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25 Feb 2021, 18:22
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CDE is a triangle with a right angle (∠CDE = 90). ACD and BCE are line segments. ∠BAC = 45. If DE = 3, CE = 4 and AB = 4*\(\sqrt{2}\), what is BC?
A. \(\frac{16}{3}\)
B. 5
C. 5*\(\sqrt{2}\)
D. 6
E. 6*\(\sqrt{2}\)
A. \(\frac{16}{3}\)
B. 5
C. 5*\(\sqrt{2}\)
D. 6
E. 6*\(\sqrt{2}\)
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25 Feb 2021, 20:01
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twobagels
Now CDE and ABC are two triangles.
“Are these two triangles similar” : No
“Are these two triangles connected to each other in any way” : No
If you are in this situation, analyse the figure and ask yourself whether a line drawn somewhere can help you.
More often than not it will be a perpendicular line that will get you a solution.
Here too, it is the perpendicular drawn from the point B that will help you.
Draw the perpendicular BS.
Take triangle ABS.
Angles : BSA =90, BAS =45, so ABS=45. The triangle is right angled isosceles triangle.
Sides : If hypotenuse is \(4\sqrt{2}\), the other two sides will be 4 each. => \(AB^2=BS^2+AS^2=2*BS^2........BS\sqrt{2}=AB=4\sqrt{2}.....BS=4\)
By the very look, the triangles BSC and CDE should tell you that they are similar : one angle is 90° in each of them and one angle is set of opposite angles.
Take triangles BSC and CDE.
Both are similar so sides will be in the same ratio.
We know the perpendiculars : BS=4 and DE=3, so ratio of sides =\(\frac{BC}{CE}=\frac{BS}{DE}...........\frac{BC}{4}=\frac{4}{3}\)
\(BC=\frac{4}{3}*4=\frac{16}{3}\)
A
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