**Samacheer Kalvi 10th Maths Book Back Solution:**

Tamil Nadu 10th Maths Book Back Answers **Unit 4 – Geometry** Ex 4.2 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 4.2 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:

**Chapter 4**

**Exercise 4.2 Geometry**

1.In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC

(i) If ADDB=34 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:**

x = 1, −12 ⇒ x = 1

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 = 35 cm and QC = 15 cm, find AD.

**Solution:
**Any line parallel to the parallel sides of a trapezium dives the non-parallel sides proportionally.

∴ By thales theorem, In ΔACD, we have

3.In ΔABC, 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 = 1.8cm.

**Solution:
**

∴ It is satisfied

∴ DE||BC

(ii) AB = 5.6 cm,

AD = 1.4 cm,

AC = 7.2 cm,

AE = 1.8 cm.

If ABAD=ACAE is satisfied then BC || DE

5.61.4=7.21.8

5.6 × 1.8 = 1.4 × 7.2

10.08 = 10.08

L.H.S = R.H.S

∴ It is satisfied

∴ DE||BC

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

**Solution:**

In the figure PQ || BC, PR ||CD.

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:**

In ∆CRQ and ∆CBA

∠CRQ = ∠CBA (as RQ || AB)

∠CQR = ∠CAB (as RQ || AB)

⇒ 72 – 12a = 6a

⇒ 18a = 72

a = 4

Side of rhombus PQ, RB = 4 cm, 4 cm.

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 AEED=BFFC

**Solution:**

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

**Solution:
**

8.In a ∆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:**

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:**

AB = 5 cm,

AC = 10 cm,

BD = 1.5 cm,

CD = 3.5 cm,

10.In figure ∠QPC = 90°, PS is its bisector. If ST⊥PR, prove that ST × (PQ + PR) = PQ × PR.

**Solution:**

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:**

By angle bisector theorem in ∆ABC,

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

**Solution:**

Construction:

Step (1) Draw a line segment PQ = 4.5 cm

Step (2) At P, draw PE such that ∠QPE = 35°.

Step (3) At P, draw PF such that ∠EPF = 90°.

Step (4) Draw ⊥^{r} bisector to PQ which intersects PF at O.

Step (5) With O centre OP as raidus draw a circle.

Step (6) From G mark arcs of 6 cm on the circle.

Mark them as R and S.

Step (7) Join PR and RQ.

Step (8) PQR is the required triangle.

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.

**Solution:**

Construction:

Step (1) Draw a line segment QR = 5 cm.

Step (2) At Q, draw QE such that ∠RQE = 40°.

Step (3) At Q, draw QF such that ∠EQF = 90°.

Step (4) Draw perpendicular bisector to QR, which intersects QF at O.

Step (5) With O as centre and OQ as raidus, draw a circle.

Step (6) From G mark arcs of radius 4.4 cm on the circle. Mark them as P and P’.

Step (7) Join PQ and PR.

Step (8) PQR is the required triangle.

Step(9) From P draw a line PN which is ⊥^{r} to LR. LR meets PN at M.

Step (10) The length of the altitude is PM = 2.2 cm.

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.

**Solution:**

Construction:

Steps (1) Draw QR = 6.5 cm.

Steps (2) Draw ∠RQE = 60°.

Steps (3) Draw ∠FQE = 90°.

Steps (4) Draw ⊥^{r} bisector to QR.

Steps (5) The ⊥^{r} bisector meets QF at O.

Steps (6) Draw a circle with O as centre and OQ as raidus.

Steps (7) Mark an arc of 4.5 cm from G on the ⊥^{r} bisector. Such that it meets LM at N.

Steps (8) Draw PP’ || QR through N.

Steps (9) It meets the circle at P, P’.

Steps (10) Join PQ and PR.

Steps (11) ∆PQR is the required triangle.

15.Construct a ∆ABC such that AB = 5.5 cm, C = 25° and the altitude from C to AB is 4 cm.

**Solution:
**Construction:

Step (1) Draw AB¯¯¯¯¯¯¯ = 5.5 cm

Step (2) Draw ∠BAE = 25°

Step (3) Draw ∠FAE = 90°

Step (4) Draw ⊥

^{r}bisector to AB.

Step (5) The ⊥

^{r}bisector meets AF at O.

Step (6) Draw a circle with O as centre and OA as radius.

Step (7) Mark an arc of length 4 cm from G on the ⊥

^{r}bisector and name as N.

Step (8) Draw CC

^{1}|| AB through N.

Step (9) Join AC & BC.

Step (10) ∆ABC is the required triangle.

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.

**Solution:**

Construction:

Steps (1) Draw a line segment BC = 5.6 cm.

Steps (2) At B, draw BE such that ∠CBE = 60°.

Steps (3) At B draw BF such that ∠EBF = 90°.

Steps (4) Draw ⊥^{r} bisector to BC, which intersects BF at 0.

Steps (5) With O as centre and OB as radius draw a circle.

Steps (6) From C, mark an arc of 4 cm on BC at D.

Steps (7) The ⊥^{r} bisector intersects the circle at I. Join ID.

Steps (8) ID produced meets the circle at A.

Now join AB and AC. ∆ABC is the required triangle.

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.

**Solution:**

Steps (1) Draw a line segment PQ = 6.8 cm

Steps (2) At P, draw PE such that ∠QPE = 50°.

Steps (3) At P, draw PF such that ∠FPE = 90°.

Step (4) Draw ⊥^{r} bisector to PQ, which intersects PF at 0.

Step (5) With O as centre and OP as radius draw a circle.

Step (6) From P mark an arc of 5.2 cm on PQ at D.

Step (7) The ⊥^{r} bisector intersects the circle at I. Join ID.

Step (8) ID produced meets the circle at R. Now join PR & QR. ∆PQR is the required triangle.

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