Enlargement and similarity

$k = \dfrac{15}{10} = 1.5$ . $EF = 1.5 \times 8 = 12$ cm. $FD = 1.5 \times 6 = 9$ cm.
$340 \div 200 = 1.7$ m.
Area ratio $= \dfrac{200}{50} = 4$ . Side ratio $= \sqrt{4} = 2$ .
$\dfrac{h}{2} = \dfrac{12}{3}$ , so $h = 8$ m.
Vector from centre to point: $(5-1, -2-0) = (4, -2)$ . Multiply by $3$ : $(12, -6)$ . Image $= (1+12, 0-6) = (13, -6)$ .
$k = \dfrac{12}{4} = 3$ . Volume ratio $= 3^3 = 27$ . The larger sphere’s volume is $27$ times greater.
Any circle can be mapped to any other by a single enlargement (centred at any point) with scale factor equal to the ratio of the radii. Since all angles in a circle are determined by its curvature and all circles have the same shape, they are always similar.
Drawing dimensions: $12 \times \dfrac{1}{100} = 0.12$ m $= 12$ cm, and $8 \times \dfrac{1}{100} = 0.08$ m $= 8$ cm. Area $= 12 \times 8 = 96$ cm $^2$ .

A scale factor of $k = -1$ produces an image that is the same size but on the opposite side of the centre, with each point reflected through the centre. The result is identical to a $180°$ rotation about the centre — a negative scale factor combines enlargement with a half-turn.
Let two similar figures have corresponding sides in ratio $k:1$ . Divide each figure into the same small unit squares (or use the same triangulation). Each unit in the larger figure has sides $k$ times as long, so its area is $k^2$ times as large. Summing over all units, total area scales by $k^2$ .
Height ratio $= \dfrac{25}{10} = 2.5$ . Volume ratio $= 2.5^3 = 15.625$ . Larger volume $= 200 \times 15.625 = 3125$ cm $^3$ .
Scale factor from model to real: $k = 50$ . Real area $= 48 \times 50^2 = 48 \times 2500 = 120\,000$ cm $^2 = 12$ m $^2$ .
Centre is $A(0,0)$ , so multiply coordinates by $\dfrac{1}{2}$ . $A' = (0,0)$ , $B' = (3,0)$ , $C' = (1.5, 2)$ . $AB = 6 \to A'B' = 3$ . $AC = \sqrt{9+16} = 5 \to A'C' = \sqrt{2.25+4} = 2.5$ . $BC = \sqrt{9+16} = 5 \to B'C' = \sqrt{2.25+4} = 2.5$ . All sides halved.

Surface area ratio $= 4:25$ , so side ratio $= \sqrt{4}:\sqrt{25} = 2:5$ . Volume ratio $= 2^3:5^3 = 8:125$ . Larger volume $= 80 \times \dfrac{125}{8} = 1250$ cm $^3$ .
New area $= k^2 \times$ original area $= 3 \times$ original area, so $k^2 = 3$ and $k = \sqrt{3}$ .
$P'_x - C_x = k(P_x - C_x)$ : $17 - 2 = k(8 - 2)$ , so $15 = 6k$ , giving $k = 2.5$ . Check: $P'_y - C_y = 10.5 - 3 = 7.5 = 2.5 \times (6 - 3) = 2.5 \times 3 = 7.5$ . Confirmed $k = 2.5$ .
Since the solids are similar with corresponding length ratio $\dfrac{l_1}{l_2}$ , volumes scale as $\left(\dfrac{l_1}{l_2}\right)^3$ . With the same material (same density $\rho$ ), mass $= \rho \times V$ , so $\dfrac{m_1}{m_2} = \dfrac{\rho V_1}{\rho V_2} = \dfrac{V_1}{V_2} = \left(\dfrac{l_1}{l_2}\right)^3$ .

Tier 1