A square in a triangle

Geometry Level pending

In the figure A B C \triangle ABC is a right triangle with legs A B = 6 AB = 6 and A C = 8 AC = 8 . A square is drawn as shown, with a side along B C BC and corners on A B AB and A C AC . Find the length of the side of the square.

None of the others 120 27 \frac {120}{27} 120 37 \frac {120}{37} 19 \sqrt {19} 57 2 \sqrt{57}2 9 2 \frac 92

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4 solutions

Chew-Seong Cheong
May 28, 2020

We note the right A B C \triangle ABC is the popular right triangle with integer side lengths of ratio 3 : 4 : 5 3:4:5 or 0.6 : 0.8 : 1 0.6:0.8:1 . We also note that A I J \triangle AIJ and B I H \triangle BIH are similar to A B C \triangle ABC .

Let the side length of the square be a a , then A I = 0.6 a AI = 0.6 a and I B = a 0.8 IB = \dfrac a{0.8} . Since:

A I + I B = A B 0.6 a + a 0.8 = 6 0.48 a + a 0.8 = 6 a = 4.80 1.48 = 120 37 \begin{aligned} AI + IB & = AB \\ 0.6a + \frac a{0.8} & = 6 \\ \frac {0.48a+a}{0.8} & = 6 \\ \implies a & = \frac {4.80}{1.48} = \boxed{\frac {120}{37}} \end{aligned}

Let A C B = α \angle {ACB}=α . Then tan α = 3 4 \tan α=\dfrac{3}{4} .

Let the length of each side of the square be a a . Then

( 4 3 + 1 + 3 4 ) a = 6 2 + 8 2 = 10 \left (\dfrac{4}{3}+1+\dfrac{3}{4}\right ) a=\sqrt {6^2+8^2}=10

a = 120 37 \implies a=\boxed {\dfrac{120}{37}} .

Nice Mr. Alak

Mahdi Raza - 1 year ago
Marvin Kalngan
Jun 1, 2020

I have a derived formula about problems like this. Let x x be the side length of the square. Then,

x = b h b + h x = \dfrac{bh}{b+h} where: b b is the side of the triangle where the side of the square coincides and h h is the perpendicular distance from b b to the opposite vertex.

Note that A B C \triangle ABC is the popular pythagorean triple, 3 4 5 3-4-5 right triangle. So the length of B C BC is 10 10 . Now we need to find h h ,

sin A B C = A C B C = h A B \sin \angle ABC = \dfrac{AC}{BC}=\dfrac{h}{AB}

8 10 = h 6 \dfrac{8}{10}=\dfrac{h}{6}

h = 24 5 h = \dfrac{24}{5}

Now, substitute.

x = b h b + h = 10 × 24 5 10 + 24 5 = 120 37 x = \dfrac{bh}{b+h}=\dfrac{10 \times \dfrac{24}{5}}{10 + \dfrac{24}{5}}=\color{#3D99F6}\boxed{\dfrac{120}{37}} a n s w e r \color{#69047E}\boxed{answer}

Hassan Abdulla
May 28, 2020

B C = 10 (pythagorean theorem) l e t H G = x = H I = G J A B C H B I A B H B = A C H I H B = 3 4 x A B C G C J A B G C = A C G J G C = 4 3 x B C = B H + H G + G C 10 = 37 12 x x = 120 37 \begin{aligned} & BC = 10 {\color{#D61F06} \text{ (pythagorean theorem)}} \\ & let \ HG = x {\color{#D61F06} =HI=GJ}\\ & \triangle ABC \sim \triangle HBI \Rightarrow \frac{AB}{HB} = \frac{AC}{HI}\Rightarrow HB=\frac{3}{4} x \\ & \triangle ABC \sim \triangle GCJ \Rightarrow \frac{AB}{GC} = \frac{AC}{GJ}\Rightarrow GC=\frac{4}{3} x \\ & BC = BH + HG + GC \Rightarrow 10 = \frac{37}{12} x \Rightarrow x=\frac{120}{37} \end{aligned}

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