\(\text{In the given figure, }AX : XB = 3 : 5\\\text{ and }XY || BC.\)
\(\text{If XY = 18 cm, then BC = ___ .}\)
\(\text{In the given figure, }AX : XB = 3 : 5\\\text{ and }XY || BC.\)
\(\text{If XY = 18 cm, then BC = ___ .}\)
\(\text{We are given that in the }\Delta ABC, \\AX : XB = 3 : 5 \text{ and }XY || BC.\)
\(\Rightarrow \frac{XB}{AX}=\frac{5}{3}\)
\(\Rightarrow \frac{XB}{AX}+1=\frac{5}{3}+1\)
\(\Rightarrow \frac{XB+AX}{AX}=\frac{5+3}{3}\)
\(\Rightarrow \frac{AB}{AX}=\frac{8}{3}\)
\(\Rightarrow \frac{AX}{AB}=\frac{3}{8}\)
\(\text{Now in }\Delta AXY \text{and }\Delta ABC,\)
\(\angle AXY =\angle ABC\\\text{ (Corresponding angles)}\)
\(\angle XAY =\angle BAC \\\text{(Common in both triangles)}\)
\(\therefore \Delta AXY \sim \Delta ABC \\\text{(By AA similarity criterion)}\)
\(\Rightarrow \frac{AX}{AB}=\frac{XY}{BC}\)
\(\Rightarrow \frac{3}{8}=\frac{18}{BC}\)
\(\therefore BC =\frac{18 \times 8}{3}~=~48~ cm\)
\(\text{We are given that in the }\Delta ABC, \\AX : XB = 3 : 5 \text{ and }XY || BC.\)
\(\Rightarrow \frac{XB}{AX}=\frac{5}{3}\)
\(\Rightarrow \frac{XB}{AX}+1=\frac{5}{3}+1\)
\(\Rightarrow \frac{XB+AX}{AX}=\frac{5+3}{3}\)
\(\Rightarrow \frac{AB}{AX}=\frac{8}{3}\)
\(\Rightarrow \frac{AX}{AB}=\frac{3}{8}\)
\(\text{Now in }\Delta AXY \text{and }\Delta ABC,\)
\(\angle AXY =\angle ABC\\\text{ (Corresponding angles)}\)
\(\angle XAY =\angle BAC \\\text{(Common in both triangles)}\)
\(\therefore \Delta AXY \sim \Delta ABC \\\text{(By AA similarity criterion)}\)
\(\Rightarrow \frac{AX}{AB}=\frac{XY}{BC}\)
\(\Rightarrow \frac{3}{8}=\frac{18}{BC}\)
\(\therefore BC =\frac{18 \times 8}{3}~=~48~ cm\)
