Question

In triangle abc, bc=2 , ac=3 and median ad and be are perpendicular. Find area(abc)

Answer 1

The medians AD and BE of ∆ABC are perpendicular. what is the length of AB if BC = 3 cm and AC = 4 cm?

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3 ANSWERS



Ale Schild, Future Mathematician

Answered Feb 7, 2016

I'll assume that DD and EE are midpoints of BCBC and ACAC respectively.

AC=2AE=4AC=2AE=4   so   AE=2AE=2

BC=2BD=3BC=2BD=3   so   BD=1.5BD=1.5

Let PP be the intersection point of ADAD and BE.BE. Note that PP is the centroid.

Because ADAD and BEBE are medians, we know that:

BP=2PEBP=2PE   and   AP=2PDAP=2PD

I just used a property that says that the centroid divides each median in the ratio 2:1. The proof isn't hard, you can find several videos explaining it (for example: Khan Academy).

In order to simplify the notation:

BP=2nBP=2n   ,   PE=nPE=n   ,   AP=2mAP=2m   ,   PD=mPD=m   and   AB=x


△EPA△EPA   ,   △BPD△BPD   and   △APE△APE are right angled at PP. Now we can use the Pythagorean Theorem:

(I)(I): (2m)2+n2=22=4(2m)2+n2=22=4

(II)(II): (2n)2+m2=1.52=2.25(2n)2+m2=1.52=2.25

(III)(III): (2m)2+(2n)2=x2(2m)2+(2n)2=x2

You can solve (I)(I) and (II)(II) to find the values of mm and nn. I solved 4(II)−(I)4(II)−(I) to find n, there are many different methods. So now we know that:

n2=13n2=13   and   m2=1112m2=1112

Let's solve (III)(III):

x2=(2m)2+(2n)2=4m2+4n2x2=(2m)2+(2n)2=4m2+4n2

x2=4⋅1112+4⋅13=113+43=153=5x2=4⋅1112+4⋅13=113+43=153=5

There are two solutions:

x=±5–√x=±5

But ABAB is a segment, the length of a segment is always positive:

x=AB=5–√x=AB=5

And we are done