Similar shapes

The bases are $BC = 6$ and $EF = 1.5$, so the scale factor is $1.5 \div 6 = \frac{1}{4}$.

$DE = 8 \times \frac{1}{4} = 2$ cm

For similar shapes every length is multiplied by the same scale factor.

$12 \div 8 = 1.5$ but $10 \div 6 = 1.67$, so the rectangles are not similar. Jim is wrong.

AC = 5 matches DF = 20, so the scale factor is $20 \div 5 = 4$.

(a) EF matches BC: $4 \times 4 = 16$ cm

(b) AB matches DE: $22 \div 4 = 5.5$ cm

(a) $BC = 15$ matches $QR = 10$, so the scale factor is $\frac{10}{15} = \frac{2}{3}$

$PR = 9 \times \frac{2}{3} = 6$ cm

(b) $GK = 10 - 4 = 6$ cm

The triangles are congruent, so $EG = GK = 6$ and $GF = HG = 4$

$EF = 6 - 4 = 2$ cm

$AB = 8.4$ matches $DE = 12.6$.

Scale factor: $12.6 \div 8.4 = 1.5$.

(a) $DF = 6.4 \times 1.5 = 9.6$ cm.

(b) $CB = 15 \div 1.5 = 10$ cm.

Triangles ABE and ACD are similar with scale factor $\frac{4.2}{6.3} = \frac{2}{3}$.

$AB = \frac{2}{3} \times 10.8 = 7.2$, so $BC = 10.8 - 7.2 = 3.6$ cm

Area $= \frac{4.2 + 6.3}{2} \times 3.6 = 18.9$ cm²

$AC = 21.6 + 58.4 = 80$ cm and $AB = 32 + 22 = 54$ cm.

In triangle $AED$, $AD = 32$ matches $AC = 80$, and $AE = 21.6$ matches $AB = 54$.

$80 \div 32 = 2.5$ and $54 \div 21.6 = 2.5$

The ratios are equal and angle $A$ is common, so the triangles are similar.

$EB = \sqrt{6^2 - 2^2} = \sqrt{32} = 5.656...$

Triangles $DEB$ and $ACB$ are similar with scale factor $\frac{AB}{DB} = \frac{9 + 6}{6} = 2.5$

$CB = 2.5 \times \sqrt{32} = 14.142...$, so $14.14$ cm

Spheres are similar. Volume ratio $A : B = 64 : 125$, so radius ratio $= \sqrt[3]{64} : \sqrt[3]{125} = 4 : 5$

Radius of C is half of B's: $4 : 2.5$ for A : C

Area ratio $= 4^2 : 2.5^2 = 16 : 6.25 = 64 : 25$

Length scale factor $= \frac{5}{2} = 2.5$, so the volume scale factor is $2.5^3 = \frac{125}{8}$

Volume of B $= 12 \times \frac{125}{8} = 1.5 \times 125 = 187.5$ cm³

(a) Radius of B is $0.9 \times$ radius of A, so area of B is $0.9^2 = 0.81$ times area of A.

Area A : area B $= 1 : 0.81 = 100 : 81$

(b) $e^2 = 1.44 f^2$, so $e = \sqrt{1.44} f = 1.2 f$.

$e : f = 1.2 : 1 = 6 : 5$

Triangles $ABD$ and $CBA$ are similar (both right-angled and they share angle $B$).

So $\frac{CB}{AB} = \frac{AB}{DB}$

$CB = \frac{13 \times 13}{5} = \frac{169}{5} = 33.8$ cm

Area scale factor $= 253.5 \div 6 = 42.25$

Length scale factor $= \sqrt{42.25} = 6.5$

Volume scale factor $= 6.5^3 = 274.625$

Clay needed $= 2 \times 274.625 = 549.25$ kg

$549.25 \div 10 = 54.925$, so he must buy 55 bags.

Mass is proportional to volume, so height ratio L : M $= \sqrt[3]{64} : \sqrt[3]{125} = 4 : 5$.

Area ratio M : P $= 144 : 16$, so height ratio $= \sqrt{144} : \sqrt{16} = 12 : 4 = 3 : 1$.

Make M the same: $4:5 = 12:15$ and $3:1 = 15:5$.

So L : M : P $= 12 : 15 : 5$.

$AB : AC : AD = 1 : (1+2) : (1+2+3) = 1 : 3 : 6$

Similar triangles, so areas are in the ratio of the squares: $1 : 9 : 36$.

BCFG $= 9 - 1 = 8$; CDEF $= 36 - 9 = 27$.

So the ratio is $1 : 8 : 27$.

Volume scale factor $= 1587.762 \div 58.806 = 27$

Length scale factor $= \sqrt[3]{27} = 3$

Height of B's triangle: $\frac{2 \times 43.74}{8.1} = 10.8$

$h = 10.8 \div 3 = 3.6$

Volume ratio A : B $= 125 : 27$, so radius ratio $= \sqrt[3]{125} : \sqrt[3]{27} = 5 : 3$.

B : C $= 1 : 2 = 3 : 6$, so A : C radii $= 5 : 6$.

Surface areas go with the square: $25 : 36$.

A regular hexagon of side $x$ is six equilateral triangles.

One triangle $= \frac{1}{2}x^2\sin 60^\circ = \frac{\sqrt{3}}{4}x^2$, so hexagon $= \frac{3\sqrt{3}}{2}x^2$

The enlarged hexagon has side $px$, area $\frac{3\sqrt{3}}{2}p^2x^2$

Shaded $= \frac{3\sqrt{3}}{2}p^2x^2 - \frac{3\sqrt{3}}{2}x^2 = \frac{3\sqrt{3}}{2}(p^2 - 1)x^2$

Area ratio $4 : 9$, so length ratio $\sqrt{4} : \sqrt{9} = 2 : 3$

Volume ratio $2^3 : 3^3 = 8 : 27$

$27 \div 8 = 3.375$

Three fills are not enough, so Tyler fills A 4 times.

Area ratio A:B = 4:25, so length ratio A:B $= \sqrt{4}:\sqrt{25} = 2:5$

Volume ratio B:C = 27:64, so length ratio B:C $= \sqrt[3]{27}:\sqrt[3]{64} = 3:4$

Make B match: A:B = 6:15 and B:C = 15:20, so A:B:C = 6:15:20

A:C = 6:20 = 3:10

Area ratio $3 : 4$, so length ratio $\sqrt{3} : \sqrt{4} = \sqrt{3} : 2$

Length scale factor from B to A: $\frac{\sqrt{3}}{2} = 0.8660$

Volume scale factor: $0.8660^3 = 0.6495$

Volume of A $= 10 \times 0.6495 = 6.50$ cm³

Let $BC = 4$ and $CD = 21$, so $BD = 25$ and $AD = BD = 25$.

Triangles ABC and DAB are similar, so $\frac{AB}{BC} = \frac{DB}{AB}$

$AB^2 = 4 \times 25 = 100$, so $AB = 10$

$AB : AD = 10 : 25 = 2 : 5$