# 10th Maths Book Back Geometry Ex 4.1

## Samacheer Kalvi 10th Maths Book Back Solution:

Tamil Nadu 10th Maths Book Back Answers Unit 4 – Geometry Ex 4.1 are provided on this page. Samacheer Kalvi Maths Book Back Solutions/ Guide available for all Units. TN Samacheer Kalvi 10th Maths Book consists of 8 Units and each unit book back solutions given below topics wise with Questions and Answers. The complete Samacheer Kalvi Books Back Answers/Solutions are available on our site.

The Samacheer kalvi 10th Maths solutions are useful to enhance your skills. Candidates who prepared for the Competitive and board exams 10th Maths Book Back Answers in English and Tamil Medium. The 10th Maths Unit 4 Geometry consist of 5 units. Each Unit Book Back Answers provide topic-wise on this page. 10th Maths Book Back Answers are prepared according to the latest syllabus. The 10th Maths Book Back Relations and Functions Ex 4.1 Answers in English.

### 10th Maths Book Back Answers/Solutions:

TN Samacheer Kalvi 10th Maths Chapter 4 Book Back Exercise given below. The 10th Maths Book Back Solutions Guide is uploaded below:

### Exercise 4.1 Geometry

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

Solution:  x = 156 = 2.5

2. A girl looks the reflection of the top of the lamp post on the mirror which is 6.6 m away from the foot of the lamppost. The girl whose height is 1.25 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 the same line, find the height of the lamp post?
Solution:
In the picture ∆MLN, ∆MGRare similar triangles. ∴ Height of the lamp post is 3.3 m.

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 the picture ∆ABC, ∆DEC are similar triangles. Height of a tower = 42 m

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 ∆RPQ,
RP2 + PQ2 = QR2
∴ PQ2 = QR2 – RP2 ………… (1)
In ∆TPQ,
TP2 + PQ2 = QT2
∴ PQ22 = QT2 – TP2 ………….. (2)
Equating (1) and (2) we get,
QR2 – RP2 = QT2 – TP2
RP = RT + TP
∴ QR2 – (RT + TP)2 = QT2 – TP2
∴ QR2 – RT2 – TP2 – 2RT.TP = QT2 – TP2
QR2 = QT2 + RT2 + 2RT.TP …………. (5) In ∆QSR,
QS2 + SR2 = QR2
SR2 = QR2 – SR2 ………..(3)
In ∆TSR,
ST2 + SR2 = TR2
∴ SR2 = TR2 – TS2 ………… (4)
Equating (3) and (4) we get
QR2 – SQ2 = TR2 – TS2
SQ = QT + TS
∴ QR2 – (2T + TS)2 = TR2 – TS2
QR2 – 2T2 – TS2 – 2QT.TS = TR2 – TS2
∴ 2R2 = TR2 + QT2 + 2QT.TS ………… (6)
Now equating (5) – (6), we get
QT2 +RT2 + 2RT. TP = QT2 + RT2 + 2QT.TS
∴ PT.TR = ST.TQ
Hence proved.

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:
∠A is common & ∠C = ∠E = 90°
∴ by similarity  Substituting the values of DE and AE in (1)
we can prove that 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:
∆ACB ~ ∆APQ  7.In 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 & ∆QMO, we have
∠OLP = ∠MQO (each equal to 90°)
and ∠LOP = ∠OMQ (corresponding angles)
∆LOP ~ ∆QMO (by AA criterion of similarity)

(ii) In ∆LOP & ∆PRN, we have
∠PLO = ∠NRP (each equal to 90°)
∠LPO = ∠PNR (corresponding angles)
∆LOP ~ ∆RPN

(iii) In ∆QMO & ∆RPN .
Since ∆LOP ~ ∆QMO and ∆LOP ~ ∆RPN
∠QMO ~ ∆RPN

(iv) We have
∆QMO ~ ∆RPN (using (iii))
MQRP=QORN (∵ PROQ is a square)
QR2 = MQ × RN. [RP = QO, QO = QR]

8.If ∆ABC ~ ∆DEF such that area of ∆ABC is 9 cm2 and the area of ∆DEF is 16 cm2 and BC = 2.1 cm. Find the length of EF.
Solution:
Since the area of two similar triangles is equal to the ratio of the squares of any two corresponding sides 9.Two vertical poles of heights 6 m and 3 m are erected above a horizontal ground AC. Find the value of y. Solution:
∆PAC, ∆QBC are similar 6 triangles ⇒ AC = AB + BC
= 2BC + BC
AC = 3BC
Substituting AC = 3BC in (1), we get
(AC)y = 6BC
3(BC)y = 6(BC)
y = 63 = 2m

10.Construct a triangle similar to a given triangle PQR with its sides equal to 23 of the corresponding sides of the triangle PQR (scale factor 23).
Solution:
Given a triangle PQR, we are required to construct another triangle whose sides are 35 of the corresponding sides of the triangle PQR. Steps of construction:
(1) Draw any ray QX making an acute angle with QR on the side opposite to the vertex P.
(2) Locate 3 (the greater of 2 and 3 in 23 ) points. Q1 Q2, Q2 on QX so that QQ1 = Q1Q2 = Q2Q3
(3) Join Q3R and draw a line through Q2 (the second point, 2 being smaller of 2 and 3 in 23) parallel to Q3R to intersect QR at R’.
(4) Draw line through R’ parallel to the line RP to intersect QP at P’.
The ΔP’QR’ is the required triangle each of the whose sides is 23 of the corresponding sides of 3 ΔPQR.

11. Construct a triangle similar to a given triangle LMN with its sides equal to 45 of the corresponding sides of the triangle LMN
(scale factor 45).
Solution:
Given a triangle LMN, we are required to construct another triangle whose sides are 45 of the corresponding sides of the ΔLMN. Steps of construction:
(1) Draw any ray making an acute angle to the vertex L.
(2) Locate 5 points (greater of 4 and 5 in 45 ) M1, M2, M3, M4, and M5 and MX so that MM1 = M1M2 = M2M3 = M3M4 = M4M5
(3) Join M5N and draw a line parallel to M5N through M4 (the fourth point, 4 being the smaller of 4 and 5 in 45) to intersect MN atN’.
(4) Draw a line through N1 parallel to the line NL to intersect ML and L’. Then ΔL’MN’ is the required triangle each of the whose sides is 45 of the corresponding sides of ΔLMN.

12.Construct a triangle similar to a given triangle ABC with its sides equal to 65 of the corresponding sides of the triangle ABC (scale factor 64 ).
Solution:
ΔABC is the given triangle. We are required to construct another triangle whose sides are 65 of the corresponding sides of the given triangle ABC
Steps of construction: (1) Draw any ray BX making an acute angle with BC on the opposite side to the vertex A.
(2) Locate 6 points (the greater of 6 and 5 in 65 ) B1, B2, B3, B4, B5, B6 so that BB1 = B1B2 = B2B3 = B4B5 = B5B6.
(3) Join B5 (the fifth point, 5 being smaller of 5 and 6 in 65) to C and draw a live through B6 parallel to B5C intersecting the extended line segment BC at C1.
(4) Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’.
Then ΔA’BC’ is the required triangle each of whose sides is 65 of the corresponding sides of the given triangle ABC.

13.Construct a triangle similar to a given triangle PQR with its sides equal to 73 of the corresponding sides of the triangle PQR (scale factor 73).
Solution:
Given a triangle ΔPQR. We have to construct another triangle whose sides are 73 of the corresponding sides of the given ΔPQR. Steps of construction:
(1) Draw any ray QX making an acute angle with QR on the opposite side to the vertex P.
(2) Locate 7 points (the greater of 7 and 3 in 73) Q1, Q2, Q3, Q4, Q5, Q6, and Q7 so that QQ2 = Q1Q2 = Q2Q3 = Q3Q4 = Q4Q5 = Q5Q6
= Q6Q7
(3) Join Q3 to R and draw a line segment through Q7 parallel to Q3R intersecting the extended line segment QR at R’.
(4) Draw a line segment through R’ parallel to PR intersecting the extended line segment QP at P’.
Then ΔP’QR’ is the required triangle each of whose sides is 73 of the corresponding sides of the given triangle.