You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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

10th Maths Exercise 4.2 Samacheer Kalvi Question 1.
In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC
(i) If \(\frac{\mathbf{A D}}{\mathbf{D B}}=\frac{3}{4}\) 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:
Samacheer Kalvi 10th Maths Chapter 4 Geometry Ex 4.2 1
x = 1, \(\frac{-1}{2}\) ⇒ x = 1

Ex 4.2 Class 10 Samacheer 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 = 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
10th Maths Exercise 4.2 Samacheer Kalvi Chapter 4 Geometry

10th Maths Exercise 4.2 Question 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:
Ex 4.2 Class 10 Samacheer Kalvi Chapter 4 Geometry
∴ It is satisfied
∴ DE||BC

(ii) AB = 5.6 cm,
AD = 1.4 cm,
AC = 7.2 cm,
AE = 1.8 cm.
If \(\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}}\) is satisfied then BC || DE
\(\frac{5.6}{1.4}=\frac{7.2}{1.8}\)
5.6 × 1.8 = 1.4 × 7.2
10.08 = 10.08
L.H.S = R.H.S
∴ It is satisfied
∴ DE||BC

Exercise 4.2 Class 10 Samacheer Kalvi Question 4.
In fig. if PQ || BC and PR ||CD prove that
10th Maths Exercise 4.2 Samacheer Kalvi Chapter 4 Geometry
Solution:
In the figure PQ || BC, PR ||CD.
Exercise 4.2 Class 10 Samacheer Kalvi Chapter 4 Geometry

10th Maths Geometry Exercise 4.2 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:
In ∆CRQ and ∆CBA
∠CRQ = ∠CBA (as RQ || AB)
∠CQR = ∠CAB (as RQ || AB)
10th Maths Geometry Exercise 4.2 Samacheer Kalvi Chapter 4
⇒ 72 – 12a = 6a
⇒ 18a = 72
a = 4
Side of rhombus PQ, RB = 4 cm, 4 cm.

10th Maths Geometry Exercise 4.2 12th Sum 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 \(\frac{{A E}}{{E D}}=\frac{{B F}}{{F C}}\)
Solution:
 10th Maths Geometry Exercise 4.2 12th Sum Samacheer Kalvi

10th Geometry Exercise 4.2 Question 7.
In figure DE || BC and CD || EF . Prove that AD2 = AB × AF.
10th Geometry Exercise 4.2 Samacheer Kalvi Chapter 4
Solution:
10th Geometry 4.2 Samacheer Kalvi Chapter 4 Ex 4.2

10th Geometry 4.2 Question 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:
10th Maths Exercise 4.2 13th Sum Samacheer Kalvi Chapter 4 Geometry
10th Maths Geometry Exercise 4.2 11th Sum Samacheer Kalvi Chapter 4

10th Maths Exercise 4.2 13th Sum 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:
AB = 5 cm,
AC = 10 cm,
BD = 1.5 cm,
CD = 3.5 cm,
10th Maths Geometry Exercise 4.2 14th Sum Samacheer Kalvi Chapter 4

10th Maths Geometry Exercise 4.2 11th Sum Question 10.
In figure ∠QPC = 90°, PS is its bisector. If ST⊥PR, prove that ST × (PQ + PR) = PQ × PR.
10th Maths Geometry Exercise 4.2 Solutions Samacheer Kalvi Chapter 4
Solution:
Samacheer Kalvi 10th Maths Practical Geometry Chapter 4 Ex 4.2
10th Geometry Samacheer Kalvi Maths Solutions Chapter 4 Ex 4.2

10th Maths Geometry Exercise 4.2 14th Sum 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:
By angle bisector theorem in ∆ABC,
10th Maths Ex 4.2 Samacheer Kalvi Chapter 4 Geometry

10th Maths Geometry Exercise 4.2 Solutions Question 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.
10th Maths Exercise 4.2 11th Sum Samacheer Kalvi Chapter 4 Geometry

Samacheer Kalvi 10th Maths Practical Geometry 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.
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.
10 Maths Exercise 4.2 Samacheer Kalvi Chapter 4 Geometry

10th Geometry 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.
Solution:
10th Maths Geometry Exercise 4.2 15th Sum Samacheer Kalvi Chapter 4

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.

10th Maths Ex 4.2 Question 15.
Construct a ∆ABC such that AB = 5.5 cm, C = 25° and the altitude from C to AB is 4 cm.
Solution:
10th Maths Exercise 4.2 14th Sum Samacheer Kalvi Chapter 4 Geometry
Construction:
Step (1) Draw \(\overline{\mathrm{AB}}\) = 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 CC1 || AB through N.
Step (9) Join AC & BC.
Step (10) ∆ABC is the required triangle.

10th Maths Exercise 4.2 11th Sum 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.
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°.
10th Maths 4.2 Samacheer Kalvi Solutions Chapter 4 Geometry
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.

10 Maths Exercise 4.2 Question 1 7.
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:
10th Class Maths Chapter 4 Exercise 4.2 Geometry Samacheer Kalvi
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.

Leave a Reply