Ch 6: Triangles - Exercise 6.2 Practice (New Questions)

Q1: Finding Missing Sides

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

(i) Find NR

Step 1: Apply BPT
Since MN || QR, by Basic Proportionality Theorem:
$\frac{PM}{MQ} = \frac{PN}{NR}$

Step 2: Substitute Values
$\frac{2}{4} = \frac{3}{NR}$

Step 3: Solve
$0.5 = \frac{3}{NR}$
$NR = \frac{3}{0.5} = 6$

NR = 6 cm

(ii) Find PM

Step 1: Apply BPT
$\frac{PM}{MQ} = \frac{PN}{NR}$

Step 2: Substitute Values
$\frac{PM}{5} = \frac{2.4}{6}$

Step 3: Solve
$\frac{2.4}{6} = 0.4$
$PM = 0.4 \times 5 = 2$

PM = 2 cm

Q2: Converse BPT Check

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

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

Calculate Ratios:
$\frac{DX}{XE} = \frac{4}{5} = 0.8$
$\frac{DY}{YF} = \frac{8}{10} = 0.8$

Since ratios are equal:

XY || EF

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

Calculate Ratios:
$\frac{DX}{XE} = \frac{3}{4} = 0.75$
$\frac{DY}{YF} = \frac{4}{5} = 0.8$

Since $0.75 \neq 0.8$:

XY is NOT parallel to EF

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

Find Segments:
$XE = DE - DX = 10 - 2 = 8$ cm
$YF = DF - DY = 15 - 3 = 12$ cm

Calculate Ratios:
$\frac{DX}{XE} = \frac{2}{8} = \frac{1}{4}$
$\frac{DY}{YF} = \frac{3}{12} = \frac{1}{4}$

XY || EF (Ratios are equal)

Q3: Ratio Equality Proof

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

Proof:

1. In $\Delta ABC$, it is given that $XY || BC$.
By Basic Proportionality Theorem (BPT):
$\frac{AY}{AB} = \frac{AX}{AC}$ ... (i)

2. In $\Delta ADC$, it is given that $XZ || DC$.
By BPT:
$\frac{AZ}{AD} = \frac{AX}{AC}$ ... (ii)

3. From (i) and (ii), both LHS are equal to $\frac{AX}{AC}$.
Therefore, $\frac{AY}{AB} = \frac{AZ}{AD}$.

Hence Proved.

Q4: Double Parallel Lines Proof

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

Step 1: In $\Delta PQT$

It is given that $SU || PT$.
By Basic Proportionality Theorem (BPT):
$\frac{QS}{SP} = \frac{QU}{UT}$ ... (i)

Step 2: In $\Delta PQR$

It is given that $ST || PR$.
By Basic Proportionality Theorem (BPT):
$\frac{QS}{SP} = \frac{QT}{TR}$ ... (ii)

Step 3: Conclusion

From equation (i) and (ii), the LHS ($\frac{QS}{SP}$) is the same.
Therefore, the RHS must be equal:
$\frac{QU}{UT} = \frac{QT}{TR}$

Hence Proved.

Q5: Internal Parallel Lines

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

Proof:

1. In $\Delta XOP$, $AB || XP$.
By BPT: $\frac{OA}{AX} = \frac{OB}{BP}$ ... (i)

2. In $\Delta XOQ$, $AC || XQ$.
By BPT: $\frac{OA}{AX} = \frac{OC}{CQ}$ ... (ii)

3. From (i) and (ii):
$\frac{OB}{BP} = \frac{OC}{CQ}$

4. In $\Delta OPQ$, since the sides are divided in the same ratio, by Converse of BPT:
$BC || PQ$.

Hence Proved.

Q6: 3D Prism Analogy

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

Proof:

1. In $\Delta OXY$, given $AB || XY$.
$\frac{OA}{AX} = \frac{OB}{BY}$ ... (i)

2. In $\Delta OXZ$, given $AC || XZ$.
$\frac{OA}{AX} = \frac{OC}{CZ}$ ... (ii)

3. From (i) and (ii):
$\frac{OB}{BY} = \frac{OC}{CZ}$

4. In $\Delta OYZ$, the ratio of sides is equal.
By Converse of BPT: $BC || YZ$.

Hence Proved.

Q7: Mid-Point Theorem (Using BPT)

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.

Given: $\Delta PQR$, D is mid-point of PQ (so $PD = DQ$). Line DE || QR.

To Prove: E is mid-point of PR ($PE = ER$).

Proof:

1. By BPT, since DE || QR:
$\frac{PD}{DQ} = \frac{PE}{ER}$

2. Since D is mid-point, $PD = DQ$, so $\frac{PD}{DQ} = 1$.

3. Therefore: $1 = \frac{PE}{ER} \Rightarrow PE = ER$.

Thus, E bisects PR.

Q8: Converse Mid-Point Theorem

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.

Given: X is mid-point of AB ($AX=XB$), Y is mid-point of AC ($AY=YC$).

Proof:

1. Calculate ratio of LHS:
$\frac{AX}{XB} = 1$ (Since $AX=XB$)

2. Calculate ratio of RHS:
$\frac{AY}{YC} = 1$ (Since $AY=YC$)

3. Therefore, $\frac{AX}{XB} = \frac{AY}{YC}$.

By Converse of BPT, $XY || BC$.

Q9: Trapezium Property

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

Given: $KL || NM$.

Construction: Draw a line $OZ || NM$ (meeting LM at Z).

Proof:

1. In $\Delta LNM$, $OZ || NM$. By BPT:
$\frac{LO}{ON} = \frac{LZ}{ZM}$ ... (i)

2. Since $KL || NM$ and $OZ || NM$, then $OZ || KL$.

3. In $\Delta LMK$, $OZ || KL$. By BPT:
$\frac{MO}{OK} = \frac{MZ}{ZL}$ or $\frac{OK}{MO} = \frac{LZ}{ZM}$ ... (ii)

4. From (i) and (ii): $\frac{LO}{ON} = \frac{OK}{MO}$.

Rearranging: $\frac{KO}{LO} = \frac{MO}{NO}$.

Q10: Converse Trapezium

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

Given: $\frac{PO}{QO} = \frac{RO}{SO} \Rightarrow \frac{PO}{RO} = \frac{QO}{SO}$.

Construction: Draw line $OE || PQ$ meeting PS at E.

Proof:

1. In $\Delta SPQ$, $OE || PQ$. By BPT:
$\frac{SE}{EP} = \frac{SO}{OQ}$ ... (i)

2. Given $\frac{SO}{OQ} = \frac{RO}{OP}$.
So, $\frac{SE}{EP} = \frac{RO}{OP}$ (Substituting in (i)).

3. Now look at $\Delta PSR$. We have $\frac{SE}{EP} = \frac{RO}{OP}$.
By Converse of BPT, $EO || SR$.

4. Since $OE || PQ$ (Construction) and $EO || SR$ (Proved), then $PQ || SR$.

5. Since one pair of opposite sides is parallel, PQRS is a trapezium.