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Difference between revisions of "2021 WSMO Speed Round Problems/Problem 4"

(Created page with "==Problem== A square <math>ABCD</math> with side length <math>10</math> is placed inside of a right isosceles triangle <math>XYZ</math> with <math>\angle XYZ=90^{\circ}</math>...")
 
 
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A square <math>ABCD</math> with side length <math>10</math> is placed inside of a right isosceles triangle <math>XYZ</math> with <math>\angle XYZ=90^{\circ}</math> such that <math>A</math> and <math>B</math> are on <math>XZ</math>, <math>C</math> is on <math>YZ</math>, and <math>D</math> is on <math>XY</math>. Find the area of <math>XYZ</math>.
 
A square <math>ABCD</math> with side length <math>10</math> is placed inside of a right isosceles triangle <math>XYZ</math> with <math>\angle XYZ=90^{\circ}</math> such that <math>A</math> and <math>B</math> are on <math>XZ</math>, <math>C</math> is on <math>YZ</math>, and <math>D</math> is on <math>XY</math>. Find the area of <math>XYZ</math>.
  
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==Solution 1==
+
==Solution==
 
We are given that <math>AB = BC = CD = DA = 10</math>. Then <math>YD = \frac{10}{\sqrt2} = 5\sqrt2</math>, and <math>XD = \sqrt2 \cdot AD = 10\sqrt2</math>. This means that <math>XY = 10\sqrt2+5\sqrt2 = 15\sqrt2</math>, for an answer of <math>\frac{(15\sqrt2)^2}{2} = \boxed{225}</math>.
 
We are given that <math>AB = BC = CD = DA = 10</math>. Then <math>YD = \frac{10}{\sqrt2} = 5\sqrt2</math>, and <math>XD = \sqrt2 \cdot AD = 10\sqrt2</math>. This means that <math>XY = 10\sqrt2+5\sqrt2 = 15\sqrt2</math>, for an answer of <math>\frac{(15\sqrt2)^2}{2} = \boxed{225}</math>.
  
 
~OlympusHero
 
~OlympusHero

Latest revision as of 15:27, 2 May 2025

Problem

A square $ABCD$ with side length $10$ is placed inside of a right isosceles triangle $XYZ$ with $\angle XYZ=90^{\circ}$ such that $A$ and $B$ are on $XZ$, $C$ is on $YZ$, and $D$ is on $XY$. Find the area of $XYZ$.

Solution

We are given that $AB = BC = CD = DA = 10$. Then $YD = \frac{10}{\sqrt2} = 5\sqrt2$, and $XD = \sqrt2 \cdot AD = 10\sqrt2$. This means that $XY = 10\sqrt2+5\sqrt2 = 15\sqrt2$, for an answer of $\frac{(15\sqrt2)^2}{2} = \boxed{225}$.

~OlympusHero

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