What is the Correct Way to Calculate the Area of a Triangle?

[Yazan975]

View attachment 8430

 

[MarkFL]

Let's go through this step by step...can you find the area of $$\triangle ABC$$ ?

 

[Yazan975]

Let's go through this step by step...can you find the area of $$\triangle ABC$$ ?

3200cm^2

 

[MarkFL]

3200cm^2

That's not correct...how did you arrive at that answer?

 

[Yazan975]

That's not correct...how did you arrive at that answer?

1600cm^2?

For 3200 I multiplied Base and Height but didn't divide by 2

 

[MarkFL]

1600cm^2?

For 3200 I multiplied Base and Height but didn't divide by 2

Okay, good...so what must the area of $$\triangle AED$$ be?

 

[Yazan975]

Okay, good...so what must the area of $$\triangle AED$$ be?

xy/2 cm^2

 

[MarkFL]

xy/2 cm^2

Yes, that's correct in terms of \(x\) and \(y\), but we should also be able to assign a numeric value to its area based on the fact that the areas of the two shaded regions are the same. (Thinking)

This will give us an equation...can you state it?

 

[Yazan975]

Yes, that's correct in terms of \(x\) and \(y\), but we should also be able to assign a numeric value to its area based on the fact that the areas of the two shaded regions are the same. (Thinking)

This will give us an equation...can you state it?

xy/2 cm^2 = 800
xy = 1600

Is that right?

 

[MarkFL]

xy/2 cm^2 = 800
xy = 1600

Is that right?

Excellent! (Yes)

Okay, now the next thing I notice is that within $$\triangle ABC$$ there are two similar triangles, with $$\triangle AED$$ being the smaller of the two. This means we may state:

$$\frac{y}{x}=\frac{40}{x+10}$$

Do you see where this comes from?

What I would do here is solve both equations we now have for \(y\), and equate the two results to get an equation in \(x\)...can you state this equation?

 

[Yazan975]

Excellent! (Yes)

Okay, now the next thing I notice is that within $$\triangle ABC$$ there are two similar triangles, with $$\triangle AED$$ being the smaller of the two. This means we may state:

$$\frac{y}{x}=\frac{40}{x+10}$$

Do you see where this comes from?

What I would do here is solve both equations we now have for \(y\), and equate the two results to get an equation in \(x\)...can you state this equation?

I see where it comes from but problem is I got stuck at 40x = 1600+10y
I couldn't get an equation in terms of x

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I see where it comes from but problem is I got stuck at 40x = 1600+10y
I couldn't get an equation in terms of x

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I see where it comes from but problem is I got stuck at 40x = 1600+10y
I couldn't get an equation in terms of x

 

[MarkFL]

I see where it comes from but problem is I got stuck at 40x = 1600+10y
I couldn't get an equation in terms of x

I would solve the second equation for \(y\) to obtain:

$$y=\frac{40x}{x+10}$$

And solving our first equation for \(y\) we have:

$$y=\frac{1600}{x}$$

Hence:

$$\frac{40x}{x+10}=\frac{1600}{x}$$

This results in a quadratic in \(x\)...can you write the equation in standard quadratic form?

 

[Yazan975]

Yeah, here's the formula:

x^2-40x-400=0(after dividing the equation by 40)

Hence, x=20??

Thing is my answer sheet says y= 30 and x=50

 

[MarkFL]

Yeah, here's the formula:

x^2-40x-400=0(after dividing the equation by 40)

Yes, that equation is correct.

Hence, x=20??

Thing is my answer sheet says y= 30 and x=50

I would use the quadratic formula to solve the equation, and you'll find a different value for \(x\), after discarding the negative root. Can you find this value?

I don't see how \(x\) and \(y\) can differ by 20 though, and the values for \(x\) and \(y\) given by your answer sheet cannot be correct.

 

[Wilmer]

I don't see how \(x\) and \(y\) can differ by 20 though, and the values for \(x\) and \(y\) given by your answer sheet cannot be correct.

Agree. Anyhow, the difference between x and y is not required as a given.
Looks like the math teacher that made that up "eyed" 50 and 30 !