15. A right-angled triangle with constant area is given Prove that the hypotenuse of the
triangle is least when the triangle is isosceles.
16. The lengths of the sides
Answers
Explanation:
15.Let ΔABC be a right-angled triangle:
Let AB=h
and AC=x
By Pythagoras theorem,
AB=
h
2
−x
2
1
Area (ΔABC)=
2
1
×Base×Height
A=
2
1
x
h
2
−x
2
Differentiating wrt X, we get -
dx
dA
=
2
1
h
2
−x
2
+
2
x
(
2
h
2
−x
2
(−2x)
) [By chain rule]
=
2
1
(
h
2
−x
2
h
2
−x
2
−x
2
)=
2
1
(
h
2
−x
2
h
2
−2x
2
)
To maximum the area, we put
dx
dA
=0
⇒
2
1
(
h
2
−2x
2
h
2
−2x
2
)=0
⇒x=
2
h
Area is maximum if & only if
dx
2
d
2
A
at x=
2
h
is negative
So,
dx
2
d
2
A
=
2
1
⎣
⎢
⎢
⎡
(h
2
−x
2
)
h
2
−x
2
(−4x)−(h
2
−2x
2
)
2
1
(h
2
−x
2
)
−1/2
(−2x)
⎦
⎥
⎥
⎤
[product rule]
=
2
1
[
h
2
−x
2
−4x
+
(h
2
−x
2
)
3/2
x(h
2
−h
2
)
]
dx
2
d
2
A
=
2
1
⎣
⎢
⎢
⎢
⎢
⎡
h
2
−
2
h
2
−4/
2
h
+
(h
2
−
2
h
2
)
3/2
h
h
(h
2
−h
2
)
⎦
⎥
⎥
⎥
⎥
⎤
=
2
1
[−4+0]=−2<0
∴ area is maximum when x=
2
h
Base =
h
2
−x
2
=
h
2
−
2
h
2
=
2
h
∴ΔABC= Isosceles triangle
16.
To get NO, you have to use the fact that the two triangles are similar, and have corresponding sides. Since they are similar, we know that
24 / 40 = NO / 50
Note the the base of the big triangle is 50, because (40 + 10 = 50). So now we cross multiply:
24 * 50 = 40 * NO
24 * 5 = 4 * NO
6 * 5 = NO
30 = NO
A=1/2(xy)
y=2A/x
h^2=x^2+(4A^2)/x^2
f(x)=x^2+(4A^2)/x^2
differentiating with respecr to (W.r.t) x
f'(x)=2x-(8A^2)/x^3
differentiating w.r.t x
f"(x)=2+24A^2/x^4
critical point
x=root(2A)
