Chapter P.2, Problem 66E

Solution

a.

The area of the triangle $BPQ$ .

  $\frac{7a^2}{16}$

Given:

The four points, $A\left(0,0\right),B\left(0,a\right),C\left(a,a\right),\text{and }D\left(a,0\right)$ . The point P is the midpoint of segment CD and Q is the point that is one-fourth of the way from A to D on the segment AD .

Calculation:

The plot of the points is shown below

  

Note that the distance AQ is one-fourth of AD , so the length of AQ is $\frac{1}{4}\times a=\frac{a}{4}$ . And the length of DQ is $a-\frac{a}{4}=\frac{3a}{4}$ .

Since P is the midpoint of CD , the length of CP and PD is $\frac{a}{2}$ .

The area of the square ABCD is,

  $\begin{array}{c}\text{area of square }ABCD={\left(\text{Side}\right)}^2\\ ={\left(a\right)}^2\\ =a^2\end{array}$

Area of right-triangle BAQ is,

  $\begin{array}{c}\text{area of triangle }BAQ=\frac{1}{2}\times AQ\times AB\\ =\frac{1}{2}\times \frac{a}{4}\times a\\ =\frac{a^2}{8}\end{array}$

Area of right-triangle BCP is,

  $\begin{array}{c}\text{area of triangle }BCP=\frac{1}{2}\times BC\times CP\\ =\frac{1}{2}\times a\times \frac{a}{2}\\ =\frac{a^2}{4}\end{array}$

And, area of right-triangle QDP is,

  $\begin{array}{c}\text{area of triangle }QDP=\frac{1}{2}\times QD\times DP\\ =\frac{1}{2}\times \frac{3a}{4}\times \frac{a}{2}\\ =\frac{3a^2}{16}\end{array}$

Hence the area of the triangle BPQ is,

  $\begin{array}{c}\text{area of triangle }BPQ=\text{Area of }ABCD-\text{Area of }BAQ-\text{Area of }BCP-\text{Area of }QDP\\ =a^2-\frac{a^2}{8}-\frac{a^2}{4}-\frac{3a^2}{16}\\ =\frac{16a^2-2a^2-4a^2-3a^2}{16}\\ =\frac{16a^2-9a^2}{16}\\ =\frac{7a^2}{16}\end{array}$

Thus, the area of the triangle BPQ is $\frac{7a^2}{16}$ .

b.

Compare the area of triangle $BPQ$ with the area of the square ABCD .

  $\text{Area of triangle }BPQ=\frac{7}{16}\times \text{Area of square }ABCD$

Given:

The four points, $A\left(0,0\right),B\left(0,a\right),C\left(a,a\right),\text{and }D\left(a,0\right)$ . The point P is the midpoint of segment CD and Q is the point that is one-fourth of the way from A to D on the segment AD .

Calculation:

Since the area of the square ABCD is $a^2$ and the area of the triangle BPQ is $\frac{7a^2}{16}$ , the area of the triangle can be given as the area of square with relation,

  $\begin{array}{c}\text{Area of triangle }BPQ=\frac{7}{16}{a}^2\\ =\frac{7}{16}\times \text{Area of square }ABCD\end{array}$

Thus, $\text{Area of triangle }BPQ=\frac{7}{16}\times \text{Area of square }ABCD$ .