Pythagoras' Theorem and Surds Flashcards

Flashcards covering Pythagoras' theorem applications, surds definitions and key ideas, reasoning error analysis, and perimeter problems.

What are surds, and why are they classified as irrational numbers?

Surds are numbers that have a $\sqrt{\phantom{x}}$ sign when written in simplest form. They are not whole numbers and cannot be written as a fraction. Written as a decimal, their decimal places continue forever with no repeated pattern, making them irrational numbers.

What are four examples of surds listed in the key ideas, and why are $\sqrt{3}$ and $\sqrt{24}$ not whole numbers?

Examples of surds include $\sqrt{2}$, $\sqrt{5}$, $2\sqrt{3}$, and $\sqrt{90}$. The expressions $\sqrt{3}$ and $\sqrt{24}$ are not whole numbers because taking their square root does not yield an integer.

Is $\sqrt{a^2 + b^2}$ equal to $a + b$? Explain using the example $\sqrt{3^2 + 4^2}$.

No, $\sqrt{a^2 + b^2} \neq a + b$. For example, $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$, whereas $3 + 4 = 7$.

A bushwalker travels $3\,km$ horizontally and then along a diagonal path forming a right-angled triangle with legs of $2\,km$ and $1.5\,km$. What is the total distance traveled, rounded to one decimal place?

The diagonal path length is $\sqrt{2^2 + 1.5^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5\,km$. The total distance is $3\,km + 2.5\,km = 5.5\,km$.

A $20\,cm$ straw sits inside a cylindrical glass that is $14\,cm$ high with a base diameter of $4\,cm$. What length of straw sticks above the top of the glass, rounded to two decimal places?

The diagonal distance inside the glass is $\sqrt{14^2 + 4^2} = \sqrt{196 + 16} = \sqrt{212} \approx 14.56\,cm$. The length sticking above the top is $20\,cm - 14.56\,cm = 5.44\,cm$.

What error was made in the following working: $c^2 = 3^2 + 4^2 = 7^2 = 49 \implies c = 7$?

The base numbers $3$ and $4$ were added before squaring, incorrectly evaluating $(3 + 4)^2$ instead of evaluating each square separately ($3^2 + 4^2 = 9 + 16 = 25$).

An isosceles triangle with a base of $6$ and a height of $4$ is split into two right-angled triangles. What is side length $c$ and the perimeter of the isosceles triangle?

Each right-angled triangle has a base of $3$ and height of $4$, so $c = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$. The perimeter is $5 + 5 + 6 = 16$.

A kite has a horizontal diagonal of $24\,cm$, an upper vertical height of $5\,cm$, and a lower vertical height of $16\,cm$. What is its perimeter?

The upper side length is $\sqrt{12^2 + 5^2} = \sqrt{169} = 13\,cm$. The lower side length is $\sqrt{12^2 + 16^2} = \sqrt{400} = 20\,cm$. The perimeter is $2 \times 13\,cm + 2 \times 20\,cm = 66\,cm$.

How does Pythagoras' theorem prove that a triangle with side lengths $5$, $8$, and $10$ is not a right-angled triangle?

For a right-angled triangle, $a^2 + b^2 = c^2$. Substituting the side lengths gives $5^2 + 8^2 = 25 + 64 = 89$, while $10^2 = 100$. Since $89 \neq 100$, it is not right-angled.

Can an isosceles triangle or an equilateral triangle be right-angled? Explain both cases.

An isosceles triangle can be right-angled if its two equal sides meet at a $90^\circ$ angle (having angles $45^\circ$, $45^\circ$, $90^\circ$). An equilateral triangle cannot be right-angled because all three of its angles are $60^\circ$.

What is the perimeter of an isosceles triangle with a base of $4\,cm$ and a height of $7\,cm$, correct to two decimal places?

Splitting the base gives two right-angled triangles with base $2\,cm$ and height $7\,cm$. Each slant side is $\sqrt{2^2 + 7^2} = \sqrt{53} \approx 7.28\,cm$. The perimeter is $4\,cm + 2 \times 7.28\,cm = 18.56\,cm$.

What is the hypotenuse length $c$ of a right-angled triangle with legs $19$ and $32$, rounded to two decimal places?

The hypotenuse is $c = \sqrt{19^2 + 32^2} = \sqrt{361 + 1024} = \sqrt{1385} \approx 37.22$.

A rectangular board is $1\,m$ wide and $3\,m$ high. What is the length of a cut along its diagonal, correct to the nearest cm?

The diagonal is $\sqrt{1^2 + 3^2} = \sqrt{10} \approx 3.1623\,m$, which equals $316.23\,cm$. Rounded to the nearest cm, the cut length is $316\,cm$.

What is the screen size (diagonal length) of a television that is $1.2\,m$ wide and $70\,cm$ high, rounded to the nearest cm?

Converting width to cm gives $120\,cm$. The diagonal is $\sqrt{120^2 + 70^2} = \sqrt{14400 + 4900} = \sqrt{19300} \approx 138.92\,cm$, which rounds to $139\,cm$.