The ratio of the corresponding sides of the larger and smaller ∆s are the same.

$\frac{\mathrm{|PR|}}{\mathrm{|PY|}}$ = $\frac{\mathrm{|QR|}}{\mathrm{|WY|}}$ = $\frac{\mathrm{|PQ|}}{\mathrm{|PW|}}$

|PQ| = |PW| + |WQ|

|WQ| = |PQ| - |PW|

Use the known sides to find the |PQ|

$\frac{\mathrm{|QR|}}{\mathrm{|WY|}}$ = $\frac{\mathrm{|PQ|}}{\mathrm{|PW|}}$

$\frac{\mathrm{15\; cm}}{\mathrm{6\; cm}}$ = $\frac{\mathrm{|PQ|}}{\mathrm{4\; cm}}$

Cross multiply.

|PQ| x 6 cm = 15 cm x 4 cm

Divide both sides by 6 cm

|PQ| = $\frac{\mathrm{15\; cm\; x\; 4\; cm}}{\mathrm{6\; cm}}$ = 10 cm

|WQ| = |PQ| - |PW|

|WQ| = 10 cm - 4 cm

|WQ| = 6 cm

Let |MQ| = y

Since ∆OMN is similar to ∆OPQ, the corresponding sides are proportional.

Scale factor = $\frac{\mathrm{|OP|}}{\mathrm{|ON|}}$ = $\frac{\mathrm{|OQ|}}{\mathrm{|OM|}}$ = $\frac{\mathrm{|PQ|}}{\mathrm{|MN|}}$

|OM| = 10 - |MQ|

|OM| = 10 - y

$\frac{\mathrm{8\; cm}}{\mathrm{6\; cm}}$ = $\frac{\mathrm{10\; cm}}{\mathrm{10\; -\; y}}$

$\frac{4}{3}$ = $\frac{10}{\mathrm{10\; -\; y}}$

Cross-multiply

4 (10 - y) = 3 x 10

40 - 4y = 30

-4y = 30 - 40

-4y = -10

Divide both sides by -4

y = $\frac{-10}{-4}$

y = 2.5 cm

|MQ| = 2.5 cm

The ratio of the corresponding sides of the larger and smaller ∆s are the same.

$\frac{\mathrm{|PR|}}{\mathrm{|PY|}}$ = $\frac{\mathrm{|QR|}}{\mathrm{|WY|}}$ = $\frac{\mathrm{|PQ|}}{\mathrm{|PW|}}$

|PQ| = |PW| + |WQ|

|WQ| = |PQ| - |PW|

Use the known sides to find the |PQ|

$\frac{\mathrm{|QR|}}{\mathrm{|WY|}}$ = $\frac{\mathrm{|PQ|}}{\mathrm{|PW|}}$

$\frac{\mathrm{15\; cm}}{\mathrm{6\; cm}}$ = $\frac{\mathrm{|PQ|}}{\mathrm{4\; cm}}$

Cross multiply.

|PQ| x 6 cm = 15 cm x 4 cm

Divide both sides by 6 cm

|PQ| = $\frac{\mathrm{15\; cm\; x\; 4\; cm}}{\mathrm{6\; cm}}$ = 10 cm

|WQ| = |PQ| - |PW|

|WQ| = 10 cm - 4 cm

|WQ| = 6 cm