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Properties of Triangle question

2016 · Shift 1 · Q25
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  5. /2016 · Shift 1 · Q25

Properties of Triangle question

2016 · Shift 1 · Q25

JEE AdvancedMathematicsProperties of TriangleMultiple correct+4 / −2
In a triangle ΔXYZ\Delta XYZΔXYZ, let x,y,zx, y, zx,y,z be the lengths of sides opposite to the angles X,Y,ZX, Y, ZX,Y,Z respectively, and 2s=x+y+z2s = x + y + z2s=x+y+z. If s−x4=s−y3=s−z2{{s - x} \over 4} = {{s - y} \over 3} = {{s - z} \over 2}4s−x​=3s−y​=2s−z​ and area of incircle of the triangle XYZXYZXYZ is 8π3{{8\pi } \over 3}38π​, then
  1. A
    area of the triangle XYZXYZXYZ is 666\sqrt 666​
  2. B
    the radius of circumcircle of the triangle XYZXYZXYZ is 3566{{35} \over 6}\sqrt 6635​6​
  3. C
    sin⁡X2sin⁡Y2sin⁡Z2=435\sin {X \over 2}\sin {Y \over 2}\sin {Z \over 2} = {4 \over {35}}sin2X​sin2Y​sin2Z​=354​
  4. D
    sin⁡2(X+Y2)=35{\sin ^2}\left( {{{X + Y} \over 2}} \right) = {3 \over 5}sin2(2X+Y​)=53​
View written solutionFree

Correct answer: A, C, D

  1. Use the given semiperimeter relations

Given

s−x4=s−y3=s−z2=k\frac{s-x}{4}=\frac{s-y}{3}=\frac{s-z}{2}=k4s−x​=3s−y​=2s−z​=k

for some positive constant kkk.

So,

s−x=4k,s−y=3k,s−z=2k.s-x=4k,\quad s-y=3k,\quad s-z=2k.s−x=4k,s−y=3k,s−z=2k.

Now add these three equations:

(s−x)+(s−y)+(s−z)=4k+3k+2k=9k.(s-x)+(s-y)+(s-z)=4k+3k+2k=9k.(s−x)+(s−y)+(s−z)=4k+3k+2k=9k.

But

(s−x)+(s−y)+(s−z)=3s−(x+y+z)=3s−2s=s.(s-x)+(s-y)+(s-z)=3s-(x+y+z)=3s-2s=s.(s−x)+(s−y)+(s−z)=3s−(x+y+z)=3s−2s=s.

Hence,

s=9k.s=9k.s=9k.

Therefore,

x=s−4k=5k,y=s−3k=6k,z=s−2k=7k.x=s-4k=5k,\quad y=s-3k=6k,\quad z=s-2k=7k.x=s−4k=5k,y=s−3k=6k,z=s−2k=7k.

So the sides are in the ratio

5:6:7.5:6:7.5:6:7.
  1. Use the incircle area to find the inradius

Area of incircle is

πr2=8π3.\pi r^2=\frac{8\pi}{3}.πr2=38π​.

Thus,

r2=83⇒r=83=263.r^2=\frac{8}{3} \quad\Rightarrow\quad r=\sqrt{\frac{8}{3}}=\frac{2\sqrt6}{3}.r2=38​⇒r=38​​=326​​.
  1. Find the scale factor kkk using Heron's formula

For a triangle,

Δ=rs.\Delta=rs.Δ=rs.

Also by Heron,

Δ=s(s−x)(s−y)(s−z).\Delta=\sqrt{s(s-x)(s-y)(s-z)}.Δ=s(s−x)(s−y)(s−z)​.

Substitute:

s=9k,s−x=4k,s−y=3k,s−z=2k.s=9k,\quad s-x=4k,\quad s-y=3k,\quad s-z=2k.s=9k,s−x=4k,s−y=3k,s−z=2k.

Then

Δ=(9k)(4k)(3k)(2k)=216k4=66 k2.\Delta=\sqrt{(9k)(4k)(3k)(2k)} =\sqrt{216k^4} =6\sqrt6\,k^2.Δ=(9k)(4k)(3k)(2k)​=216k4​=66​k2.

But also,

Δ=rs=263⋅9k=66 k.\Delta=rs=\frac{2\sqrt6}{3}\cdot 9k=6\sqrt6\,k.Δ=rs=326​​⋅9k=66​k.

Equating,

66 k2=66 k⇒k=16\sqrt6\,k^2=6\sqrt6\,k \quad\Rightarrow\quad k=166​k2=66​k⇒k=1

(since k>0k>0k>0).

Hence,

x=5,y=6,z=7,x=5,\quad y=6,\quad z=7,x=5,y=6,z=7,

and

s=9.s=9.s=9.
  1. Check Option A: area of triangle

Using

Δ=rs=263⋅9=66.\Delta=rs=\frac{2\sqrt6}{3}\cdot 9=6\sqrt6.Δ=rs=326​​⋅9=66​.

So Option A is correct.

  1. Check Option B: circumradius

Use

Δ=xyz4R.\Delta=\frac{xyz}{4R}.Δ=4Rxyz​.

So,

R=xyz4Δ=5⋅6⋅74⋅66=210246=3546=35624.R=\frac{xyz}{4\Delta}=\frac{5\cdot 6\cdot 7}{4\cdot 6\sqrt6} =\frac{210}{24\sqrt6} =\frac{35}{4\sqrt6} =\frac{35\sqrt6}{24}.R=4Δxyz​=4⋅66​5⋅6⋅7​=246​210​=46​35​=24356​​.

But the option says

3566.\frac{35}{6}\sqrt6.635​6​.

These are not equal.

So Option B is incorrect.

  1. Check Option C: sin⁡X2sin⁡Y2sin⁡Z2\sin \frac X2 \sin \frac Y2 \sin \frac Z2sin2X​sin2Y​sin2Z​

We use the identity

sin⁡X2sin⁡Y2sin⁡Z2=r4R.\sin\frac X2\sin\frac Y2\sin\frac Z2=\frac{r}{4R}.sin2X​sin2Y​sin2Z​=4Rr​.

We have

r=263,R=35624.r=\frac{2\sqrt6}{3},\qquad R=\frac{35\sqrt6}{24}.r=326​​,R=24356​​.

Thus,

r4R=2634⋅35624=2633566=263⋅6356=435.\frac{r}{4R} =\frac{\frac{2\sqrt6}{3}}{4\cdot \frac{35\sqrt6}{24}} =\frac{\frac{2\sqrt6}{3}}{\frac{35\sqrt6}{6}} =\frac{2\sqrt6}{3}\cdot \frac{6}{35\sqrt6} =\frac{4}{35}.4Rr​=4⋅24356​​326​​​=6356​​326​​​=326​​⋅356​6​=354​.

Hence,

sin⁡X2sin⁡Y2sin⁡Z2=435.\sin\frac X2\sin\frac Y2\sin\frac Z2=\frac{4}{35}.sin2X​sin2Y​sin2Z​=354​.

So Option C is correct.

  1. Check Option D: sin⁡2(X+Y2)\sin^2\left(\frac{X+Y}{2}\right)sin2(2X+Y​)

Since

X+Y+Z=π,X+Y+Z=\pi,X+Y+Z=π,

we get

X+Y2=π−Z2=π2−Z2.\frac{X+Y}{2}=\frac{\pi-Z}{2}=\frac\pi2-\frac Z2.2X+Y​=2π−Z​=2π​−2Z​.

Therefore,

sin⁡2(X+Y2)=cos⁡2Z2.\sin^2\left(\frac{X+Y}{2}\right)=\cos^2\frac Z2.sin2(2X+Y​)=cos22Z​.

Now use

sin⁡2Z2=(s−x)(s−y)xy.\sin^2\frac Z2=\frac{(s-x)(s-y)}{xy}.sin22Z​=xy(s−x)(s−y)​.

Here,

(s−x)=4,(s−y)=3,x=5,y=6.(s-x)=4,\quad (s-y)=3,\quad x=5,\quad y=6.(s−x)=4,(s−y)=3,x=5,y=6.

So,

sin⁡2Z2=4⋅35⋅6=1230=25.\sin^2\frac Z2=\frac{4\cdot 3}{5\cdot 6}=\frac{12}{30}=\frac25.sin22Z​=5⋅64⋅3​=3012​=52​.

Hence,

cos⁡2Z2=1−25=35.\cos^2\frac Z2=1-\frac25=\frac35.cos22Z​=1−52​=53​.

Thus,

sin⁡2(X+Y2)=35.\sin^2\left(\frac{X+Y}{2}\right)=\frac35.sin2(2X+Y​)=53​.

So Option D is correct.

  1. Final conclusion

The correct options are:

A, C, D\boxed{A,\ C,\ D}A, C, D​

which matches the stored correct answer.

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