Question

Please solve q 106 ?

Answer 1

`(3) " " 63`

Explanation:

As shown in the figure, `AD,BE and CF` are the three medians of `DeltaABC`,
draw a line `FG`, congruent and parallel to `BE`,
as `AF=FB, FG=BE and FG` // `BE`,
`=> DeltaAFG and DeltaFBE` are congruent,
`=> AG=FE, and AG` // `FE` // `DC`,
as `FE=1/2BC=DC, and AG` // `DC, => AG=DC`,
`=> CG` is congruent and parallel to `DA`
`=> DeltaCFG` is the second triangle formed of which the three side lengths are the same as the medians of `DeltaABC`
Now, `DeltaAFH and DeltaABE` are similar,
as `AF=1/2AB, => FH=1/2BE=1/2FG, => FH=HG`
similarly, `AH=1/2AE=1/4AC, => AH:AC=1:4`
Now, see Fig 2,
let `|ABC|` denote area of `ABC`,
`|AFC|=1/2|ABC|`
`|CHF|=1/2|CGF|`
`|AFH|:|AFC|=AH:AC=1:4, => |AFH|=1/4|AFC|`
`=>|FHC|=|AFC|-|AFH|=|AFC|-1/4|AFC|=3/4|AFC|`
`=> |FHC|:|AFC|=3:4`
`=> color(red)(|CFG|:|ABC|=3:4)`
Now, use Heron's formula to find `|ABC|`,
`=> |ABC|=sqrt(s(s-a)(s-b)(s-c))`
where `a=13, b=14 and c=15`,
and `s=(a+b+c)/2=(13+14+15)/2=21`
`=> |ABC|=sqrt(21(21-13)(21-14)(21-15))`
`=sqrt(21xx8xx7xx6)=sqrt7056=84`
`=> |CFG|=3/4*|ABC|=3/4*84=63 " units"^2`

Answer 2

AD, BE and CF are three medians of `Delta ABC` intersect at G, the centroid of the triangle. AD is produced up to H such that `GD=DH`. `B.H and C,H` are joined. As D is the mid point of each diagonals of the quadrilateral BGCH then it must be a parallelogram. So CG or FG are parallel to BH. F being mid point of AB , G must be mid point of AH.
Hence `AG=GH=2GD`

So `AD=AG+GD=3GD`

`=> AG=2/3AD`

Similarly `BG=2/3BE and CG=2/3CF`

Again `BH=CG and GH =AD`

Hence in `Delta BGH`

`BG=2/3BE, BH=2/3CFand GH=2/3AD`

So if each side of `DeltaBGH` be `3/2` times increased then we get the triangle formed by the lengths of the medians `AD,BE and CF`. If the area of the triangles formed by medians be `Delta_"med"`. then

`Delta_"med"=9/4Delta BGH`

[By Heron's formula it can be shown that the increase of each side of triangle by m times will increase the area of the triangle by `m^2` times]

Again `DeltaABC=3Delta BGH`

So `Delta_"med"=9/4xx1/3DeltaABC=3/4DeltaABC`

Now using Heron's formula we can get area of `Delta ABC`
we have `a=14,b=15and c=13`,

semi perimeter `s=(a+b+c)/2=(14+15+13)/2=21`
So `Delta ABC=sqrt(21xx(21-14)(21-15)(21-13))=84` sq unit

Hence `Delta_"med"=3/4DeltaABC=3/4xx84=63` sq unit.

Answer 3

I got in general `M = 3/4 A` where M is the area of the triangle made from the medians. Here we have

`16A^2=(13+14+15)(-13+14+15)(13-14+15)(13+14-15)`

`16A^2=42(16)(14)(12)` or `A^2=7(2)(3)(7)(2)(2)(2)(3)` or `A=7(4)(3)` or

`M=3/4 A =7(3)(3)=63`

Explanation:

I didn't peek at the other answers. Let's try a direct approach and figure out the lengths of the medians. Can we do it in general.

Triangle sides `a,b,c`, coordinates `O(0,0), P(a,0), Q(x,y).`

`OP^2=a^2`

`OQ^2=x^2+y^2=b^2`

`PQ^2=(x-a)^2+y^2=c^2`

`x^2 + y^2 - 2ax + a^2 = c^2`

`b^2 - 2ax + a^2 = c^2`

`x = 1/{2a} ( a^2 +b^2 - c^2 )`

`y^2 = b^2 - x^2 = {4a^2 b^2 - (a^2+b^2-c^2)^2}/{4a^2}`

I recognize the numerator as sixteen times the squared area of the triangle. Even if you didn't we can see it because we know the area of the triangle is `A=1/2 ay` or `y={2A}/a` or `y^2={4A^2}/a^2={16A^2}/{4a^2}.`

So in passing we found a formula for the area of a triangle given the sides. I'll write it out the variants in full for reference.

`16A^2 = 4a^2 b^2-(a^2+b^2-c^2)^2`

`16A^2 = (a^2+b^2+c^2)^2-2(a^4+b^4+c^4)`

`16A^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c)`

One of these is almost always better than Heron.

Back to the problem. For sanity we'll write `y` as

`y = sqrt{ {16A^2}/{4a^2} } = {2A}/a`

We have triangle

`O(0,0), P(a,0), Q(1/{2a} ( a^2 +b^2 - c^2 ), {2A}/a)`

The midpoints of the sides are

`R=1/2 OP = (a/2, 0)`

`S = 1/2 PQ = ( 1/{4a}(3a^2+b^2-c^2), A/a)`

`T=1/2 OQ = (1/{4a}(a^2+b^2-c^2), A/a)`

The median squared lengths are

#OS^2 = 1/{16a^2} ( 16A^2 + (3a^2+b^2-c^2)^2 ) #

#PT^2 = (a - (1/{4a}(a^2+b^2-c^2))^2 + A^2/a^2 = 1/(16 a^2)((3 a^2 - b^2 + c^2)^2 + 16 A^2) #

#QR^2 = (a/2 - 1/{2a} ( a^2 +b^2 - c^2 ))^2 + {4A^2}/a^2 = 1/{16a^2} \ 4(16 A^2 + (b^2 - c^2)^2) #

Let's call the area of the median triangle `M.` We can choose one of formulas above, how about

`16M^2 = 4 PT^2 QR^2 - (PT^2+QR^2-OS^2)^2`

# 16M^2 (16a^2)^2 = 4((3 a^2 - b^2 + c^2)^2 + 16 A^2)4(16 A^2 + (b^2 - c^2)^2) - ( ((3 a^2 - b^2 + c^2)^2 + 16 A^2) + 4(16 A^2 + (b^2 - c^2)^2) - ( 16A^2 + (3a^2+b^2-c^2)^2 ) = 2304 a^4 A^2 #

`M^2/A^2 = 2304/16^3 = 9/16`

`M = 3/4 A`

That simplified nicely.