prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.
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Answers
Step-by-step explanation:
The proving of the given scenario is described below.
Step-by-step explanation:
In ΔABC and ΔXYZ,
If ABC \sim∼ XYZ
When the corresponding sides of a triangle are similar, then
⇒ \frac{AB}{XY} =\frac{BC}{YZ} =\frac{AC}{XZ} =K
XY
AB
=
YZ
BC
=
XZ
AC
=K
from the above equation, we get
⇒ AB=K\times XYAB=K×XY ...(equation 1)
⇒ BC=K\times YZBC=K×YZ ...(equation 2)
⇒ AC=K\times XZAC=K×XZ ...(equation 3)
On adding the equation 1, equation 2, equation 3, we get
⇒ AB+BC+AC=K(XY+YZ+XZ)AB+BC+AC=K(XY+YZ+XZ)
⇒ \frac{AB+BC+AC}{XY+YZ+XZ}=K
XY+YZ+XZ
AB+BC+AC
=K
⇒ \frac{Perimeter \ of \ \Delta ABC}{Perimeter \ of \ \Delta XYZ}
Perimeter of ΔXYZ
Perimeter of ΔABC
⇒ \frac{AB}{XY} =\frac{BC}{YZ} =\frac{AC}{XZ}
XY
AB
=
YZ
BC
=
XZ
AC
Learn more:
https://brainly.in/question/3057349
hence solved and proved!!!