To determine:To Find:the measure of side BB' in the figure shown such that VA' = 15, A'A=20 and VB=49 Given: Figure is shown below. 12210202971.jpg

Calculation:
Since, plane of ${\displaystyle \mathrm{△}{A}^{\prime }{B}^{\prime }{C}^{\prime }}$ is similar to the plane of ${\displaystyle \mathrm{△}ABC}$
Since, ABCD is a parallelogram.
${\displaystyle \mathrm{\angle }V=\mathrm{\angle }V\therefore }$(Common)
${\displaystyle \mathrm{\angle }VAB=\mathrm{\angle }V{A}^{\prime }{B}^{\prime }\therefore }$(Corresponding angle in two similar plane)
Therefore by AA similarity, ${\displaystyle \mathrm{△}VAB\sim \mathrm{△}V{A}^{\prime }{B}^{\prime }}$
Ratio of corresponding sides in two similar triangles is equal.
${\displaystyle \frac{VA}{V{A}^{\prime }}=\frac{VC}{V{C}^{\prime }}}$
${\displaystyle \frac{V{A}^{\prime }+{\mathrm{\forall }}^{\prime }}{V{A}^{\prime }}=\frac{VB}{V{B}^{\prime }}}$
${\displaystyle \frac{15+20}{15}=\frac{49}{V{B}^{\prime }}}$
${\displaystyle V{B}^{\prime }=\frac{15\times 49}{35}}$
VB'=21
VB=BB'+VB'
49=BB+21
BB'=28
Therefore, the answer is 28.
Therefore, the answer is 42.