**10th Maths Book Back Question and Answers – Chapter 4 Exercise 4.2:**

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**Samacheer Kalvi 10th Maths Book Back Answers – Ex 4.2 Geometry**

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**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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