NUMBER

QUESTION 1 [3 marks] — Criterion A Medium

Evaluate each expression, showing full working:

a. $\sqrt{144} + \sqrt{81}$

[1]

Show Solution

$$12+9=21$$

b. $\sqrt{225} - \sqrt{49}$

[1]

Show Solution

$$15-7=8$$

c. $(\sqrt{36})^2 + 5^2$

[1]

Show Solution

$$36+25=61$$

QUESTION 2 [5 marks] — Criterion A Medium

A square garden plot has an area of $169$ m$^2$.

a. Find the side length of the plot.

[1]

Show Solution

$$\sqrt{169}=13\text{ m}$$

b. Find the perimeter of the plot, and the cost of fencing it at \$18.50 per metre.

[2]

Show Solution

Perimeter $=4\times13=52$ m. Cost $=52\times18.50=\$962$.

c. The gardener wants to double the AREA of the plot (keeping it square). Find the new side length, correct to 2 decimal places.

[2]

Show Solution

New area $=169\times2=338$ m$^2$. New side $=\sqrt{338}\approx18.38$ m.

QUESTION 3 [5 marks] — Criterion B Medium

Investigate what happens to the difference between consecutive square numbers as the numbers get larger.

a. Calculate $2^2-1^2$, $3^2-2^2$, $4^2-3^2$, and $5^2-4^2$.

[2]

Show Solution

$4-1=3$. $9-4=5$. $16-9=7$. $25-16=9$.

b. Describe the pattern in these differences (3, 5, 7, 9, ...).

[1]

Show Solution

The differences form the sequence of consecutive odd numbers, increasing by 2 each time.

c. Using the pattern, predict the difference $11^2-10^2$ without calculating both squares, then verify your prediction.

[2]

Show Solution

The pattern of differences (3,5,7,9,...) for $n^2-(n-1)^2$ equals $2n-1$. For $n=11$: $2(11)-1=21$. Verifying: $121-100=21$ — matches.

QUESTION 4 [5 marks] — Criterion C Medium

A student claims that $\sqrt{a^2+b^2}$ is the same as $a+b$.

a. Test the claim using $a=3, b=4$, showing full working for both sides.

[2]

Show Solution

$\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$. $a+b=3+4=7$. Since $5\ne7$, the claim is false.

b. Explain why $\sqrt{a^2+b^2} \ne a+b$ in general, referring to what happens when you square both expressions.

[3]

Show Solution

Squaring $a+b$ gives $(a+b)^2=a^2+2ab+b^2$, which includes an extra $2ab$ term compared to $a^2+b^2$. Since this extra term is generally not zero (unless $a=0$ or $b=0$), $\sqrt{a^2+b^2}$ and $a+b$ are generally different — the square root does not simply 'distribute' over addition.

QUESTION 5 [4 marks] — Criterion D Medium

A television screen is advertised by its diagonal length. A screen has width $80$ cm and height $45$ cm.

a. Using $\text{diagonal}^2 = \text{width}^2 + \text{height}^2$, find the diagonal length of the screen, correct to 1 decimal place.

[2]

Show Solution

$$\text{diagonal} = \sqrt{80^2+45^2} = \sqrt{6400+2025} = \sqrt{8425} \approx 91.8 \text{ cm}$$

b. TV sizes are usually rounded to the nearest inch (1 inch $\approx$ 2.54 cm). What size (in inches) would this TV be advertised as?

[2]

Show Solution

$91.8 \div 2.54 \approx 36.1$ inches, so it would be advertised as a 36-inch TV.

QUESTION 6 [7 marks] — Criterion A Hard

A square-shaped park has an area of 5,625 m$^2$. A square garden bed inside it has area exactly $\frac{1}{25}$ of the park's area.

a. Find the side length of the park.

[2]

Show Solution

$$\sqrt{5625}=75 \text{ m}$$

b. Find the area, then the side length, of the garden bed.

[3]

Show Solution

Garden area $=5625\div25=225$ m$^2$. Garden side length $=\sqrt{225}=15$ m.

c. Verify your garden side length is consistent with the park's side length, using the fact that if garden AREA is $\frac{1}{25}$ of park area, garden SIDE should be $\frac{1}{5}$ of park side (since area scales with the SQUARE of side length).

[2]

Show Solution

Park side $\div5 = 75\div5=15$ m, matching the garden side found in part (b) exactly — confirming the relationship (since $(\frac{1}{5})^2=\frac{1}{25}$, consistent with the area ratio given).

QUESTION 7 [7 marks] — Criterion B Hard

Investigate the claim that the difference between consecutive perfect squares equals the corresponding sequence of odd numbers, and use this to find a large square WITHOUT a calculator.

a. Verify that $n^2-(n-1)^2=2n-1$ algebraically.

[2]

Show Solution

$$n^2-(n-1)^2 = n^2-(n^2-2n+1) = 2n-1$$

b. Given $30^2=900$, use the identity from part (a) to find $31^2$ WITHOUT directly multiplying $31\times31$.

[2]

Show Solution

$31^2 = 30^2+(2\times31-1) = 900+61=961$.

c. Continue this technique to find $32^2$ and $33^2$, building each from the previous answer.

[3]

Show Solution

$32^2=31^2+(2\times32-1)=961+63=1024$. $33^2=32^2+(2\times33-1)=1024+65=1089$.

QUESTION 8 [5 marks] — Criterion C Hard

A student claims that $\sqrt{a^2}=a$ is always true for any number $a$.

a. Test the claim with $a=6$ (positive) and $a=-6$ (negative), calculating $\sqrt{a^2}$ in each case.

[3]

Show Solution

$a=6$: $\sqrt{6^2}=\sqrt{36}=6=a$ ? (works). $a=-6$: $\sqrt{(-6)^2}=\sqrt{36}=6$, but $a=-6$, so $\sqrt{a^2}\ne a$ here ? (fails).

b. State the CORRECT general rule connecting $\sqrt{a^2}$ to $a$, valid for both positive and negative $a$, using absolute value notation.

[2]

Show Solution

$$\sqrt{a^2} = |a|$$ (the square root of a squared number always gives the ABSOLUTE VALUE of the original number, since a square root itself is always defined to be non-negative).

QUESTION 9 [6 marks] — Criterion D Hard

A construction crew needs to brace a rectangular wall frame diagonally to prevent it from leaning. The wall frame is 4.8 m wide and 3.6 m tall.

a. Using the Pythagorean relationship (diagonal)$^2$=(width)$^2$+(height)$^2$, find the exact length of diagonal bracing needed.

[3]

Show Solution

$$\text{diagonal}=\sqrt{4.8^2+3.6^2}=\sqrt{23.04+12.96}=\sqrt{36}=6 \text{ m}$$

b. The crew has bracing material sold only in whole-metre lengths, and needs 10cm extra at each end for secure fastening. Determine the minimum length of bracing material (in whole metres) they must purchase, and explain your rounding decision.

[3]

Show Solution

Required length including fastening allowance: $6+0.1+0.1=6.2$ m. Since material is sold in whole metres, they must round UP (not to the nearest whole metre) to ensure enough material — purchasing 7 m, since 6 m would be insufficient for the 6.2 m actually needed.

QUESTION 10 [7 marks] — Criterion A Hard

A square-shaped park has an area of 5,625 m$^2$. A square garden bed inside it has area exactly $\frac{1}{25}$ of the park's area.

a. Find the side length of the park.

[2]

Show Solution

$$\sqrt{5625}=75 \text{ m}$$

b. Find the area, then the side length, of the garden bed.

[3]

Show Solution

Garden area $=5625\div25=225$ m$^2$. Garden side length $=\sqrt{225}=15$ m.

c. Verify your garden side length is consistent with the park's side length, using the fact that if garden AREA is $\frac{1}{25}$ of park area, garden SIDE should be $\frac{1}{5}$ of park side (since area scales with the SQUARE of side length).

[2]

Show Solution

Park side $\div5 = 75\div5=15$ m, matching the garden side found in part (b) exactly — confirming the relationship (since $(\frac{1}{5})^2=\frac{1}{25}$, consistent with the area ratio given).

QUESTION 11 [7 marks] — Criterion B Hard

Investigate the claim that the difference between consecutive perfect squares equals the corresponding sequence of odd numbers, and use this to find a large square WITHOUT a calculator.

a. Verify that $n^2-(n-1)^2=2n-1$ algebraically.

[2]

Show Solution

$$n^2-(n-1)^2 = n^2-(n^2-2n+1) = 2n-1$$

b. Given $30^2=900$, use the identity from part (a) to find $31^2$ WITHOUT directly multiplying $31\times31$.

[2]

Show Solution

$31^2 = 30^2+(2\times31-1) = 900+61=961$.

c. Continue this technique to find $32^2$ and $33^2$, building each from the previous answer.

[3]

Show Solution

$32^2=31^2+(2\times32-1)=961+63=1024$. $33^2=32^2+(2\times33-1)=1024+65=1089$.

QUESTION 12 [5 marks] — Criterion C Hard

A student claims that $\sqrt{a^2}=a$ is always true for any number $a$.

a. Test the claim with $a=6$ (positive) and $a=-6$ (negative), calculating $\sqrt{a^2}$ in each case.

[3]

Show Solution

$a=6$: $\sqrt{6^2}=\sqrt{36}=6=a$ ? (works). $a=-6$: $\sqrt{(-6)^2}=\sqrt{36}=6$, but $a=-6$, so $\sqrt{a^2}\ne a$ here ? (fails).

b. State the CORRECT general rule connecting $\sqrt{a^2}$ to $a$, valid for both positive and negative $a$, using absolute value notation.

[2]

Show Solution

$$\sqrt{a^2} = |a|$$ (the square root of a squared number always gives the ABSOLUTE VALUE of the original number, since a square root itself is always defined to be non-negative).

QUESTION 13 [6 marks] — Criterion D Hard

A construction crew needs to brace a rectangular wall frame diagonally to prevent it from leaning. The wall frame is 4.8 m wide and 3.6 m tall.

a. Using the Pythagorean relationship (diagonal)$^2$=(width)$^2$+(height)$^2$, find the exact length of diagonal bracing needed.

[3]

Show Solution

$$\text{diagonal}=\sqrt{4.8^2+3.6^2}=\sqrt{23.04+12.96}=\sqrt{36}=6 \text{ m}$$

b. The crew has bracing material sold only in whole-metre lengths, and needs 10cm extra at each end for secure fastening. Determine the minimum length of bracing material (in whole metres) they must purchase, and explain your rounding decision.

[3]

Show Solution

Required length including fastening allowance: $6+0.1+0.1=6.2$ m. Since material is sold in whole metres, they must round UP (not to the nearest whole metre) to ensure enough material — purchasing 7 m, since 6 m would be insufficient for the 6.2 m actually needed.

Square numbers and square roots