Largest Square

Geometry Level 4

What is the area (in cm 2 ) \text{ cm}^2) of the largest square that can be inscribed in a triangle with side lengths 5 cm 5\text{ cm} , 6 cm 6\text{ cm} , and 9 cm 9\text{ cm} rounded to 3 decimal places?


Note: Two vertices of the square must lie on one side of the triangle and the other two vertices must touch the two sides of the triangle.


The answer is 5.426.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

let A T A_T be the area of the triangle and A s A_s be the area of the square

From the figure,

A T = A S + A 1 + A 2 + A 3 A_T = A_S + A_1 + A_2 + A_3

1 2 9 h = x 2 + 1 2 x ( h − x ) + 1 2 x ( 9 − x ) \frac{1}{2}9h = x^2 + \frac{1}{2}x(h-x) + \frac{1}{2}x(9-x)

4.5 h = x 2 + 1 2 x h − 1 2 x 2 + 4.5 x − 1 2 x 2 4.5h=x^2+\frac{1}{2}xh-\frac{1}{2}x^2+4.5x-\frac{1}{2}x^2

x = 9 h 9 + h x=\frac{9h}{9+h}

Solving for h h from the figure, we have

s = 5 + 6 + 9 2 = 10 s=\frac{5+6+9}{2}=10

A = 10 ( 10 − 5 ) ( 10 − 9 ) ( 10 − 6 ) = 10 2 A = \sqrt{10(10-5)(10-9)(10-6)}=10\sqrt{2}

10 2 = 1 2 ( 9 ) ( h ) 10\sqrt{2}=\frac{1}{2}(9)(h)

h = 20 9 2 h=\frac{20}{9}\sqrt{2}

Solving for x x , we have

x = 9 ( 20 9 ) 2 9 + 20 9 2 = 2.329323683 x=\dfrac{9(\frac{20}{9})\sqrt{2}}{9+\frac{20}{9}\sqrt{2}}=2.329323683

Finally, the area is

x 2 = x^2 = 5.426 c m 2 \boxed{\color{#D61F06} 5.426~cm^2}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

What you've shown is that the largest square which has a side on the edge of length 9, has an area of 5.426. However, fow do you know that the largest square must have a side on the edge of length 9?

Calvin Lin Staff - 4 years, 2 months ago

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The angle between the sides 5 and 6 cm long is obtuse, 109. 6 ∘ 109.6^\circ , therefore a square against either the 5 cm side or the 6 cm side would not have its two remaining vertices on the triangle as required in the statement of the problem. They would also be smaller, 5.07 area for square against 6 cm side, 4.994 for the square against 5 cm side.

Marta Reece - 4 years, 2 months ago

Did the same way.

Niranjan Khanderia - 3 years, 1 month ago

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