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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.2

Question 1.

In ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC

(i) If \(\frac { AD }{ DB } \) = \(\frac { 3 }{ 4 } \) and AC = 15 cm find AE.

(ii) If AD = 8x – 7 , DB = 5x – 3 , AE = 4x – 3 and EC = 3x – 1, find the value of x.

Solution:

(i) Let AE be x

∴ EC = 15 – x

In ∆ABC we have DE || BC

By Basic proportionality theorem, we have

\(\frac { AD }{ DB } \) = \(\frac { AE }{ EC } \)

\(\frac { 3 }{ 4 } \) = \(\frac { x }{ 15-x } \)

4x = 3 (15 – x)

4x = 45 – 3x

7x = 45 ⇒ x = \(\frac { 45 }{ 7 } \) = 6.43

The value of x = 6.43

(ii) Given AD = 8x – 7; BD = 5x – 3; AE = 4x – 3; EC = 3x – 1

In ∆ABC we have DE || BC

By Basic proportionality theorem

\(\frac { AD }{ DB } \) = \(\frac { AE }{ EC } \)

\(\frac { 8x-7 }{ 5x-3 } \) = \(\frac { 4x-3 }{ 3x-1 } \)

(8x – 7) (3x – 1) = (4x – 3) (5x – 3)

24x^{2} – 8x – 21x + 7 = 20x^{2} – 12x – 15x + 9

24x^{2} – 20x^{2} – 29x + 27x + 7 – 9 = 0

4x^{2} – 2x – 2 = 0

2x^{2} – x – 1 = 0 (Divided by 2)

2x^{2} – 2x + x – 1 = 0

2x(x -1) + 1 (x – 1) = 0

(x – 1) (2x + 1) = 0

x – 1 = 0 or 2x + 1 = 0

x = 1 or 2x = -1 ⇒ x = – \(\frac { 1 }{ 2 } \) (Negative value will be omitted)

The value of x = 1

Question 2.

ABCD is a trapezium in which AB || DC and P,Q are points on AD and BC respectively, such that PQ || DC if PD = 18 cm, BQ

Solution:

Join AC intersecting PQ at S.

Let AP be x

∴ AD = x + 18

In the ∆ABC, QS || AB

By basic proportionality theorem.

\(\frac { AS }{ SC } \) = \(\frac { BQ }{ QC } \)

\(\frac { AS }{ SC } \) = \(\frac { 35 }{ 15 } \) ………(1)

In the ∆ACD; PS || DC

By basic proportionality theorem.

\(\frac { AS }{ SC } \) = \(\frac { AP }{ PD } \)

\(\frac { AS }{ SC } \) = \(\frac { x }{ 18 } \) ………..(2)

From (1) and (2) we get

\(\frac { 35 }{ 15 } \) = \(\frac { x }{ 18 } \)

15x = 35 × 18 ⇒ x = \(\frac{35 \times 18}{15}\) = 42

AD = AP + PD

= 42 + 18 = 60

The value of AD = 60 cm

Question 3.

In ∆ABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE || BC.

(i) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm.

(ii) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm.

Solution:

(i) Here AB = 12 cm; BD =12 – 8 = 4 cm; AE =12 cm; EC = 18 – 12 = 6 cm

∴ \(\frac { AD }{ DB } \) = \(\frac { 8 }{ 4 } \) = 2

\(\frac { AE }{ EC } \) = \(\frac { 12 }{ 6 } \) = 2

\(\frac { AD }{ DB } \) = \(\frac { AE }{ EC } \)

By converse of basic proportionality theorem DE || BC

(ii) Here AB = 5.6 cm; AD = 1.4 cm;

BD = AB – AD

= 5.6 – 1.4 = 4.2

AC = 7.2 cm; AE = 1.8 cm

EC = AC – AE

= 7.2 – 1.8

EC = 5.4 cm

\(\frac { AD }{ DB } \) = \(\frac { 1.4 }{ 4.2 } \) = \(\frac { 1 }{ 3 } \)

\(\frac { AE }{ EC } \) = \(\frac { 1.8 }{ 5.4 } \) = \(\frac { 1 }{ 3 } \)

\(\frac { AE }{ EC } \) = \(\frac { AD }{ DB } \)

By converse of basic proportionality theorem DE || BC

Question 4.

In fig. if PQ || BC and BC and PR || CD prove that

(i) \(\frac { AR }{ AD } \) = \(\frac { AQ }{ AB } \)

(ii) \(\frac { QB }{ AQ } \) = \(\frac { DR }{ AR } \)

Solution:

(i) In ∆ABC, We have PQ || BC

By basic proportionality theorem

\(\frac { AQ }{ AB } \) = \(\frac { AP }{ AC } \) ……(1)

In ∆ACD, We have PR || CD

basic proportionality theorem

\(\frac { AP }{ AC } \) = \(\frac { AR }{ AD } \) ………..(2)

From (1) and (2) we get

\(\frac { AQ }{ AB } \) = \(\frac { AR }{ AD } \) (or) \(\frac { AR }{ AD } \) = \(\frac { AQ }{ AB } \)

(ii) In ∆ABC, PQ || BC (Given)

By basic proportionality theorem

\(\frac { AP }{ PC } \) = \(\frac { AQ }{ QB } \) ………..(1)

In ∆ADC, PR || CD (Given)

By basic proportionality theorem

\(\frac { AP }{ PC } \) = \(\frac { AR }{ RD } \) ………(2)

From (1) and (2) we get

\(\frac { AQ }{ QB } \) = \(\frac { AP }{ RD } \) (or) \(\frac { QB }{ AQ } \) = \(\frac { RD }{ AR } \)

Question 5.

Rhombus PQRB is inscribed in ∆ABC such that ∠B is one of its angle. P, Q and R lie on AB, AC and BC respectively. If AB = 12 cm and BC = 6 cm, find the sides PQ, RB of the rhombus.

Solution:

Let the side of the rhombus be “x”. Since PQRB is a Rhombus PQ || BC

By basic proportionality theorem

\(\frac { AP }{ AB } \) = \(\frac { PQ }{ BC } \) ⇒ \(\frac { 12-x }{ BC } \) = \(\frac { x }{ 6 } \)

12x = 6 (12 – x)

12x = 72 – 6x

12x + 6x = 72

18x = 72 ⇒ x = \(\frac { 72 }{ 18 } \) = 4

Side of a rhombus = 4 cm

PQ = RB = 4 cm

Question 6.

In trapezium ABCD, AB || DC , E and F are points on non-parallel sides AD and BC respectively, such that EF || AB.

Show that = \(\frac { AE }{ ED } \) = \(\frac { BF }{ FC } \)

Solution:

Given: ABCD is a trapezium AB || DC

E and F are the points on the side AD and BC

EF || AB

To Prove: \(\frac { AE }{ ED } \) = \(\frac { BF }{ FC } \)

Construction: Join AC intersecting AC at P

Proof:

In ∆ABC, PF || AB (Given)

By basic proportionality theorem

\(\frac { AP }{ PC } \) = \(\frac { BF }{ FC } \) ………..(1)

In the ∆ACD, PE || CD (Given)

By basic Proportionality theorem

\(\frac { AP }{ PC } \) = \(\frac { AE }{ ED } \) …………..(2)

From (1) and (2) we get

\(\frac { AE }{ ED } \) = \(\frac { BF }{ FC } \)

Question 7.

In figure DE || BC and CD || EE Prove that AD^{2} = AB × AF.

Solution:

Given: In ∆ABC, DE || BC and CD || EF

To Prove: AD^{2} = AB × AF

Proof: In ∆ABC, DE || BC (Given)

By basic proportionality theorem

\(\frac { AB }{ AD } \) = \(\frac { AC }{ AE } \) ……….. (1)

In ∆ADC; FE || DC (Given)

By basic Proportionality theorem

\(\frac { AD }{ AF } \) = \(\frac { AC }{ AE } \) ……..(2)

From (1) and (2) we get

\(\frac { AB }{ AD } \) = \(\frac { AD }{ AF } \)

AD^{2} = AB × AF

Hence it is proved

Question 8.

In ∆ABC, AD is the bisector of ∠A meeting side BC at D, if AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.

Solution:

In ∆AABC AD is the internal bisector of ∠A

Given BC = 6 cm

Let BD = x ∴ DC = 6 – x cm

By Angle bisector theorem

\(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \)

\(\frac { x }{ 6-x } \) = \(\frac { 10 }{ 14 } \)

14x = 60 – 10x

24x = 60

x = \(\frac { 60 }{ 24 } \) = \(\frac { 10 }{ 4 } \) = 2.5

BD = 2.5 cm;

DC = 6 – x ⇒ 2.5 = 3.5 cm

Question 9.

Check whether AD is bisector of ∠A of ∆ABC in each of the following,

(i) AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm.

(ii) AB = 4 cm, AC 6 cm, BD = 1.6 cm and CD = 2.4 cm.

Solution:

(i) In ∆ABC, AB = 5 cm, AC = 10 cm, BD = 1.5 cm, CD = 3.5 cm

\(\frac { BD }{ DC } \) = \(\frac { 1.5 }{ 3.5 } \) = \(\frac { 15 }{ 35 } \) = \(\frac { 3 }{ 7 } \)

\(\frac { AB }{ AC } \) = \(\frac { 5 }{ 10 } \) = \(\frac { 1 }{ 2 } \)

\(\frac { BD }{ DC } \) ≠ \(\frac { AB }{ AC } \)

∴ AD is not a bisector of ∠A.

(ii) In ∆ABC, AB = 4 cm, AC = 6 cm, BD = 1.6 cm, CD = 2.4 cm

\(\frac { BD }{ DC } \) = \(\frac { 1.6 }{ 2.4 } \) = \(\frac { 16 }{ 24 } \) = \(\frac { 2 }{ 3 } \)

\(\frac { AB }{ AC } \) = \(\frac { 4 }{ 6 } \) = \(\frac { 2 }{ 3 } \)

∴ \(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \)

By angle bisector theorem; AD is the internal bisector of ∠A

Question 10.

In figure ∠QPR = 90°, PS is its bisector.

If ST ⊥ PR, prove that ST × (PQ + PR) = PQ × PR.

Solution:

Given: ∠QPR = 90°; PS is the bisector of ∠P. ST ⊥ ∠PR

To prove: ST × (PQ + PR) = PQ × PR

Proof: In ∆ PQR, PS is the bisector of ∠P.

∴ \(\frac { PQ }{ QR } \) = \(\frac { QS }{ SR } \)

Adding (1) on both side

1 + \(\frac { PQ }{ QR } \) = 1 + \(\frac { QS }{ SR } \)

\(\frac { PR+PQ }{ PR } \) = \(\frac { SR+QS }{ SR } \)

\(\frac { PQ+PR }{ PR } \) = \(\frac { QR }{ SR } \) ……….(1)

In ∆ RST And ∆ RQP

∠SRT = ∠QRP = ∠R (Common)

∴ ∠QRP = ∠STR = 90°

(By AA similarity) ∆ RST ~ RQP

\(\frac { SR }{ QR } \) = \(\frac { ST }{ PQ } \)

\(\frac { QR }{ SR } \) = \(\frac { PQ }{ ST } \) ……..(2)

From (1) and (2) we get

\(\frac { PQ+PR }{ PR } \) = \(\frac { PQ }{ ST } \)

ST (PQ + PR) = PQ × PR

Question 11.

ABCD is a quadrilateral in which AB = AD, the bisector of ∠BAC and ∠CAD intersect the sides BC and CD at the points E and F respectively. Prove that EF || BD.

Solution:

ABCD is a quadrilateral. AB = AD.

AE and AF are the internal bisector of ∠BAC and ∠DAC.

To prove: EF || BD.

Construction: Join EF and BD

Proof: In ∆ ABC, AE is the internal bisector of ∠BAC.

By Angle bisector theorem, we have,

∴ \(\frac { AB }{ AC } \) = \(\frac { BE }{ EC } \) ………(1)

In ∆ ADC, AF is the internal bisector of ∠DAC

By Angle bisector theorem, we have,

\(\frac { AD }{ AC } \) = \(\frac { DF }{ FC } \)

∴ \(\frac { AB }{ AC } \) = \(\frac { DF }{ FC } \) (AB = AD given) ………(2)

From (1) and (2), we get,

\(\frac { BE }{ EC } \) = \(\frac { DF }{ FC } \)

Hence in ∆ BCD,

BD || EF (by converse of BPT)

Question 12.

Construct a ∆PQR which the base PQ = 4.5 cm, R = 35° and the median from R to RG is 6 cm.

Answer:

Steps of construction

- Draw a line segment PQ = 4.5 cm
- At P, draw PE such that ∠QPE = 60°
- At P, draw PF such that ∠EPF = 90°
- Draw the perpendicular bisect to PQ, which intersects PF at O and PQ at G.
- With O as centre and OP as radius draw a circle.
- From G mark arcs of radius 5.8 cm on the circle. Mark them at R and S
- Join PR and RQ. PQR is the required triangle.

Question 13.

Construct a ∆PQR in which QR = 5 cm, ∠P = 40° and the median PG from P to QR is 4.4 cm. Find the length of the altitude from P to QR.

Answer:

Steps of construction

- Draw a line segment RQ = 5 cm.
- At R draw RE such that ∠QRE = 40°
- At R, draw RF such that ∠ERF = 90°
- Draw the perpendicular bisector to RQ, which intersects RF at O and RQ at G.
- With O as centre and OP as radius draw a circle.
- From G mark arcs of radius 4.4 cm on the circle. Mark them as P and S.
- Join PR and PQ. Then ∆PQR is the required triangle.
- From P draw a line PN which is perpendicular to RQ it meets at N.
- Measure the altitude PN.

PN = 2.2 cm.

Question 14.

Construct a ∆PQR such that QR = 6.5 cm, ∠P = 60° and the altitude from P to QR is of length 4.5 cm.

Answer:

Steps of construction

- Draw a line segment QR = 6.5 cm.
- At Q draw QE such that ∠RQE = 60°.
- At Q, draw QF such that ∠EQF = 90°.
- Draw the perpendicular of QR which intersects QF at O and QR at G.
- With O as centre and OQ as radius draw a circle.
- X Y intersects QR at G. On X Y, from G mark an arc at M. Such that GM = 4.5 cm.
- Draw AB through M which is parallel to QR.
- AB Meets the circle at P and S.
- join QP and RP.

PQR is the required triangle.

Question 15.

Construct a ∆ABC such that AB = 5.5 cm, ∠C = 25° and the altitude from C to AB is

Answer:

Steps of construction

- Draw a line segment AB = 5.5 cm.
- At A draw AE such that ∠BAE = 25°.
- At A draw AF such that ∠EAF = 90°.
- Draw the perpendicular bisector of AB which intersects AF at O and AB at G.
- With O as centre and OB as radius draw a circle.
- X Y intersects AB at G. On X Y, from G mark an arc at M. Such that GM = 4 cm.
- Through M draw a line parallel to AB intersect the circle at C and D.
- Join AC and BC.

ABC is the required triangle.

Question 16.

Draw a triangle ABC of base BC = 5.6 cm, ∠A = 40° and the bisector of ∠A meets BC at D such that CD = 4 cm.

Answer:

Steps of construction

- Draw a line segment BC = 5.6 cm.
- At B draw BE such that ∠CBE = 40°.
- At B draw BF such that ∠EBF = 90°.
- Draw the perpendicular bisector to BC which intersects BF at O and BC at G.
- With O as centre and OB as radius draw a circle.
- From C mark an arc of 4 cm on CB at D.
- The perpendicular bisector intersects the circle at I. Joint ID.
- ID produced meets the circle at A. Now Join AB and AC.

This ABC is the required triangle.

Question 17.

Draw ∆PQR such that PQ = 6.8 cm, vertical angle is 50° and the bisector of the vertical angle meets the base at D where PD = 5.2 cm

Answer:

Steps of construction

- Draw a line segment PQ = 6.8 cm.
- At P draw PE such that ∠QPE = 50°.
- At P draw PF such that ∠EPF = 90°.
- Draw the perpendicular bisector to PQ which intersects PF at O and PQ at G.
- With O as centre and OP as radius draw a circle.
- From P mark an arc of 5.2 cm on PQ at D.
- The perpendicular bisector intersects the circle at I. Join ID.
- ID produced meets the circle at A. Now Joint PR and QR. This PQR is the required triangle.