The area of a square with right triangle inside it

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Homework Help Overview

The discussion revolves around finding the area of a square that contains a right triangle inside it. The problem involves basic trigonometric relations and the properties of similar triangles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to determine the length of a side of the square by first analyzing the right triangle within it. Some participants suggest exploring the similarity of triangles and establishing relationships between the sides of the square and the triangle.

Discussion Status

The discussion includes various attempts to approach the problem, with some participants offering insights into the relationships between the angles and sides involved. There appears to be a productive exchange of ideas, although no consensus has been reached on a specific method.

Contextual Notes

There is mention of a broken link to a figure that is essential for understanding the problem, which may have impacted the clarity of the discussion. Additionally, the original poster has provided specific measurements and angles related to the triangle.

JasonHathaway
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Homework Statement



I need to find the area of the square in the following figure:

image.jpg

Homework Equations


Basic Trig relations.

The Attempt at a Solution


I aimed to find the length of BC, but first I had to find the unknowns of the right triangle CDE, which are EC=5m, <DCE=36.86ْ , <DEC=53.13ْ .

Then I thought that I can use some trig relation for angles to get <BDC and BCD respectively (since the angles of the square are all 90ْ).

And that's where I am stuck. Any help?
 
Last edited:
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figure shows as a broken link
 
Just fixed it. Sorry about that.
 
Hm ... interesting problem. It seems as though there's likely to be some trick to it that you are not seeing, but whatever it is, I'm not seeing it either.
 
The upper two right triangles AED and BDC are similar. Look at the angles.

squaretriangle.JPG
 
Last edited:
Call BC = s, DB = y, and AD = s-y. It's easy to get two equations in s and y.
 
I got it now. Thank you everyone for your help. I really appreciate it.
 

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