Answer:
BC = 30.7 units (nearest tenth)
Step-by-step explanation:
As ∠AQR = ∠CQR then ∠AQB = ∠CQD
This means that ΔABQ ~ ΔCDQ
Therefore, the side lengths of two similar triangles are proportional.
[tex]\begin{aligned} \sf \dfrac{AB}{CD} & =\sf \dfrac{BQ}{QD}\\\sf \implies \dfrac{32}{19.2} & =\sf \dfrac{15.0}{QD}\\\sf \implies QD & = \sf 9 \end{aligned}[/tex]
Pythagoras' Theorem: [tex]\sf a^2+b^2=c^2[/tex]
(where a and b are the legs, and c is the hypotenuse, of a right triangle)
Given:
- a = BD = BQ + QD = 15 + 9 = 24
- b = CD = 19.2
- c = BC
Substituting values into the formula:
[tex]\begin{aligned}\sf 24^2+19.2^2 & = \sf BC^2\\\sf BC^2 & = \sf 944.64\\\sf BC & = \sf \pm\sqrt{944.64}\\\sf BC & = \sf 30.7 \ (nearest \ tenth)\end{aligned}[/tex]
(since distance is positive only)
