Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Find the magnitude of vertex angle of isosceles triangle of given area A such that radius r of circle inscribed into the triangle is maximum

Find the magnitude of vertex angle of isosceles triangle of given area A such that radius r of circle inscribed into the triangle is maximum

Grade:12

1 Answers

Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
7 years ago
Dear student,

Here is help with a solution to finding angles of a triangle having
maximum area if it is drawn within a circle so that the triangle's
vertices just touch the circle's perimeter (i.e., a triangle inscribed
within a circle).
First draw an isosceles triangle within a circle having an arbitrary
radius equal to R:
.
68-1495_Sol.JPG
The angle at the top of the triangle is assigned to Phi. One half of
this angle is Phi/2. The two equal angles at the base of the triangle
are assigned to Theta. The two equal side lengths of this triangle are
assigned to L. The base of the triangle is assigned to B. The height
of the triangle is assigned to H.
Now, it is well known that the Area (A) of any triangle is:
A = (1/2)BH.
To find the angles associated with the inscribed triangle (described
above) having a maximum area, one may first re-express the area
formula above in terms involving only Phi (or Phi/2) and R.
You should be able to show that:
L/2 = Rcos(Phi/2), so
L = 2Rcos(Phi/2), and
B = 2[2Rcos(Phi/2)sin(Phi/2)] = 4Rcos(Phi/2)sin(Phi/2), and
H = 2Rcos(Phi/2)cos(Phi/2) = 2Rcos^2(Phi/2).
Therefore:
A = (1/2)BH = (1/2)[4Rcos(Phi/2)sin(Phi/2)][2Rcos^2(Phi/2)], or
A = 4R^2[cos^3(Phi/2)sin(Phi/2)] .
Now, one may take the first derivative of A with respect to Phi [i.e.,
dA/dPhi] by holding R constant because the angles associated with the
inscribed triangle having a maximum area are independent of the
circle's size. If [dA/dPhi] is set to equal zero, one may solve for
the value of Phi that satisfies this condition. Once that value of Phi
is known, one may solve for Theta because: Phi + 2Theta = 180. Then,
all the angles associated with the triangle having a maximum area will
be found.
Upon simplification, one should find that:
[dA/dPhi] = 2R^2cos^2(Phi/2)[cos^2(Phi/2) - 3sin^2(Phi/2)].
Setting: [cos^2(Phi/2) - 3sin^2(Phi/2)] = 0, one can show that:
Phi/2 = 30 deg, so:
Phi = 60 deg, therefore,
2Theta = 120 deg, and,
Theta = 60 deg.
Therefore, the inscribed triangle having a maximum area is an
equilateral triangle
Regards
Sumit

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free