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Straight Lines and Pair of Straight Lines question

2021 · 18 Mar · Shift 2 · Q34
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  5. /2021 · 18 Mar · Shift 2 · Q34

Straight Lines and Pair of Straight Lines question

2021 · 18 Mar · Shift 2 · Q34

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Let the centroid of an equilateral triangle ABC be at the origin. Let one of the sides of the equilateral triangle be along the straight line x + y = 3. If R and r be the radius of circumcircle and incircle respectively of Δ\DeltaΔ ABC, then (R + r) is equal to :
  1. A
    727\sqrt 272​
  2. B
    92{9 \over {\sqrt 2 }}2​9​
  3. C
    222\sqrt 222​
  4. D
    323\sqrt 232​
View written solutionFree

Correct answer: B

  1. Use the centroid/incenter property of an equilateral triangle

In an equilateral triangle, the centroid, circumcenter, and incenter all coincide.

Given that the centroid is at the origin, the incenter and circumcenter are also at the origin.

So, the distance from the origin to any side of the triangle equals the inradius rrr.

  1. Find the distance from the origin to the given side

One side lies on the line x+y=3 ⇒ x+y−3=0.x+y=3 \,\Rightarrow\, x+y-3=0.x+y=3⇒x+y−3=0.

Distance of the origin (0,0)(0,0)(0,0) from this line is r=∣0+0−3∣12+12=32.r=\frac{|0+0-3|}{\sqrt{1^2+1^2}}=\frac{3}{\sqrt{2}}.r=12+12​∣0+0−3∣​=2​3​.

Thus, r=32.r=\frac{3}{\sqrt{2}}.r=2​3​.

  1. Relate circumradius and inradius for an equilateral triangle

For an equilateral triangle of side aaa, R=a3,r=a23.R=\frac{a}{\sqrt{3}}, \qquad r=\frac{a}{2\sqrt{3}}.R=3​a​,r=23​a​. Hence, R=2r.R=2r.R=2r.

So, R+r=2r+r=3r.R+r=2r+r=3r.R+r=2r+r=3r.

Substitute r=32r=\frac{3}{\sqrt{2}}r=2​3​: R+r=3⋅32=92.R+r=3\cdot \frac{3}{\sqrt{2}}=\frac{9}{\sqrt{2}}.R+r=3⋅2​3​=2​9​.

  1. Match with the options

R+r=92.R+r=\frac{9}{\sqrt{2}}.R+r=2​9​.

So the correct option is B.

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