Question

Ab is a diameter of a circle c is a point on the circle, show that the area of the triangle abc is maximum if its an isosceles triangle

Answer 1

ABCΔABC is a right angled
triangle, right angled at CC
(∵(∵ angle in a semicircle is 90∘)90∘)
Hence the area of the ΔABCΔABC is
A=12×AC×BCA=12×AC×BC
Let DD be the diameter and BC=xBC=x
∴AC=D2−x2−−−−−−−√∴AC=D2−x2
∴A=12=D2−x2−−−−−−−√×x∴A=12=D2−x2×x
Step 2
Differentiating w.r.t xx we get,
dAdx=12[D2−x2−−−−−−−√.1+x12D2−x2−−−−−−√(−2x)]dAdx=12[D2−x2.1+x12D2−x2(−2x)]
=12[D2−x2−x2D2−x2−−−−−−√]=12[D2−2x2D2−x2−−−−−−√]=12[D2−x2−x2D2−x2]=12[D2−2x2D2−x2]
To find the maximum or minimum area
dAdx=0dAdx=0
i.e., 12[D2−2x2D2−x2−−−−−−√]=012[D2−2x2D2−x2]=0
⇒D2=2x2⇒D2=2x2
D=x2–√D=x2
Step 3
To prove that the triangle has maximum area let us differentiate dAdxdAdx again
d2Adx2=D2−x2−−−−−−√(−4x)−(D2−2x2)12D2−x2√(−2x)(D2−x2)d2Adx2=D2−x2(−4x)−(D2−2x2)12D2−x2(−2x)(D2−x2)
d2Adx2=D2−x2−−−−−−√[−3x]D2−x2<0d2Adx2=D2−x2[−3x]D2−x2<0
hence this proves that the triangle has maximum value.
AC2=D2−x2AC2=D2−x2
=2x2−x2=2x2−x2
AC2=x2AC2=x2
⇒AC=x⇒AC=x
AC=BC=xAC=BC=x
Hence the triangle has maximum value when it is an isosceles triangle.