Finding Gradient: Tips & Techniques

[Yazan975]

View attachment 8410

 

[MarkFL]

What type of plane figure is the given area?

 

[Yazan975]

What type of plane figure is the given area?

It does not specify

 

[MarkFL]

It does not specify

We can see that it is a trapezoid. A formula for the area \(A\) of a trapezoid is:

\(\displaystyle A=\frac{h}{2}(B+b)\)

where:

\(\displaystyle h\) is the height (we see is is 5 units)

\(\displaystyle B\) is the "big base" (this is unknown)

\(\displaystyle b\) is the "little base" (we see this is 2 units)

So, plugging everything we know into the area formula, we obtain:

\(\displaystyle \frac{35}{2}=\frac{5}{2}(B+2)\)

Solve this for \(B\)...what do you get?

 

[Yazan975]

We can see that it is a trapezoid. A formula for the area \(A\) of a trapezoid is:

\(\displaystyle A=\frac{h}{2}(B+b)\)

where:

\(\displaystyle h\) is the height (we see is is 5 units)

\(\displaystyle B\) is the "big base" (this is unknown)

\(\displaystyle b\) is the "little base" (we see this is 2 units)

So, plugging everything we know into the area formula, we obtain:

\(\displaystyle \frac{35}{2}=\frac{5}{2}(B+2)\)

Solve this for \(B\)...what do you get?

Thanks! Big help. I got the answer

 

[MarkFL]

Thanks! Big help. I got the answer

For the benefit of others who may read this thread, I will complete the problem. This will make the thread more useful (hint hint).

I posted:

\(\displaystyle \frac{35}{2}=\frac{5}{2}(B+2)\)

Multiply through by \(\dfrac{2}{5}\):

\(\displaystyle 7=B+2\)

Subtract through by 2 and arrange as:

\(\displaystyle B=5\)

From this, we may conclude that the point \((5,5)\) is on the line, and we also know \((0,2)\) is on the line (the \(y\)-intercept), and so the slope \(m\) of the line is:

\(\displaystyle m=\frac{5-2}{5-0}=\frac{3}{5}\)

Armed with the slope and intercept, we may give the equation of the line as:

\(\displaystyle y=\frac{3}{5}x+2\)