# 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

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