Respuesta :

fieryanswererft

Answer:

BC = 30.73

Here,

[tex]\sf \frac{AB}{BQ} = \frac{CD}{DQ}[/tex]

so first solve for QD

[tex]\sf \hookrightarrow \frac{32}{15} = \frac{19.2}{DQ}[/tex]

[tex]\sf \hookrightarrow 32(DQ)} =19.2(15)[/tex]

[tex]\sf \hookrightarrow 32(DQ)} =288[/tex]

[tex]\sf \hookrightarrow DQ =9[/tex]

  • Hence, QD = 9

Now! using Pythagoras theorem,

  • CD² + BD² = BC²
  • 19.2² + (9+15)² = BC²
  • BC = √368.64+576
  • BC = 30.73499634
  • BC = 30.73 ( rounded to nearest hundredth )
semsee45

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)