If you start at vertex A and use the "shortest route" algorithm, what would be the second path to
1 answer:
Answer:
Shortest distance=|AC|/.
Kindly find the attached for the figure
Step-by-step explanation:
This problem can be addressed using right-angled,
Let have right angle triangle with the shortest distance=hypotenuse;
hypotenuse=|AC|
using pythagoras theorem, we have
|AC|^2=|AB|^2+|BC|^2
Let |AB|=|BC|
|AC|^2=2|AB|^2
|AB|=|AC|/
Shortest distance=|AC|/
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Answer:
The correct options are;
1) ΔBCD is similar to ΔBSR
2) BR/RD = BS/SC
3) (BR)(SC) = (RD)(BS)
Step-by-step explanation:
1) Given that RS is parallel to DC, we have;
∠BDC = ∠BRS (Angles on the same side of transversal)
Similarly;
∠BCD = ∠BSR (Angles on the same side of transversal)
∠CBD = ∠CBD = (Reflexive property)
Therefore;
ΔBCD ~ ΔBSR Angle, Angle Angle (AAA) rule of congruency
2) Whereby ΔBCD ~ ΔBSR, we therefore have;
BC/BS = BD/BR → (BS + SC)/BS = (BR + RD)/BR = 1 + SC/BS = RD/BR + 1
1 + SC/BS = 1 + RD/BR = SC/BS = 1 + BR/RD - 1
SC/BS = RD/BR
Inverting both sides
BR/RD = BS/SC
3) From BR/RD = BS/SC the above we have by cross multiplication;
BR/RD = BS/SC gives;
BR × SC = RD × BR → (BR)(SC) = (RD)(BR).
