The smallest circle touching all the sides of triangle is called its incircle. The centre of this circle is called the incentre. The radius of the incircle of the triangle ABC is denoted by r. Note that the incentre lies on the point of intersection of internal angular bisectors of its angles. Again the circle touching the sides BC and the two sides AB and AC produced of a triangle is called escribed circle opposite to the angle A. The radius of this circle is denoted by r 1 . The centre of this circle is called excentre. Similarly r 2 , r 3 be the radii of the escribed circles opposite to the angles B and C respectively. The point of intersection of altitudes AK, BL and CM of a triangle is called orthocentre of the triangle ABC. The triangle KLM is called PEDAL triangle of the triangle ABC. It is interesting to note that the triangle ABC is pedal triangle of triangle I 1 , I 2 , I 3 fromed by joining three excentres I 1 , I 2 , I 3 and therefore the incentre I is the orthocentre of the triangle I 1 I 2 I 3 . We will denote the circumcentre and centroid by O and G respectively. The circumradius of Δ ABC is denoted by R. It is well known that r 1 = 
, where Δ is area of triangle.
(i) In any triangle, r 1 must be equal to
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Ans.
(i)
Sol. In an equilateral triangle r 1 = 
= 
= 
and R = 
= 
⇒ r 1 = 
.
We note that choices , and are not even true in an equilateral triangle since they respectively become 
, 
and 
but the expression given in choice becomes 
. ⇒ Option is correct.
(ii)
Sol. Note that HKCL is concyclic quadrilateral and since ∠ ACH = 90° – A, ∠ LKA = 90° – A
Again HMBK is a cyclic quadrilateral
⇒ ∠ MKA = ∠ MBL = 90° – A
∠ LKA = 180° – 2A
⇒ Choice is correct.
(iii)
Sol. In an equilateral triangle feet of the altitudes will be mid points of the sides.
⇒ Side of the pedal triangle KLM = 
⇒ R ′ (circumradius) of pedal triangle
= 
= 
⇒ Choice is correct and other choice are automatically ruled out.
(iv)
Sol. Let ABC be an equilateral then a = b = c= a and R
= 
= 
.
If the triangle ABC is equilateral then pedal triangles
KLM be also equilateral with side 
.
⇒ Area Δ KLM = 
= 
Perimeter Δ KLM = 
⇒ In-radius of Δ KLM = 
= 
= 
Now observe the table:
Choice value of expression given in choice for equilateral triangle
(D ) 
⇒ choice is correct.
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems