Selina Textbook Solutions: Loci | Mathematics Class 10 ICSE PDF Download

 Page 1


Loci (Locus & Its Constructions) 
 
Exercise 16A 
Question 1. 
Given— PQ is perpendicular bisector of side AB of the triangle ABC. 
 
 
 
Prove— Q is equidistant from A and B. 
 
Solution: 
 
 
Construction: Join AQ 
Proof: In ?AQP and ?BQP, 
AP = BP (given) 
 
?QPA = ?QPB (Each = 90 ) 
PQ = PQ (Common) 
 
By Side-Angle-Side criterion of congruence, we have 
?AQP ? ?BQP (SAS postulate) 
 
The corresponding parts of the triangle are congruent 
Page 2


Loci (Locus & Its Constructions) 
 
Exercise 16A 
Question 1. 
Given— PQ is perpendicular bisector of side AB of the triangle ABC. 
 
 
 
Prove— Q is equidistant from A and B. 
 
Solution: 
 
 
Construction: Join AQ 
Proof: In ?AQP and ?BQP, 
AP = BP (given) 
 
?QPA = ?QPB (Each = 90 ) 
PQ = PQ (Common) 
 
By Side-Angle-Side criterion of congruence, we have 
?AQP ? ?BQP (SAS postulate) 
 
The corresponding parts of the triangle are congruent 
? AQ = BQ (CPCT) 
Hence Q is equidistant from A and B. 
Question 2. 
Given— CP is bisector of angle C of ? ABC. 
 
 
Prove— P is equidistant from AC and BC. 
 
Solution: 
 
 
 
Page 3


Loci (Locus & Its Constructions) 
 
Exercise 16A 
Question 1. 
Given— PQ is perpendicular bisector of side AB of the triangle ABC. 
 
 
 
Prove— Q is equidistant from A and B. 
 
Solution: 
 
 
Construction: Join AQ 
Proof: In ?AQP and ?BQP, 
AP = BP (given) 
 
?QPA = ?QPB (Each = 90 ) 
PQ = PQ (Common) 
 
By Side-Angle-Side criterion of congruence, we have 
?AQP ? ?BQP (SAS postulate) 
 
The corresponding parts of the triangle are congruent 
? AQ = BQ (CPCT) 
Hence Q is equidistant from A and B. 
Question 2. 
Given— CP is bisector of angle C of ? ABC. 
 
 
Prove— P is equidistant from AC and BC. 
 
Solution: 
 
 
 
Question 3. 
Given— AX bisects angle BAG and PQ is perpendicular bisector of AC which meets AX 
at point Y. 
 
Prove— 
(i) X is equidistant from AB and AC. 
(ii) Y is equidistant from A and C. 
 
Solution: 
 
Page 4


Loci (Locus & Its Constructions) 
 
Exercise 16A 
Question 1. 
Given— PQ is perpendicular bisector of side AB of the triangle ABC. 
 
 
 
Prove— Q is equidistant from A and B. 
 
Solution: 
 
 
Construction: Join AQ 
Proof: In ?AQP and ?BQP, 
AP = BP (given) 
 
?QPA = ?QPB (Each = 90 ) 
PQ = PQ (Common) 
 
By Side-Angle-Side criterion of congruence, we have 
?AQP ? ?BQP (SAS postulate) 
 
The corresponding parts of the triangle are congruent 
? AQ = BQ (CPCT) 
Hence Q is equidistant from A and B. 
Question 2. 
Given— CP is bisector of angle C of ? ABC. 
 
 
Prove— P is equidistant from AC and BC. 
 
Solution: 
 
 
 
Question 3. 
Given— AX bisects angle BAG and PQ is perpendicular bisector of AC which meets AX 
at point Y. 
 
Prove— 
(i) X is equidistant from AB and AC. 
(ii) Y is equidistant from A and C. 
 
Solution: 
 
 
Question 4. 
Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw 
perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant 
from B and C. 
 
Solution: 
Given: In triangle ABC, AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm 
 
 
 
Steps of Construction: 
i) Draw a line segment BC = 6.3 cm 
ii) With centre B and radius 4.2 cm, draw an arc. 
 
iii) With centre C and radius 5 cm, draw another arc which intersects the first arc at A. 
iv) Join AB and AC. 
?ABC is the required triangle. 
 
v) Again with centre B and C and radius greater than  draw arcs which intersects 
each other at L and M. 
 
vi) Join LM intersecting AC at D and BC at E. 
vii) Join DB. 
Page 5


Loci (Locus & Its Constructions) 
 
Exercise 16A 
Question 1. 
Given— PQ is perpendicular bisector of side AB of the triangle ABC. 
 
 
 
Prove— Q is equidistant from A and B. 
 
Solution: 
 
 
Construction: Join AQ 
Proof: In ?AQP and ?BQP, 
AP = BP (given) 
 
?QPA = ?QPB (Each = 90 ) 
PQ = PQ (Common) 
 
By Side-Angle-Side criterion of congruence, we have 
?AQP ? ?BQP (SAS postulate) 
 
The corresponding parts of the triangle are congruent 
? AQ = BQ (CPCT) 
Hence Q is equidistant from A and B. 
Question 2. 
Given— CP is bisector of angle C of ? ABC. 
 
 
Prove— P is equidistant from AC and BC. 
 
Solution: 
 
 
 
Question 3. 
Given— AX bisects angle BAG and PQ is perpendicular bisector of AC which meets AX 
at point Y. 
 
Prove— 
(i) X is equidistant from AB and AC. 
(ii) Y is equidistant from A and C. 
 
Solution: 
 
 
Question 4. 
Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw 
perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant 
from B and C. 
 
Solution: 
Given: In triangle ABC, AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm 
 
 
 
Steps of Construction: 
i) Draw a line segment BC = 6.3 cm 
ii) With centre B and radius 4.2 cm, draw an arc. 
 
iii) With centre C and radius 5 cm, draw another arc which intersects the first arc at A. 
iv) Join AB and AC. 
?ABC is the required triangle. 
 
v) Again with centre B and C and radius greater than  draw arcs which intersects 
each other at L and M. 
 
vi) Join LM intersecting AC at D and BC at E. 
vii) Join DB. 
 
Hence, D is equidistant from B and C. 
Question 5. 
In each of the given figures: PA = PH and QA = QB.