Chapter 10.3, Problem 6CE

To determine

By considering given figure, find the following relation between

1. Lines XY and RS.
2. Lines KJ and RS.
3. Lines KJ and XY.
4. XYJK is which type of quadrilateral?
5. Is $XG=GJ$?
6. Is $RG=\frac{2}{3}RJ$?

Answer to Problem 6CE

1. $XY=\frac{1}{2}RS$ .
2. $KJ=\frac{1}{2}RS$ .
3. $XY=KJ$ .
4. XYJK is a parallelogram.
5. Centroid of X is G and X is the mid-point of the line RG.
6. $RG=\frac{2}{3}RJ$ .

Explanation of Solution

Given:

RJ and SK are the medians of the triangle RST; X, Y are mid-points of RG and SG.

Calculation:

a.

From the theorem of mid-segment,

X, Y are the mid-points of RG and SG.

Hence, $XY=\frac{1}{2}RS$ .

b.

In triangle RST,

RJ and SK are the mid-points of the triangle RST and K, J are the mid-points of RT and RS.

From the theorem of mid-segment,

  $KJ=\frac{1}{2}RS$ .

c.

From (a) and (b),

  $XY=\frac{1}{2}RS,$ and

  $KJ=\frac{1}{2}RS$ .

Then, $XY=KJ$ .

d.

In quadrilateral XYJK,

XY is parallel to KJ, and $XY=KJ.$ Also, KX is parallel to JY.

Hence, XYJK is a parallelogram.

e.

From the given figure, $XG=GJ$ ,

  $\begin{array}{l}RG=\frac{2}{3}RJ,\text{ and}\\ GJ=\frac{1}{3}RJ.\end{array}$

Now, since the mid-point of RG is x,

  $\begin{array}{l}XG=\frac{1}{2}RG,\\ \Rightarrow XG=\frac{1}{2}\left(\frac{2}{3}RJ\right),\\ \Rightarrow XG=\frac{1}{3}RJ,\\ \text{SO},XG=GJ.\end{array}$

Hence, the centroid of X is G, and X is the mid-point of the line RG.

f.

Since J is the mid-point of the line TS, and from theorem 10.4,

  $RG=\frac{2}{3}RJ$ .