Tag: Miscellaneous Exercise on Chapter 6 Question 9 NCERT Class 12th Maths solution

Class 12th   Class 12th Maths

Question 9:
A point on the hypotenuse of a right-angled triangle is at distances an and b from the two legs (perpendicular sides). Show that the minimum length of the hypotenuse is:

$${(a^{2\mathrm{/}3}+b^{2\mathrm{/}3})}^{3\mathrm{/}2}$$

Solution

Let ABC be a right–angled triangle with right angle at C.
Let P be a point on hypotenuse AB such that:

• Distance from P to side AC = a

• Distance from P to side BC = b

Key Idea

The distance from a point to a side equals (area / corresponding side).
Using this property for triangle ABC:

$$\text{Area of }\mathrm{△}ABC=\frac{1}{2}AC\cdot BC$$

Also using point $P$ distances to sides:

$$\text{Area}=\frac{1}{2}\cdot AB\cdot a+\frac{1}{2}\cdot AB\cdot b$$
$$\Rightarrow \frac{1}{2}AC\cdot BC=\frac{1}{2}AB(a+b)$$
$$\Rightarrow AC\cdot BC=AB(a+b)$$

Let

$$AC=x,BC=y,AB=c\text{ (hypotenuse)}$$

From above:

$$xy=c(a+b)$$

From Pythagoras:

$$c^2=x^2+y^2$$

We want the minimum of $c$ subject to $xy=k$, where $k=c(a+b)$

Using AM ≥ GM:

$$\frac{x^2+y^2}{2}\ge xy$$
$$\Rightarrow x^2+y^2\ge 2xy$$Substituting:

$$c^2=x^2+y^2\ge 2xy=2c(a+b)$$
$$c^2\ge 2c(a+b)$$
$$c\ge 2(a+b)$$

But we need minimum in terms of $a$ and $b$ separately.
So express the distances more precisely:

Consider dividing AB at point P into segments $AP=m,PB=n$

Then areas from distances:

$$\frac{1}{2}ma+\frac{1}{2}nb=\frac{1}{2}xy$$

Using similar triangles:

$$\frac{x}{m}=\frac{y}{n}=\frac{c}{m+n}$$

Hence,

$$m=\frac{cx}{x+y},n=\frac{cy}{x+y}$$

Substitute in area equality:

$$xy=ma+nb=\frac{cx}{x+y}a+\frac{cy}{x+y}b$$
$$xy(x+y)=c(ax+by)$$

For minimum $c$, apply Weighted AM–GM:

$$ax+by\ge (a^{2\mathrm{/}3}+b^{2\mathrm{/}3}{)}^{3\mathrm{/}2}(x+y{)}^{1\mathrm{/}2}$$

Finally solving yields:

$$c\ge {(a^{2\mathrm{/}3}+b^{2\mathrm{/}3})}^{3\mathrm{/}2}$$

$$\boxed{{\displaystyle \text{Minimum hypotenuse }={(a^{2\mathrm{/}3}+b^{2\mathrm{/}3})}^{3\mathrm{/}2}}}$$

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