Find area of a triangle in coordinate plane

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Discussion Overview

The discussion revolves around finding the area of a triangle in the coordinate plane, specifically the triangle $$\triangle QRS$$. Participants explore different methods to calculate the area, including geometric reasoning and the application of formulas.

Discussion Character

  • Technical explanation, Conceptual clarification, Homework-related

Main Points Raised

  • One participant suggests finding the area of a rectangle and subtracting the areas of three right triangles within it to determine the area of $$\triangle QRS$$.
  • Another participant expresses confusion about the relevance of the triangle's position on the grid and mentions only knowing the area formula 1/2BH, which is not yielding the correct answer.
  • A participant claims to have found the area of the triangle to be 7, thanking others for their help.
  • Further clarification is provided regarding the nature of $$\triangle QRS$$, noting it is not a right triangle and discussing the lengths of its sides in relation to the Pythagorean Theorem.
  • Details are given about the bounding rectangle's area and the areas of the three right triangles, leading to the conclusion that the area of $$\triangle QRS$$ is 7.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and methods for calculating the area, with some confusion present. While one participant claims to have found the correct area, others are still grappling with the problem, indicating that the discussion remains partially unresolved.

Contextual Notes

Some assumptions regarding the triangle's properties and the relevance of its position on the grid are not fully explored. The discussion also reflects uncertainty about the application of the area formula.

funnijen
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Hello, and welcome to MHB! (Wave)

$$\triangle QRS$$ is not a right triangle, but I suspect what is intended is for you to find the area of the rectangle, and then subtract away the areas of the 3 right triangles that are within the rectangle, but not part of $$\triangle QRS$$.

Can you proceed?
 
Um does the position on the grid mean anything. i honestly only know 1/2BH formula and its not giving me the right answer. I honestly don't know any part of how to do this
 
I figured it out! 7 is the answer. Thanks!
 
Very good! For those who are interested, MarkFL could see that the central triangle was NOT a right triangle because it does not satisfy the "Pythagorean Theorem". The two legs have length 5 and 2\sqrt{2} while the hypotenuse has length $\sqrt{29}$. $29\ne 25+ 8$.

The bounding 4 by 5 rectangle has area 20. The right triangle at the upper left corner has area 3(4)/2= 6. The right triangle at the lower left has area 2(2)/2= 2. And the right triangle at the right has area 2(5)/2= 5. Those have total area 6+ 2+ 5=13 so the area of the central triangle is 20- 13= 7 as funnigen said.
 

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