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04 May 2020, 02:27
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
63% (02:52) correct
37%
(02:23)
wrong
based on 57
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In the figure above, angle ABC is right angle and AC = 100 cm. If AD = DE = EF = FC, what is the value of BD^2 + BE^2 + BF^2?
A. 12,500 cm^2
B. 10,500 cm^2
C. 10,000 cm^2
D. 8,750 cm^2
E. 5,000 cm^2
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Updated on: 05 May 2020, 04:11
Originally posted by freedom128 on 04 May 2020, 04:07.
Last edited by freedom128 on 05 May 2020, 04:11, edited 1 time in total.
Last edited by freedom128 on 05 May 2020, 04:11, edited 1 time in total.
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To simplify calculation, let ABC be right-angled isosceles triangle (AB=BC and angle A = angle C = 45 deg). For clarity and detailed solution, attached is my self-explanatory sketch.
AB=BC,
AE=EC,
angle A = angle C = 45 deg.
Thus, angle E = 90 deg.
BE^2+EA^2=AB^2
--> BE^2 = 50^2 = 2500
AD = DE = EF = FC = 25cm
angle E = 90 deg
Thus,
--> BD^2 = BE^2+ED^2 = 50^2+25^2 = 3125
--> BF^2 = BE^2+EF^2 = 50^2+25^2 = 3125
Finally,
BE^2 + BD^2 + BF^2
= 2500 + 3125 + 3125
= 8750 cm^2
FINAL ANSWER IS (D)
Posted from my mobile device
AB=BC,
AE=EC,
angle A = angle C = 45 deg.
Thus, angle E = 90 deg.
BE^2+EA^2=AB^2
--> BE^2 = 50^2 = 2500
AD = DE = EF = FC = 25cm
angle E = 90 deg
Thus,
--> BD^2 = BE^2+ED^2 = 50^2+25^2 = 3125
--> BF^2 = BE^2+EF^2 = 50^2+25^2 = 3125
Finally,
BE^2 + BD^2 + BF^2
= 2500 + 3125 + 3125
= 8750 cm^2
FINAL ANSWER IS (D)
Posted from my mobile device
04 May 2020, 05:34
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Quote:
AD,DE,EF,FC=100/4=25
middle segment is the right-triangle's median
thus, this segment is 1/2*hypotenuse = 100/2 = 50
and it is the perpendicular of the hypotenuse, makes a 90o
now we have another right triangle: BED and BEF
BDˆ2=BEˆ2+DEˆ2=50ˆ2+25ˆ2=2500+625=3125
BFˆ2=BEˆ2+EFˆ2=3125
BDˆ2+BFˆ2+BEˆ2=3125+3125+2500=8750
Ans (D)
