Q6. - Geometry

Question

Given: $\overline{RJ}$and $\overline{SK}$are medians of $▵RST;$

                       $X$and $Y$are the midpoints of $\overline{RG}$ and $\overline{SG.}$

1. How are $\overline{XY}$ and $\overline{RS}$ related? Why?
2. How are $\overline{KJ}$ and $\overline{RS}$ related? Why?
3. How are $\overline{KJ}$ and $\overline{XY}$ related? Why?
4. What special kind of quadrilateral is $XYJK?$ why?
5. Why does $XG=GJ?$
6. Explain why  $RG=\frac{2}{3}RJ.$

Step-by-Step Solution

Answer

1. $XY=\frac{1}{2}RS.$
2. $XJ=\frac{1}{2}RS.$
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. $RG=\frac{2}{3}RJ.$

Step 1. Given information:

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

Step 2. Concept used:

 We use basic geometric and angle concept.                                  

Step 3. Applying the concept:

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,

$RG=\frac{2}{3}RJ$.