Question

Show that a triangle inscribed in circle equilateral triangle has largest perimeter

Answer 1


Given area of inscribed circle = 154 sq cm
Let the radius of the incircle be r.
⇒ Area of this circle = πr2 = 154
(22/7) × r2 = 154
⇒ r2 = 154 × (7/22) = 49
∴ r = 7 cm
Recall that incentre of a circle is the point of intersection of the angular bisectors.
Given ABC is an equilateral triangle and AD = h be the altitude.
Hence these bisectors are also the altitudes and medians whose point of intersection divides the medians in the ratio 2 : 1
∠ADB = 90° and OD = (1/3) AD
That is r = (h/3)
Þ h = 3r = 3 × 7 = 21 cm
Let each side of the triangle be a, then
Altitude of an equilateral triangle is (√3/2) times its side
That is h = (√3a/2)

∴ a = 14√3 cm
We know that perimeter of an equilateral triangle = 3a
= 3 × 14 √3 = 42√3
= 42 × 1.73 = 72.66 cm

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Syeda , SubjectMatterExpert

Member since Jan 25 2017

Answer.

Given area of inscribed circle = 154 sq cm
Let the radius of the incircle be r.
⇒ Area of this circle = πr2 = 154
(22/7) × r2 = 154
⇒ r2 = 154 × (7/22) = 49
∴ r = 7 cm
Recall that incentre of a circle is the point of intersection of the angular bisectors.
Given ABC is an equilateral triangle and AD = h be the altitude.
Hence these bisectors are also the altitudes and medians whose point of intersection divides the medians in the ratio 2 : 1
∠ADB = 90° and OD = (1/3) AD
That is r = (h/3)
Þ h = 3r = 3 × 7 = 21 cm
Let each side of the triangle be a, then
Altitude of an equilateral triangle is (√3/2) times its side
That is h = (√3a/2)

∴ a = 14√3 cm
We know that perimeter of an equilateral triangle = 3a
= 3 × 14 √3 = 42√3
= 42 × 1.73 = 72.66 cm    

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