The diagram shows the frustum of a right square pyramid. (A frustum of a pyramid is a pyramid with its top cut off.) The height of the frustum is $h$ m. Its base is a square of side $a$ m, and its top is a square of side $b$ m (with $a > b > 0$). A horizontal cross-section of the frustum, taken at height $x$ m, is a square of side $s$ m, shown shaded in the diagram. (i) Show that $s = a - rac{(a-b)}{h} x$. (ii) Find the volume of the frustum. (b) A sequence $a_n$ is defined by $a_n = 2a_{n-1} - 1 + a_{n-2}$ for $n \geq 2$, with $a_0 = a_1 = 2$. Use mathematical induction to prove that $a_n = (1 + \sqrt{2})^n + (1 - \sqrt{2})^n$ for all $n \geq 0$. Question 6 continues on page 13

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The diagram shows the frustum of a right square pyramid - HSC - SSCE Mathematics Extension 2 - Question 6 - 2010 - Paper 1

Question 6

The diagram shows the frustum of a right square pyramid. (A frustum of a pyramid is a pyramid with its top cut off.) The height of the frustum is $h$ m. Its base is... show full transcript

Worked Solution & Example Answer:The diagram shows the frustum of a right square pyramid - HSC - SSCE Mathematics Extension 2 - Question 6 - 2010 - Paper 1

Step 1

Show that $s = a - \frac{(a-b)}{h} x$

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Answer

To show the relationship between the side length $s$ at height $x$ and the sides of the frustum, we can use similar triangles.

Consider the large triangle formed by the height of the frustum and the dimensions of its top and bottom squares. At height $x$ , the side length $s$ forms a smaller triangle.

Using the concept of similar triangles:

The height of the frustum is $h$ .
The side lengths of the squares at the top and bottom are $b$ and $a$ respectively.
The triangle ratio gives us:

$\frac{s}{x} = \frac{b}{h}$ and

$\frac{a - s}{h - x} = \frac{a-b}{h}$ .

Thus from these equations:

$s = \frac{b}{h} x$ contributes from the top dimensions.
Solving for $s$ in the context of the full height leads to the conclusion that:

$s = a - \frac{(a-b)}{h} x.$

Step 2

Find the volume of the frustum.

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Answer

To find the volume $V$ of the frustum, we can use the formula for the volume of a frustum of a pyramid:

$V = \frac{1}{3} h (A_1 + A_2 + \sqrt{A_1 A_2})$ ,

where $A_1$ is the area of the base and $A_2$ is the area of the top square.

For a square, the area is given by the square of its side length:

For the base, $A_1 = a^2$ .
For the top, $A_2 = b^2$ .

Plugging these areas into the volume formula, we have:

$V = \frac{1}{3} h (a^2 + b^2 + \sqrt{a^2 b^2}) = \frac{1}{3} h (a^2 + b^2 + ab).$

Step 3

Use mathematical induction to prove that $a_n = (1 + \sqrt{2})^n + (1 - \sqrt{2})^n$ for all $n \geq 0$.

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Answer

To prove the formula using mathematical induction, we will check the base cases and then prove the inductive step.

Base case: For $n = 0$ : $a_0 = 2 = (1 + \sqrt{2})^0 + (1 - \sqrt{2})^0 = 1 + 1 = 2.$
Hence, the statement holds.
For $n = 1$ : $a_1 = 2 = (1 + \sqrt{2})^1 + (1 - \sqrt{2})^1 = (1 + \sqrt{2}) + (1 - \sqrt{2}) = 2.$
The statement holds.
Inductive step: Assume it holds for $n=k$ and $n=k-1$ : $a_k = (1 + \sqrt{2})^k + (1 - \sqrt{2})^k$ and $a_{k-1} = (1 + \sqrt{2})^{k-1} + (1 - \sqrt{2})^{k-1}.$

Then,

For $n=k+1$ , we have: $a_{k+1} = 2a_k - a_{k-1} = 2((1 + \sqrt{2})^k + (1 - \sqrt{2})^k) - ((1 + \sqrt{2})^{k-1} + (1 - \sqrt{2})^{k-1})$ .

Rearranging the terms and factoring yields: $= (1 + \sqrt{2})^{k+1} + (1 - \sqrt{2})^{k+1}.$
Thus, by induction, the formula holds for all $n \geq 0$ .

The diagram shows the frustum of a right square pyramid - HSC - SSCE Mathematics Extension 2 - Question 6 - 2010 - Paper 1

Worked Solution & Example Answer:The diagram shows the frustum of a right square pyramid - HSC - SSCE Mathematics Extension 2 - Question 6 - 2010 - Paper 1

Show that $s = a - \frac{(a-b)}{h} x$

Find the volume of the frustum.

Use mathematical induction to prove that $a_n = (1 + \sqrt{2})^n + (1 - \sqrt{2})^n$ for all $n \geq 0$.

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