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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.1

Question 1.

Check whether the which triangles are similar and find the value of x.

Solution:

(i) In ∆ABC and ∆AED

\(\frac { AB }{ AD } \) = \(\frac { AC }{ AE } \)

\(\frac { 8 }{ 3 } \) = \(\frac{11}{\frac{2}{2}}\)

\(\frac { 8 }{ 3 } \) = \(\frac { 11 }{ 4 } \) ⇒ 32 ≠ 33

∴ The two triangles are not similar.

(ii) In ∆ABC and ∆PQC

∠PQC = 70°

∠ABC = ∠PQC = 70°

∠ACB = ∠PCQ (common)

∆ABC ~ ∆PQC

\(\frac { 5 }{ X } \) = \(\frac { 6 }{ 3 } \)

6x = 15

x = \(\frac { 15 }{ 6 } \) = \(\frac { 5 }{ 2 } \)

∴ x = 2.5

∆ ABC and ∆PQC are similar. The value of x = 2.5

Question 2.

A girl looks the reflection of the top of the lamp post on the mirror which is 66 m away from the foot of the lamppost. The girl whose height is 12.5 m is standing 2.5 m away from the mirror. Assuming the mirror is placed on the ground facing the sky and the girl, mirror and the lamppost are in a same line, find the height of the lamp post.

Solution:

Let the height of the tower ED be “x” m. In ∆ABC and ∆EDC.

∠ABC = ∠CED = 90° (vertical Pole)

∠ACB = ∠ECD (Laws of reflection)

∆ ABC ~ ∆DEC

\(\frac { AB }{ DE } \) = \(\frac { BC }{ EC } \)

\(\frac { 1.5 }{ x } \) = \(\frac { 0.4 }{ 87.6 } \)

x = \(\frac{1.5 \times 87.6}{0.4}\) = \(\frac{1.5 \times 876}{4}\)

= 1.5 × 219 = 328.5

The height of the Lamp Post = 328.5 m

Question 3.

A vertical stick of length 6 m casts a shadow 400 cm long on the ground and at the same time a tower casts a shadow 28 m long. Using similarity, find the height of the tower.

Solution:

In ∆ABC and ∆PQR,

∠ABC = ∠PQR = 90° (Vertical Stick)

∠ACB = ∠PRQ (Same time casts shadow)

∆BCA ~ ∆QRP

\(\frac { AB }{ PQ } \) = \(\frac { BC }{ QR } \)

\(\frac { 6 }{ x } \) = \(\frac { 4 }{ 28 } \)

4x = 6 × 28 ⇒ x = \(\frac{6 \times 28}{4}\) = 42

Length of the lamp post = 42m

Question 4.

Two triangles QPR and QSR, right angled at P and S respectively are drawn on the same base QR and on the same side of QR. If PR and SQ intersect at T, prove that PT × TR = ST × TQ.

Solution:

In ∆PQT and ∆STR we have

∠P = ∠S = 90° (Given)

∠PTQ = ∠STR (Vertically opposite angle)

By AA similarity

∆PTQ ~ ∆STR we get

\(\frac { PT }{ ST } \) = \(\frac { TQ }{ TR } \)

PT × TR = ST × TQ

Hence it is proved.

Question 5.

In the adjacent figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE.

Solution:

In ∆ABC and ∆ADE

∠ACB = ∠AED = 90°

∠A = ∠A (common)

∴ ∆ABC ~ ∆ADE (By AA similarity)

\(\frac { BC }{ DE } \) = \(\frac { AB }{ AD } \) = \(\frac { AC }{ AE } \)

\(\frac { 12 }{ DE } \) = \(\frac { 13 }{ 3 } \) = \(\frac { 5 }{ AE } \)

In ∆ABC, AB^{2} = BC^{2} + AC^{2}

= 122 + 52 = 144 + 25 = 169

AB = \(\sqrt { 169 }\) = 13

Consider, \(\frac { 13 }{ 3 } \) = \(\frac { 5 }{ AE } \)

∴ AE = \(\frac{5 \times 3}{13}\) = \(\frac { 15 }{ 13 } \)

AE = \(\frac { 15 }{ 13 } \) and DE = \(\frac { 36 }{ 13 } \)

Consider, \(\frac { 12 }{ DE } \) = \(\frac { 13 }{ 3 } \)

DE = \(\frac{12 \times 3}{13}\) = \(\frac { 36 }{ 13 } \)

Question 6.

In the adjacent figure, ∆ACB ~ ∆APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ.

Solution:

Given ∆ACB ~ ∆APQ

\(\frac { AC }{ AP } \) = \(\frac { BC }{ PQ } \) = \(\frac { AB }{ AQ } \)

\(\frac { AC }{ 2.8 } \) = \(\frac { 8 }{ 4 } \) = \(\frac { 6.5 }{ AQ } \)

Consider \(\frac { AC }{ 2.8 } \) = \(\frac { 8 }{ 4 } \)

4 AC = 8 × 2.8

AC = \(\frac{8 \times 2.8}{4}\) = 5.6 cm

Consider \(\frac { 8 }{ 4 } \) = \(\frac { 6.5 }{ AQ } \)

8 AQ = 4 × 6.5

AQ = \(\frac{4 \times 6.5}{8}\) = 3.25 cm

Length of AC = 5.6 cm; Length of AQ = 3.25 cm

Question 7.

If figure OPRQ is a square and ∠MLN = 90°. Prove that

(i) ∆LOP ~ ∆QMO

(ii) ∆LOP ~ ∆RPN

(iii) ∆QMO ~ ∆RPN

(iv) QR2 = MQ × RN.

Solution:

(i) In ∆LOP and ∆QMO

∠OLP = ∠OQM = 90°

∠LOP = ∠OMQ (Since OQRP is a square OP || MN)

∴ ∆LOP~ ∆QMO (By AA similarity)

(ii) In ∆LOP and ∆RPN

∠OLP = ∠PRN = 90°

∠LPO = ∠PNR (OP || MN) .

∴ ∆LOP ~ ∆RPN (By AA similarity)

(iii) In ∆QMO and ∆RPN

∠MQO = ∠NRP = 90°

∠RPN = ∠QOM (OP || MN)

∴ ∆QMO ~ ∆RPN (By AA similarity)

(iv) We have ∆QMO ~ ∆RPN

\(\frac { MQ }{ PR } \) = \(\frac { QO }{ RN } \)

\(\frac { MQ }{ QR } \) = \(\frac { QR }{ RN } \)

QR^{2} = MQ × RN

Hence it is proved.

Question 8.

If ∆ABC ~ ∆DEF such that area of ∆ABC is 9cm^{2} and the area of ∆DEF is 16 cm^{2} and BC = 2.1 cm. Find the length of EF.

Solution:

Given ∆ABC ~ ∆DEF

\(\frac { 9 }{ 16 } \) = \(\frac{(2.1)^{2}}{\mathrm{E} \mathrm{F}^{2}}\)

(\(\frac { 3 }{ 4 } \))^{2} = (\(\frac { 2.1 }{ EF } \))^{2}

\(\frac { 3 }{ 4 } \) = \(\frac { 2.1 }{ EF } \)

EF = \(\frac{4 \times 2.1}{3}\) = 2.8 cm

Legth of EF = 2.8 cm

Question 9.

Two vertical poles of heights 6 m and 3 m are erected above a horizontal ground AC. Find the value of y.

Solution:

In the ∆PAC and ∆BQC

∠PAC = ∠QBC = 90°

∠C is common

∆PAC ~ QBC

\(\frac { AP }{ BQ } \) = \(\frac { AC }{ BC } \)

\(\frac { 6 }{ y } \) = \(\frac { AC }{ BC } \)

∴ \(\frac { BC }{ AC } \) = \(\frac { y }{ 6 } \) …..(1)

In the ∆ACR and ∆QBC

∠ACR = ∠QBC = 90°

∠A is common

∆ACR ~ ABQ

\(\frac { RC }{ QB } \) = \(\frac { AC }{ AB } \)

\(\frac { 3 }{ y } \) = \(\frac { AC }{ AB } \)

\(\frac { AB }{ AC } \) = \(\frac { y }{ 3 } \) ……..(2)

By adding (1) and (2)

\(\frac { BC }{ AC } \) + \(\frac { AB }{ AC } \) = \(\frac { y }{ 6 } \) + \(\frac { y }{ 3 } \)

1 = \(\frac{3 y+6 y}{18}\)

9y = 18 ⇒ y = \(\frac { 18 }{ 9 } \) = 2

The Value of y = 2m

Question 10.

Construct a triangle similar to a given triangle PQR with its sides equal to \(\frac { 2 }{ 3 } \) of the corresponding sides of the triangle PQR (scale factor \(\frac { 2 }{ 3 } \) ).

Solution:

Given ∆PQR, we are required to construct another triangle whose sides are \(\frac { 2 }{ 3 } \) of the corresponding sides of the ∆PQR

Steps of construction:

(i) Construct a ∆PQR with any measurement.

(ii) Draw a ray QX making an acute angle with QR on the side opposite to the vertex P.

(iii) Locate 3 points Q_{1}, Q_{2} and Q_{3} on QX.

So that QQ_{1} = Q_{1}Q_{2} = Q_{2}Q_{3}

(iv) Join Q_{3} R and draw a line through Q_{2} parallel to Q_{3} R to intersect QR at R’.

(v) Draw a line through R’ parallel to the line RP to intersect QP at P’. Then ∆ P’QR’ is the required triangle.

Question 11.

Construct a triangle similar to a given triangle LMN with its sides equal to \(\frac { 4 }{ 5 } \) of the corresponding sides of the triangle LMN (scale factor \(\frac { 4 }{ 5 } \) ).

Solution:

Given a triangle LMN, we are required to construct another ∆ whose sides are \(\frac { 4 }{ 5 } \) of the corresponding sides of the ∆LMN.

Steps of Construction:

- Construct a ∆LMN with any measurement.
- Draw a ray MX making an acute angle with MN on the side opposite to the vertex L.
- Locate 5 Points Q
_{1}, Q_{2}, Q_{3}, Q_{4}, Q_{5}on MX.

So that MQ_{1}= Q_{1}Q_{2}= Q_{2}Q_{3}= Q_{3}Q_{4}= Q_{4}Q_{5} - Join Q
_{5}N and draw a line through Q_{4}. Parallel to Q_{5}N to intersect MN at N’. - Draw a line through N’ parallel to the line LN to intersect ML at L’.

∴ ∆L’ MN’ is the required triangle.

Question 12.

Construct a triangle similar to a given triangle ABC with its sides equal to \(\frac { 6 }{ 5 } \) of the corresponding sides of the triangle ABC (scale factor \(\frac { 6 }{ 4 } \)).

Solution:

Given triangle ∆ABC, we are required to construct another triangle whose sides are \(\frac { 6 }{ 5 } \) of the corresponding sides of the ∆ABC.

Steps of construction

(i) Construct an ∆ABC with any measurement.

(ii) Draw a ray BX making an acute angle with BC.

(iii) Locate 6 points Q_{1}, Q_{2}, Q_{3}, Q_{4}, Q_{5}, Q_{6} on BX such that

BQ_{1} = Q_{1}Q_{2} = Q_{2}Q_{3} = Q_{3}Q_{4} = Q_{5}Q_{6}

(iv) Join Q_{5} to C and draw a line through Q_{6} parallel to Q_{5} C intersecting the extended line BC at C’.

(v) Draw a line through C’ parallel to AC intersecting the extended line segment AB at A’.

∆A’BC’ is the required triangle.

Question 13.

Construct a triangle similar to a given triangle PQR with its sides equal to \(\frac { 7 }{ 3 } \) of the corresponding sides of the triangle PQR (scale factor \(\frac { 7 }{ 3 } \)).

Solution:

Given triangle ABC, we are required to construct another triangle whose sides are \(\frac { 7 }{ 3 } \) of the corresponding sides of the ∆ABC.

Steps of construction

(i) Construct a ∆PQR with any measurement.

(ii) Draw a ray QX making an acute angle with QR on the side opposite to the vertex P.

(iii) Locate 7 points Q_{1}, Q_{2}, Q_{3}, Q_{4}, Q_{5}, Q_{6}, Q_{7} on QX.

So that

QQ_{1} = Q_{1}Q_{2} = Q_{2}Q_{3} = Q_{3}Q_{4} = Q_{5}Q_{6} = Q_{6}Q_{7}

(iv) Join Q_{3} to R and draw a line through Q_{7} parallel to Q_{3}R intersecting the extended line segment QR at R’.

(v) Draw a line through parallel to RP.

Intersecting the extended line segment QP at P’.

∴ ∆P’QR’ is the required triangle.