AB // CD // EF // GH
AX = XY = YH
AC = 1.5 cm
To find:
The value of AG
Solution:
Since CD // GH .... (given)
∴ CX // GH ... (from the given figure)
Considering Δ ACX and Δ AGH, we get
∠CAX = ∠GAH ..... [common angle]
∠CAX = ∠AGH ...... [corresponding angles]
∴ Δ ACX ~ Δ AGH .... [By AA similarity]
Now, we know that the corresponding sides of two similar triangles are proportional to each other.
So, from similar Δ ACX & Δ AGH, we get
∴ \frac{AC}{AG} = \frac{AX}{AH}
AG
AC
=
AH
AX
\implies \frac{AC}{AG} = \frac{AX}{AX\: +\: XY \:+\: YH}⟹
AG
AC
=
AX+XY+YH
AX
∵ AX = XY = YH (given)
\implies \frac{AC}{AG} = \frac{AX}{AX\: +\: AX \:+\: AX}⟹
AG
AC
=
AX+AX+AX
AX
\implies \frac{AC}{AG} = \frac{AX}{3AX}⟹
AG
AC
=
3AX
AX
\implies \frac{AC}{AG} = \frac{1}{3}⟹
AG
AC
=
3
1
substituting AC = 1.5 cm
\implies \frac{1.5}{AG} = \frac{1}{3}⟹
AG
1.5
=
3
1
on cross-multiplication
\implies AG = 1.5 imes 3⟹AG=1.5×3
\implies \bold{AG = 4.5\:cm}⟹AG=4.5cm
Thus, \boxed{\bold{\underline{AG = 4.5\:cm}}}
AG=4.5cm
.
