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:

1

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:

2

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:

3

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 }

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