Rectangle is inscribed in a right angle triangle ABC. Show that the area of the largest search is a half the area of the triangle?

Let the right angle triangle ABC have sides of length a, b, and c, where c is the hypotenuse. Let the rectangle be inscribed in the triangle such that one side of the rectangle is along the hypotenuse c and the other side is along one of the other sides of the triangle, say side a.

Let the sides of the rectangle along the hypotenuse and side a be x and y respectively. Since the rectangle is inscribed in the triangle, the sides of the rectangle are parallel to the sides of the triangle.

Now, the area of the rectangle is given by A = x*y.

Since the sides of the rectangle are parallel to the sides of the triangle, the sides of the rectangle are proportional to the sides of the triangle. Therefore, we can write:

x = k*a
y = k*b

where k is a constant of proportionality.

Substitute these values of x and y in the area formula of the rectangle:

A = k*a*k*b = k^2*a*b

Now, we know that the area of the triangle is given by:

Area of triangle = 1/2 * base * height = 1/2 * a * b

Since the rectangle is inscribed in the triangle, the area of the rectangle is less than or equal to the area of the triangle. Therefore, we have:

A <= 1/2 * a * b

Substitute the value of A in the above inequality:

k^2*a*b <= 1/2 * a * b

k^2 <= 1/2

k <= 1/sqrt(2)

Therefore, the maximum area of the rectangle is when k = 1/sqrt(2), which gives the area of the rectangle as:

A = (1/sqrt(2))^2 * a * b = 1/2 * a * b

Hence, the area of the largest rectangle inscribed in the right angle triangle ABC is half the area of the triangle.

Let us assume that the rectangle of the largest area has sides parallel to the sides containing the right angle of the triangle ABC.

Let AB=c,BC=a,CA=b are the sides of right triangle and PQ=x and QR=y are the sides of rectangle.

Using Pythagoras theorem, we have
$$a^2+b^2=c^2$$

Area of rectangle,
$$R=xy$$

and area of triangle,
$$T=\frac{1}{2}ab$$

Now,
$$R=xy$$

$$=\frac{(c^2-a^2+b^2)(c^2-b^2+a^2)}{4}$$

$$=\frac{(c^4+a^4+b^4-a^2b^2-b^2c^2-c^2a^2)}{4}$$

$$=\frac{(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)}{4}$$

$$=\frac{c^4-2(a^2b^2+b^2c^2+c^2a^2)}{4}\quad\because (a^2+b^2+c^2)^2=a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)$$

$$=\frac{1}{4}[c^4-2(a^2b^2+b^2c^2+c^2a^2)]\tag1$$

As
$$a^4+b^4+c^4=a^2b^2+b^2c^2+c^2a^2\tag2$$

From (1) and (2), we get
$$R=\frac{1}{4}c^4-\frac{1}{2}(a^2b^2+b^2c^2+c^2a^2)$$

$$=\frac{1}{4}c^4-\frac{1}{2}\left(c^4-\frac{1}{2}c^4\right)$$

$$=\frac{1}{4}c^4-\frac{1}{4}c^4$$

$$=\frac{1}{4}c^4$$

$$=\frac{1}{4}(a^2+b^2+c^2)^2$$

$$=\frac{1}{4}(2T)^2\quad\because a^2+b^2+c^2=2T$$

$$=\frac{T^2}{2}$$

$$R=\frac{1}{2}T$$

Hence, the area of the largest rectangle is a half the area of the triangle.