Let f, g, h be the length of the perpendicular from the circumcenter of the $\vartriangle ABC$ in the sides a, b and c respectively. If $\dfrac{a}{f} + \dfrac{b}{g} + \dfrac{c}{h} = \lambda \dfrac{{abc}}{{fgh}}$, then the values of $\lambda $ is-
$
(a){\text{ }}\dfrac{1}{4} \\
(b){\text{ }}\dfrac{1}{2} \\
(c){\text{ 1}} \\
(d){\text{ 2}} \\
$
Last updated date: 25th Mar 2023
•
Total views: 307.2k
•
Views today: 5.84k
Answer
307.2k+ views
Hint – In this question f, g, h are the length of perpendicular from the circumcenter of the given triangle. So the circumcenter is the center of a triangle’s circumcircle. It is basically the intersection of perpendicular bisectors. Use the property that the angle at the circumcenter of the triangle is double the angle of the vertex along with a basic trigonometric formula to find the value of $\lambda $.
“Complete step-by-step answer:”
Given equation
$ \Rightarrow \dfrac{a}{f} + \dfrac{b}{g} + \dfrac{c}{h} = \lambda \dfrac{{abc}}{{fgh}}$
Let O be the circumcenter of the triangle ABC as shown in figure.
And we all know that the angle at the circumcenter of the triangle is double the angle of the vertex.
$\therefore \angle BOC = 2A,{\text{ }}\angle BOA = 2C,\;\angle COA = 2B$
Now OD, OE and OF are the perpendiculars from the circumcenter on the sides BC, CA and AB having lengths a, b and c respectively.
$ \Rightarrow \angle DOC = \dfrac{{\angle BOC}}{2} = A$
Similarly,
$\angle COE = \dfrac{{\angle COA}}{2} = B,{\text{ }}\angle BOF = \dfrac{{\angle BOA}}{2} = C$
And the length of the perpendiculars OD, OE and OF are f, g and h respectively (see figure).
Since OD is perpendicular to BC,
$\therefore DC = \dfrac{{BC}}{2} = \dfrac{a}{2}$.
Similarly,
$\therefore EC = \dfrac{{AC}}{2} = \dfrac{b}{2},{\text{ FB}} = \dfrac{{AB}}{2} = \dfrac{c}{2}$
Now in triangle ADC
$\tan A = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \dfrac{{DC}}{{OD}} = \dfrac{{\dfrac{a}{2}}}{f} = \dfrac{a}{{2f}}$……………………… (1)
Similarly,
$\tan B = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \dfrac{{EC}}{{OE}} = \dfrac{{\dfrac{b}{2}}}{g} = \dfrac{b}{{2g}}$…………………….. (2)
$\tan C = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \dfrac{{BF}}{{OF}} = \dfrac{{\dfrac{c}{2}}}{h} = \dfrac{c}{{2h}}$ ……………………….. (3)
Now add equation (1), (2) and (3) we have
$\dfrac{a}{{2f}} + \dfrac{b}{{2g}} + \dfrac{c}{{2h}} = \tan A + \tan B + \tan C$…………………………. (4)
Now as we know that in a triangle sum of all angles is equal to 180 degree.
$ \Rightarrow A + B + C = {180^0}$
Then $\tan A + \tan B + \tan C = \tan A\tan B\tan C$
So use this property in equation (4) we have,
$ \Rightarrow \dfrac{a}{{2f}} + \dfrac{b}{{2g}} + \dfrac{c}{{2h}} = \tan A\tan B\tan C$
Now from equation (1), (2) and (3) we have
$ \Rightarrow \dfrac{a}{{2f}} + \dfrac{b}{{2g}} + \dfrac{c}{{2h}} = \left( {\dfrac{a}{{2f}}} \right)\left( {\dfrac{b}{{2g}}} \right)\left( {\dfrac{c}{{2h}}} \right)$
Now simplify the above equation we have,
$ \Rightarrow \dfrac{1}{2}\left( {\dfrac{a}{f} + \dfrac{b}{g} + \dfrac{c}{h}} \right) = \dfrac{1}{8}\left( {\dfrac{a}{f}} \right)\left( {\dfrac{b}{g}} \right)\left( {\dfrac{c}{h}} \right)$
$ \Rightarrow \dfrac{a}{f} + \dfrac{b}{g} + \dfrac{c}{h} = \dfrac{1}{4}\dfrac{{abc}}{{fgh}}$
Now compare this equation from given equation we have,
$\lambda = \dfrac{1}{4}$
Hence option (a) is correct.
Note – Whenever we face such types of problems the key concept is firstly to have a diagrammatic representation of the information provided in the question statement. The basic understanding of circumcenter along with the various trigonometric ratios formulas help in getting the right track to reach the answer.
“Complete step-by-step answer:”
Given equation
$ \Rightarrow \dfrac{a}{f} + \dfrac{b}{g} + \dfrac{c}{h} = \lambda \dfrac{{abc}}{{fgh}}$

Let O be the circumcenter of the triangle ABC as shown in figure.
And we all know that the angle at the circumcenter of the triangle is double the angle of the vertex.
$\therefore \angle BOC = 2A,{\text{ }}\angle BOA = 2C,\;\angle COA = 2B$
Now OD, OE and OF are the perpendiculars from the circumcenter on the sides BC, CA and AB having lengths a, b and c respectively.
$ \Rightarrow \angle DOC = \dfrac{{\angle BOC}}{2} = A$
Similarly,
$\angle COE = \dfrac{{\angle COA}}{2} = B,{\text{ }}\angle BOF = \dfrac{{\angle BOA}}{2} = C$
And the length of the perpendiculars OD, OE and OF are f, g and h respectively (see figure).
Since OD is perpendicular to BC,
$\therefore DC = \dfrac{{BC}}{2} = \dfrac{a}{2}$.
Similarly,
$\therefore EC = \dfrac{{AC}}{2} = \dfrac{b}{2},{\text{ FB}} = \dfrac{{AB}}{2} = \dfrac{c}{2}$
Now in triangle ADC
$\tan A = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \dfrac{{DC}}{{OD}} = \dfrac{{\dfrac{a}{2}}}{f} = \dfrac{a}{{2f}}$……………………… (1)
Similarly,
$\tan B = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \dfrac{{EC}}{{OE}} = \dfrac{{\dfrac{b}{2}}}{g} = \dfrac{b}{{2g}}$…………………….. (2)
$\tan C = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}} = \dfrac{{BF}}{{OF}} = \dfrac{{\dfrac{c}{2}}}{h} = \dfrac{c}{{2h}}$ ……………………….. (3)
Now add equation (1), (2) and (3) we have
$\dfrac{a}{{2f}} + \dfrac{b}{{2g}} + \dfrac{c}{{2h}} = \tan A + \tan B + \tan C$…………………………. (4)
Now as we know that in a triangle sum of all angles is equal to 180 degree.
$ \Rightarrow A + B + C = {180^0}$
Then $\tan A + \tan B + \tan C = \tan A\tan B\tan C$
So use this property in equation (4) we have,
$ \Rightarrow \dfrac{a}{{2f}} + \dfrac{b}{{2g}} + \dfrac{c}{{2h}} = \tan A\tan B\tan C$
Now from equation (1), (2) and (3) we have
$ \Rightarrow \dfrac{a}{{2f}} + \dfrac{b}{{2g}} + \dfrac{c}{{2h}} = \left( {\dfrac{a}{{2f}}} \right)\left( {\dfrac{b}{{2g}}} \right)\left( {\dfrac{c}{{2h}}} \right)$
Now simplify the above equation we have,
$ \Rightarrow \dfrac{1}{2}\left( {\dfrac{a}{f} + \dfrac{b}{g} + \dfrac{c}{h}} \right) = \dfrac{1}{8}\left( {\dfrac{a}{f}} \right)\left( {\dfrac{b}{g}} \right)\left( {\dfrac{c}{h}} \right)$
$ \Rightarrow \dfrac{a}{f} + \dfrac{b}{g} + \dfrac{c}{h} = \dfrac{1}{4}\dfrac{{abc}}{{fgh}}$
Now compare this equation from given equation we have,
$\lambda = \dfrac{1}{4}$
Hence option (a) is correct.
Note – Whenever we face such types of problems the key concept is firstly to have a diagrammatic representation of the information provided in the question statement. The basic understanding of circumcenter along with the various trigonometric ratios formulas help in getting the right track to reach the answer.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

A Short Paragraph on our Country India
