Area squaring

Geometry Level 3

The area of the largest square which can be inscribed in a right angled triangle with legs of 4units and 8 units is? Give the answer to the nearest 2 decimal places.


The answer is 7.11.

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Tanveen Dhingra by tanveen dhingra , 19, India

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3 solutions

Archiet Dev
Feb 11, 2016

In the given Figure, T r i a n g l e Triangle A D E ADE ~ T r i a n g l e Triangle A B C ABC ,

By AA similarity,

a n g l e A B C angle ABC = a n g l e A D E angle ADE = 90 degree

a n g l e B A C angle BAC = a n g l e D A E angleDAE

So, Let sides of Square be x x

( D E = E F = B F = B D = x DE=EF=BF=BD=x )

D E B C \frac{DE}{BC} = A D A B \frac{AD}{AB}

x 8 \frac{x}{8} = 4 − x 4 \frac{4-x}{4}

(x=8-2x)

x = x= 8 3 \frac{8}{3}

So, area of square = [ 8 3 \frac{8}{3} ]^2

= 7.11 =7.11

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Ivan Martinez
Jan 8, 2015

Put into the Cartesian coordinate system, the hypothenuse will have the line y= ( \frac{-x}{2})\= 4. As a square must be formed, y+x, So answer the equationy= ( \frac{-y}{2})\= 4

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Nishant Kumar
Jan 8, 2015

If the legs of the right triangle have lengths a and b, show that the side of the indicated square (inscribed square, largest inscribed square with sides parallel to the legs, however you want to describe it!) has length ab/(a+b).

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Could you elaborate your solution

Rajdeep Dhingra - 6 years, 5 months ago

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Position the right triangle on an x y xy -grid so that its vertical leg lies on the positive y y -axis and the horizontal leg lies on the positive x x -axis. Suppose the vertical leg has length a a and the horizontal leg has length b b . Then the hypotenuse lies on the line y = − a b x + a y = -\frac{a}{b}x + a .

The largest square that can be inscribed in the triangle will then have its upper right vertex lying on the hypotenuse at a point ( k , k ) (k,k) , and thus

k = − a b k + a ⟹ k ( 1 + a b ) = a ⟹ k = a b a + b . k = -\frac{a}{b}k + a \Longrightarrow k(1 + \dfrac{a}{b}) = a \Longrightarrow k = \dfrac{ab}{a + b}.

The area of the square is then k 2 = ( a b a + b ) 2 k^{2} = (\dfrac{ab}{a + b})^{2} .

(Note that because of the symmetry of the formula it does not matter which leg of the triangle we choose to be the vertical and which to be the horizontal.)

So with, say, a = 4 , b = 8 a = 4, b = 8 , we have an area of k 2 = ( 32 12 ) 2 = 64 9 k^{2} = (\frac{32}{12})^{2} = \frac{64}{9} .

Brian Charlesworth - 6 years, 5 months ago

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Thank you for the solution.

Rajdeep Dhingra - 6 years, 5 months ago

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