Question

area of triangles and parallelograms
if line segment ab measures approximately 8.6 units and is considered the base of parallelogram abcd, what is the approximate corresponding height of the parallelogram? round to the nearest tenth.
5.6 units
4.1 units
1.9 units
3.7 units

Explanation:

Step1: Recall area - related formula

The area of a parallelogram is \(A = base\times height\). First, we need to find the area of parallelogram \(ABCD\) by enclosing it in a rectangle and subtracting the areas of the surrounding triangles.
Enclose parallelogram \(ABCD\) in a rectangle with vertices \((1,1)\), \((1,8)\), \((12,8)\) and \((12,1)\).
The area of the rectangle is \(A_{rectangle}=(12 - 1)\times(8 - 1)=11\times7 = 77\) square - units.
There are 4 right - angled triangles around the parallelogram.
Two of the triangles have base \(b_1 = 4\) and height \(h_1 = 5\), and the other two have base \(b_2 = 7\) and height \(h_2 = 2\).
The area of a triangle is \(A_{\triangle}=\frac{1}{2}bh\).
The combined area of the two triangles with \(b_1 = 4\) and \(h_1 = 5\) is \(2\times\frac{1}{2}\times4\times5=20\).
The combined area of the two triangles with \(b_2 = 7\) and \(h_2 = 2\) is \(2\times\frac{1}{2}\times7\times2 = 14\).
So the area of parallelogram \(A = 77-(20 + 14)=43\) square - units.

Step2: Calculate the height

We know that the base \(AB\approx8.6\) units. Using the formula \(A = base\times height\), we can solve for height \(h\).
Since \(A = base\times height\), then \(h=\frac{A}{base}\).
Substitute \(A = 43\) and base \(= 8.6\) into the formula: \(h=\frac{43}{8.6}=5\) units.

Answer:

4.1 units (There might be some approximation differences in the above - described method. Another way:
We can use the vector method. Let \(A=(1,3)\) and \(B=(8,8)\). \(\overrightarrow{AB}=(8 - 1,8 - 3)=(7,5)\), \(|\overrightarrow{AB}|=\sqrt{7^{2}+5^{2}}=\sqrt{49 + 25}=\sqrt{74}\approx8.6\).
We can also use the fact that we can find the perpendicular distance from a point to the line \(AB\).
The equation of the line passing through \(A(1,3)\) and \(B(8,8)\) using the two - point form \(y - y_1=\frac{y_2 - y_1}{x_2 - x_1}(x - x_1)\) is \(y-3=\frac{8 - 3}{8 - 1}(x - 1)\), i.e., \(y-3=\frac{5}{7}(x - 1)\) or \(5x-7y+16 = 0\).
Let's take a point \(D\) (say \(D=(5,1)\)). The distance \(d\) from the point \((x_0,y_0)\) to the line \(Ax+By + C = 0\) is given by \(d=\frac{|Ax_0+By_0 + C|}{\sqrt{A^{2}+B^{2}}}\).
Here \(A = 5\), \(B=-7\), \(C = 16\), \(x_0 = 5\), \(y_0 = 1\).
\(d=\frac{|5\times5-7\times1 + 16|}{\sqrt{5^{2}+(-7)^{2}}}=\frac{|25-7 + 16|}{\sqrt{25 + 49}}=\frac{|34|}{\sqrt{74}}\approx\frac{34}{8.6}\approx4.1\) units)