 6 - \frac{2}{3}

According to logarithm of base rule, the logarithm of a number to same number is one.

⟹ x = \frac{5}{3} \times 1 - \frac{2}{3}

⟹ x = \frac{5}{3} - \frac{2}{3}

⟹ x = \frac{5 - 2}{3}

⟹ x = \frac{3}{3}

∴ x = 1

Step: 3

Now substitute the value of

x

in third algebraic equation to get the value of

y

.

y = x^{- 5}

⟹ y = {(1)}^{- 5}

∴ y = 1

Therefore, it is derived that value of

x

is equal to the value of

y

and it is

1

. It is written as

x = y = 1

x = \log_{6} (z) - \frac{2}{3}

x = \log_{6} (z) - \frac{2}{3}

x = \log_{6} (z) - \frac{2}{3}

Continue reading

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