Triangles – Exercise 6.2

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This page provides comprehensive Triangles – Exercise 6.2. Free NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6-2. Step-by-step explained answers for CBSE Board exams. Download PDF and practice now.

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Basic Proportionality Theorem & Converse

1. In the figures below, MN || QR. Find NR in (i) and PM in (ii).

(i)

(ii)

2. X and Y are points on sides DE and DF respectively of a $\Delta DEF$. State whether XY || EF in each case.

(i) $DX = 4$ cm, $XE = 5$ cm, $DY = 8$ cm, $YF = 10$ cm

(ii) $DX = 3$ cm, $XE = 4$ cm, $DY = 4$ cm, $YF = 5$ cm

(iii) $DE = 10$ cm, $DF = 15$ cm, $DX = 2$ cm, $DY = 3$ cm

Question 1
(i) Since $MN || QR$, by BPT: $\frac{PM}{MQ} = \frac{PN}{NR}$.
$\frac{2}{4} = \frac{3}{NR} \Rightarrow 0.5 = \frac{3}{NR} \Rightarrow NR = 6$ cm.
(ii) $\frac{PM}{5} = \frac{2.4}{6} \Rightarrow PM = 5 \times 0.4 = 2$ cm.

Question 2
(i) $\frac{DX}{XE} = \frac{4}{5} = 0.8$. $\frac{DY}{YF} = \frac{8}{10} = 0.8$. Ratios are equal, so XY || EF.
(ii) $\frac{DX}{XE} = \frac{3}{4} = 0.75$. $\frac{DY}{YF} = \frac{4}{5} = 0.8$. Ratios not equal, so XY is NOT parallel to EF.
(iii) $XE = 10-2=8$. $YF = 15-3=12$. $\frac{DX}{XE} = \frac{2}{8} = 0.25$. $\frac{DY}{YF} = \frac{3}{12} = 0.25$. Ratios equal, so XY || EF.

Applying BPT in Figures

3. In the figure below, if $XY || BC$ and $XZ || DC$, prove that $\frac{AY}{AB} = \frac{AZ}{AD}$.

4. In the figure, $SU || PT$ and $ST || PR$. Prove that $\frac{QU}{UT} = \frac{QT}{TR}$.

5. In the figure, $AB || XP$ and $AC || XQ$. Show that $BC || PQ$.

6. In the figure, A, B and C are points on $OX, OY$ and $OZ$ respectively such that $AB || XY$ and $AC || XZ$. Show that $BC || YZ$.

3. In $\Delta ABC$, $XY || BC \Rightarrow \frac{AY}{AB} = \frac{AX}{AC}$ (BPT).
In $\Delta ADC$, $XZ || DC \Rightarrow \frac{AZ}{AD} = \frac{AX}{AC}$ (BPT).
From both, $\frac{AY}{AB} = \frac{AZ}{AD}$.

4. In $\Delta PQT$, $SU || PT \Rightarrow \frac{QS}{SP} = \frac{QU}{UT}$.
In $\Delta PQR$, $ST || PR \Rightarrow \frac{QS}{SP} = \frac{QT}{TR}$.
Therefore, $\frac{QU}{UT} = \frac{QT}{TR}$.

5. In $\Delta XOP$, $AB || XP \Rightarrow \frac{OA}{AX} = \frac{OB}{BP}$.
In $\Delta XOQ$, $AC || XQ \Rightarrow \frac{OA}{AX} = \frac{OC}{CQ}$.
So, $\frac{OB}{BP} = \frac{OC}{CQ}$. By Converse BPT in $\Delta OPQ$, $BC || PQ$.

6. In $\Delta OXY$, $AB || XY \Rightarrow \frac{OA}{AX} = \frac{OB}{BY}$.
In $\Delta OXZ$, $AC || XZ \Rightarrow \frac{OA}{AX} = \frac{OC}{CZ}$.
Thus, $\frac{OB}{BY} = \frac{OC}{CZ}$. By Converse BPT in $\Delta OYZ$, $BC || YZ$.

Advanced Theorems (Mid-Point & Trapezium)

7. Using Theorem 6.1 (BPT), prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

8. Using Theorem 6.2 (Converse BPT), prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

9. KLMN is a trapezium with $KL || NM$. Diagonals intersect at O. Show that $\frac{KO}{LO} = \frac{MO}{NO}$.

10. The diagonals of a quadrilateral PQRS intersect at O such that $\frac{PO}{QO} = \frac{RO}{SO}$. Show that PQRS is a trapezium.

7. Given D is mid-point of PQ ($PD=DQ$) and $DE || QR$.
By BPT: $\frac{PD}{DQ} = \frac{PE}{ER}$.
Since $PD=DQ$, ratio is 1. So $1 = \frac{PE}{ER} \Rightarrow PE = ER$.
Thus, E bisects PR.

8. Given X, Y are mid-points. $\frac{AX}{XB} = 1$ and $\frac{AY}{YC} = 1$.
Ratios are equal. By Converse BPT, $XY || BC$.

9. Draw $OZ || NM$.
In $\Delta LNM$, $OZ || NM \Rightarrow \frac{LO}{ON} = \frac{LZ}{ZM}$.
Since $KL || NM$, $OZ || KL$. In $\Delta LMK$, $\frac{MO}{OK} = \frac{MZ}{ZL}$.
So $\frac{LO}{ON} = \frac{OK}{MO} \Rightarrow \frac{KO}{LO} = \frac{MO}{NO}$.

10. Given $\frac{PO}{QO} = \frac{RO}{SO} \Rightarrow \frac{PO}{RO} = \frac{QO}{SO}$.
Draw $OE || PQ$. In $\Delta SPQ$, $\frac{SE}{EP} = \frac{SO}{OQ}$.
Substitute given ratio: $\frac{SE}{EP} = \frac{RO}{OP}$.
In $\Delta PSR$, by Converse BPT, $EO || SR$.
Since $OE || PQ$ and $EO || SR$, then $PQ || SR$. Thus, PQRS is a trapezium.

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