## If a right angled Isosceles Triangle has its equal sides as 10 units what is the largest square that can be fit into it?

Question

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## Answers ( )

Step-by-step explanation:Let us try a right angled triangle of sides AB = 30 m and CB = 40 m. The hypotenuse is AC = 50 m.

Let BDEF be the largest square of side, x m. Triangles ADE and EFC are similar. Hence, AD/DE = EF/FC, or (30-x)/x = x/(40-x), or

(40-x)(30-x) = x^2 or

x^2–70x+1200 = x^2, or

70x = 1200, of

x = 1200/70 = 17.14 m.

The area of the square = 17.14^2 = 293.877551 sq m, while the area of the triangle is 600 sq m.

Areas of other right angled triangles can be worked out similarly.

We can generalize the answer. side of the square = product of the two legs / sum of the two legs.

Thus a RAT of 5–12–13 sides will have the largest square of side (12*13)/(12+13) = 6.24 units.

An isosceles RAT will have the largest square of side √2 * √2/[√2+√2] = 2/2√2

= 1/√2 or √2/2 = 0.707 units.

Let us try a right angled triangle of sides AB = 30 m and CB = 40 m. The hypotenuse is AC = 50 m.

Let BDEF be the largest square of side, x m. Triangles ADE and EFC are similar. Hence, AD/DE = EF/FC, or (30-x)/x = x/(40-x), or

(40-x)(30-x) = x^2 or

x^2–70x+1200 = x^2, or

70x = 1200, of

x = 1200/70 = 17.14 m.

The area of the square = 17.14^2 = 293.877551 sq m, while the area of the triangle is 600 sq m.

Areas of other right angled triangles can be worked out similarly.

We can generalize the answer. side of the square = product of the two legs / sum of the two legs.

Thus a RAT of 5–12–13 sides will have the largest square of side (12*13)/(12+13) = 6.24 units.

An isosceles RAT will have the largest square of side √2 * √2/[√2+√2] = 2/2√2

= 1/√2 or √2/2 = 0.707 units.