In the figure ΔA¹BC¹ is constructed similar to the ΔABC with the
scale factor
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Answer:
Draw a line segment AB of 4.6cm
At B draw an angle of 60
∘
With center B and radius 5.1cm draw an arch which intersect line of angle at C.
Join BC.
At A draw an angle BAX of any measure.
Starting from A cuts five equal parts on AX.
Join X
5
A
Through X
4
, Draw X
4
Q∥X
5
B.
Through Q, draw AP∥BC
Therefore,
ΔPAQ∼ΔCAB
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