Chapter 7.3, Problem 36WE

To determine

To conclude: The given figure.

Explanation of Solution

Given information:

  

In a proportion, first term and last term $a:b=c:d$ are called extremes.

Also,middle terms from proportion are referred to means.

The product of extremes equals to the product of the means is called Means-extremes Property of proportions:

  $\frac{a}{b}=\frac{c}{d}$ is equivalent to $ad=bc$ .

Property of proportions:

(1) If $\frac{a}{b}=\frac{c}{d}$ is equivalent to:

  $\begin{array}{l}ad=bc\dots \dots (1)\\ \frac{a}{c}=\frac{b}{d}\dots \dots (2)\\ \frac{b}{a}=\frac{d}{c}\dots \dots (3)\\ \frac{a+b}{b}=\frac{c+d}{d}\dots \dots (4)\end{array}$

(2) If $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\dots ,$ then $\frac{a+c+e+\dots }{b+d+f+\dots }=\frac{a}{b}=\dots \dots \dots (5)$

Now, consider a figure as shown in the textbook.

There are two quadrilateral PQRS and TNZJ.

Use proportion equation. Check whether corresponding sides are similar or not.

  $\begin{array}{c}\frac{PQ}{TN}=\frac{QR}{NZ}=\frac{RS}{ZJ}=\frac{SP}{JT}\\ \frac{2b}{3b}=\frac{2c}{3c}=\frac{2d}{3d}=\frac{2a}{3a}\\ \frac{2}{3}=\frac{2}{3}=\frac{2}{3}=\frac{2}{3}\end{array}$

Therefore, both the quadrilaterals are similar.

Now, take account two triangles, POS and TOJ.

Use proportion theorem. Check whether corresponding sides are similar or not.

  $\begin{array}{l}\frac{PO}{TO}=\frac{OS}{OJ}=\frac{SP}{JT}\hfill \\ \frac{10}{15}=\frac{10}{15}=\frac{2a}{3a}\hfill \\ \frac{2}{3}=\frac{2}{3}=\frac{2}{3}\hfill \end{array}$

Therefore, both the triangles are similar.

Now, take account two triangles, QOR and NOZ.

Use proportion equation (5). Check whether corresponding sides are similar or not.

  $\begin{array}{cc}\frac{QO}{NO}& =\frac{OR}{OJ}=\frac{RQ}{JN}\\ \frac{2b+10}{3b+15}& =\frac{2d+10}{3d+15}=\frac{2c}{3c}\end{array}$

Now use the proportion equation, ratio of their perimeter is equal to the ratio of corresponding sides for two similar polygons.

Here, there is an addition of $2b+10$ that can be written separately.

Thus,

  $\begin{array}{cc}\frac{QO}{NO}& =\frac{OR}{OJ}=\frac{RQ}{JN}\\ \frac{2b+10}{3b+15}& =\frac{2d+10}{3d+15}=\frac{2c}{3c}\\ \frac{2b}{3b}& =\frac{10}{15}=\frac{2d}{3d}=\frac{10}{15}=\frac{2c}{3c}\\ \frac{2}{3}& =\frac{2}{3}=\frac{2}{3}=\frac{2}{3}=\frac{2}{3}\end{array}$

Therefore, these two triangles are also similar.

Hence, the conclusion is that two similar quadrilateral, PQRS ~TNZJ, two small similar triangles, POS~TOJ and two bigger similar triangles, QOR ~ NOZ.