Sequences & SeriesmediumFree

Point on Centroid Line: h is the AM of h1, h2 | JEE

JEE Maths question with a full step-by-step solution.

Question
Through the centroid of an equilateral triangle a line parallel to the base is drawn. On this line an arbitrary interior point PP is taken. Let hh be the distance of PP from the base, and h1,h2h_1, h_2 its distances from the other two sides. Then
Ahh is the H.M. of h1,h2h_1, h_2
Bhh is the G.M. of h1,h2h_1, h_2
Chh is the A.M. of h1,h2h_1, h_2correct
DNone of these
Solution
Step 1: For an interior point, [△ABC]=[△PBC]+[△PCA]+[△PAB][\triangle ABC]=[\triangle PBC]+[\triangle PCA]+[\triangle PAB]. With equal sides this gives
h+h1+h2=altitude.h+h_1+h_2=\text{altitude}.
 Step 2: The centroid divides the altitude 1:21:2 from the base, so any point on the parallel line through it has h=13(altitude)h=\frac13(\text{altitude}), i.e. altitude =3h.=3h. Step 3:
h+h1+h2=3h ⇒ h1+h2=2h ⇒ h=h1+h22,h+h_1+h_2=3h\ \Rightarrow\ h_1+h_2=2h\ \Rightarrow\ h=\frac{h_1+h_2}{2},
 the A.M. of h1h_1 and h2.h_2. Correct answer: (3)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Sequences & Series · medium
Let α,β\alpha, \beta be the roots of x2−x+p=0x^2 - x + p = 0 and γ,δ\gamma, \delta be the roots of x2−4x+q=0x^2 - 4x + q = 0, where p,q∈Zp, q \in \mathbb{Z}. If α,β,γ,δ\alpha, \beta, \gamma, \delta are in G.P., then ∣p+q∣|p + q| equals:
Sequences & Series · easy
a,b,c,da, b, c, d are in increasing G.P. If the A.M. between aa and bb is 66 and the A.M. between cc and dd is 5454, then the A.M. of aa and dd is
Sequences & Series · medium
If the sum of the first 10 terms of the series 11+14⋅4+21+24⋅4+31+34⋅4+⋯\dfrac{1}{1+1^4\cdot4}+\dfrac{2}{1+2^4\cdot4}+\dfrac{3}{1+3^4\cdot4}+\cdots is mn\dfrac{m}{n} with gcd⁡(m,n)=1\gcd(m,n)=1, then m+nm+n is equal to:
Sequences & Series · medium
Let a1,a2,…,a4001a_1, a_2, \ldots, a_{4001} be in A.P. such that 1a1a2+1a2a3+⋯+1a4000a4001=10\frac{1}{a_1 a_2}+\frac{1}{a_2 a_3}+\cdots+\frac{1}{a_{4000} a_{4001}}=10 and a2+a4000=50a_2+a_{4000}=50. Then ∣a1−a4001∣|a_1-a_{4001}| equals

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath