Question

Prove that the area of a right angled triangle of given hypotenuse is maximum when the

triangle is isosceles

Answer 1

Toolbox:

Area of the triangle A=12A=12×xh2−x2−−−−−−√×xh2−x2

Step 1:

Let hh be the hypotenuse of the right angled triangle,and let xx be its altitude.



∴∴ Base of the triangle =h2−x2−−−−−−√h2−x2

The area of the triangle is A=12A=12×h2−x2−−−−−−√×h2−x2

Step 2:

Now differentiating w.r.t xx we get,

dAdx=12dAdx=12[1.h2−x2−−−−−−√+x2[1.h2−x2+x2(h2−x2)−12(−2x)](h2−x2)−12(−2x)]

=12=12[h2−x2−−−−−−√−x2h2−x2][h2−x2−x2h2−x2]

=12=12[h2−2x2h2−x2−−−−−√][h2−2x2h2−x2]

For maximum or minimum,we have

dAdxdAdx=0=0

(i.e) 12[h2−2x2h2−x2−−−−−√]12[h2−2x2h2−x2]=0=0

⇒h2=2x2⇒h2=2x2

∴x=h2–√∴x=h2

Step 3:

Again differentiating w.r.t xx

d2Adx2=12d2Adx2=12[−4x1h2−x2−−−−−√[−4x1h2−x2+(h2−2x2)(−−12)+(h2−2x2)(−−12)(h2−x2)−32(−2x)](h2−x2)−32(−2x)]

=12[−4xh2−x2−−−−−√+x(h2−2x2)(h2−x2)32]=12[−4xh2−x2+x(h2−2x2)(h2−x2)32]

When x=h2–√x=h2

d2Adx2=12[−4(h/2–√)h2−h2/2−−−−−−−√+x(h2−2(h/2–√)2)(h2−(h/2–√)2)32]d2Adx2=12[−4(h/2)h2−h2/2+x(h2−2(h/2)2)(h2−(h/2)2)32]

Step 4:

On simplifying we get,

d2Adx2d2Adx2=−2<0=−2<0

Thus AA is maximum when x=h2–√x=h2

∴∴ Base =h2−h22−−−−−−−√h2−h22

⇒h2–√⇒h2

∴A∴A is maximum when the triangle is isosceles