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Ambarish Gupta

· started a discussion

· 1 Months ago

No need to do all this stuff (Question: 117114.0 )

As we know very well:
sum of square of all sides of an equilateral triangle is equal to 4 times of sum of square of medians.

and here we know very well in equilateral triangle perpendicular and median is same

Question:
Let ABC be an equilateral triangle. Let BE be perpendicular to CA meeting CA at E, then AB2 + BC2  + CA2 is equal to:
Options:
A) 2BE2
B) 3BE2
C) 4BE2
D) 6BE2
Solution:
Ans: (c) Let ABC be an equilateral triangle and BE is \(\perp \) to AC.

\(\therefore\) AB = BC = CA.

\(\therefore\) In right angled \(\triangle\)ABE

AB2 = BE2 + AE2.............. (i)

now in right angled \(\triangle\)BEC


BC2 = BE2 + EC2.............. (ii)

adding (i) and (ii)

AB2 + BC2 = 2BE2 + AE2 + EC2

AB2 + BC2 = 2BE2 + \(\left ( \cfrac {1}{2} AC\right) ^2\) + \(\left ( \cfrac {1}{2} AC\right) ^2\).

AB2 + BC2 =2BE2 + \(\cfrac{2}{4}\)AC2

2AB2 + 2BC2 = 4BE2 + AC2

2AB2 + BC2 + AC2 - AC2 = 4BE2                   (\(\because\)AB = BC = CA)

AB2 + BC2 + CA2 = 4BE2

Ambarish Gupta

· commented

· 1 Months ago

and here we know very well in an equilateral triangle perpendicular and median are same.
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