# RD Sharma Solutions Class 10 Constructions Exercise 11.1

### RD Sharma Class 10 Solutions Chapter 11 Ex 11.1 PDF Free Download

#### Exercise 11.1

Q.1: Determine a point which divides a line segment of length 12 cm internally in the ratio of 2:3. Also, justify your construction.

Solution:

Steps of Construction:

1. Draw a line segment AB of 12 cm

2. Through the points A and B draw two parallel line on the opposite side of AB

3. Cut 2 equal parts on AX and 3 equal parts on BY such that $AX_{1}=X_{1}X_{2}$ and $BX_{1}=Y_{1}Y_{2} =Y_{2}Y_{3}$.

4. Join $X_{2}Y_{3}$ which intersects AB at P$∴ \frac{AP}{PB}=\frac{2}{3}$.

Justification:

In $\Delta AX_{2}P$ and $\Delta BY_{3}P$, we have

$\angle APX_{2}=\angle BPY_{3}$ { Because they are vertically opposite angle}

$\angle X_{2}AP=\angle Y_{3}BP$  { Because they are alternate interior angles }

$\Delta AX_{2}P$ $\Delta BY_{3}P$  { Because AA similarity }

∴ $\frac{AP}{BP}=\frac{AX_{2}}{BY_{3}}=\frac{2}{3}$  { Because of C.P.C.T }

Q.2: Divide a line segment of length 9 cm internally in the ratio 4:3. Also, give justification for the construction.

Solution:

Steps of construction:

1. Draw a line segment AB of 9 cm

2. Through the points, A and B, draw two parallel lines AX and BY on the opposite side of AB

3. Cut 4 equal parts on AX and 3 equal parts on BY such that: $AX_{1}=X_{1}X_{2}=X_{2}X_{3}=X_{3}X_{4}$ and $BY_{1}=Y_{1}Y_{2}=Y_{2}Y_{3}$

4. Join $X_{4}Y_{3}$ which intersects AB at P

$∴ \frac{AP}{PB}=\frac{4}{3}$

Justification:

In $\Delta APX_{4}$ and $\Delta BPY_{3}$, we have

$\angle APX_{4}=\angle BPY_{3}$ { Because they are vertically opposite angles }

$\angle PAX_{4} =\angle PBY_{3}$  { Because they are alternate interior angle}

$\Delta APX_{4}$ $\Delta BPY_{3}$  { Because AA similarity }

$∴ \frac{PA}{PB}=\frac{AX_{4}}{BY_{3}}=\frac{4}{3}$  { Because of C.P.C.T }

Q.3: Divide a line segment of length 14 cm internally in the ratio 2:5. Also, give justification for the construction.

Solution:

Steps of construction:

(i) Draw a line segment AB of 14 cm

(ii) Through the points A and B, draw two parallel lines AX and BY on the opposite side of AB

(iii)  Starting from A, Cut 2 equal parts on AX and starting from B, cut 5 equal parts on BY such that:

$AX_{1}=X_{1}X_{2}$ and $BY_{1}=Y_{1}Y_{2}=Y_{2}Y_{3}=Y_{3}Y_{4}=Y_{4}Y_{5}$

(iv) Join $X_{2}Y_{5}$ which intersects AB at P

$∴ \frac{AP}{PB}=\frac{2}{5}$

Justification:

In $\Delta APX_{2}$ and $\Delta BPY_{5}$, we have

$\angle APX_{2}=\angle BPY_{5}$  { Because they are vertically opposite angles }

$\angle PAX_{2} =\angle PBY_{5}$  { Because they are alternate interior angles }

Then, $\Delta APX_{2}$ $\Delta BPY_{5}$ { Because AA similarity }

$∴ \frac{AP}{PB}=\frac{AX_{2}}{BY_{5}}=\frac{2}{5}$ { Because of C.P.C.T }