Selina Concise Class 9 Maths Chapter 9 Triangles Exercise 9A Solutions
EXERCISE – 9A
(1) Which of the following pairs of triangles are congruent? In each case, stat the condition of congruency:
(a) In △ABC and △ DEF , AB = DE, BC = EF and ∠B = ∠ E
(b) In △ABC and △DEF, ∠B = ∠E = 90°, AC = DF and BC = EF
(c) In △ABC and △ QRP, AB = QR, ∠B = ∠R and ∠C = ∠P
(d) △PQR, AB = PQ, AC = PR and BC = QR
(e) In △ABC and △PQR, BC = QR, ∠A = 90°, ∠C = ∠R = 40° and ∠Q = 50°
Solution :
(i) Firstly we have to draw two triangles ABC and △DEF
In △ABC, and △DEF
AB = DE and BC = EF (∵Given)
Also, ∠B = ∠E
Two sides and one angle is equal from given information.
∴ By side angle side (SAS) test of congruency.
∴ △ABC and DEF are congruent to each other.
Therefore, △ABC ≅ △DEF
(ii) Firstly we have to draw two triangles △ABC and DEF
In △ABC and △DEF.
∠B = ∠E = 90°, AC = DF and BC = EF (∵ Given)
AC = DF (∵ Hypotenus)
∴ △ABC ≅ △DEF (∵ by Right angle Hypotenus side (RHS) congruency)
(iii) Firstly we have to draw two triangles △ABC and △QRP
In △ABC and △PQR,
AB = QR (∵ Given)
∠B = ∠R (∵ Given)
∠C = ∠P (∵ Given)
One side and two angles are equal by given information.
∴ by angle angle side (AAS) test of congruency.
∴ △ABC and △QRP are congruent to each other.
Therefore, △ABC ≅ △QRP
(iv) Firstly we have to draw triangles of △ABC and △PQR
In △ABC and △PQR,
AB = PQ (∵ Given)
AC = PQ (∵ Given)
BC = QR (∵ Given)
Three sides are equal from given information.
by side-side-side (SSS) test of congruency.
∴ △ABC and △PQR are congruent to each other
Therefore, △ABC ≅ △PQR.
(v) Firstly we have to draw two triangles of △ABC and △PQR.
In △ABC and △PQR,
BC = QR (∵ Given)
∠A = 90° (∵ Given)
∠Q = 50° = ∠B (∵ Given)
∠C = ∠R = 40° (∵ Given)
Two angles and one side are equal from given information.
∴ by angle side angle (ASA) test of congruency.
∴ △ABC and △PQR are congruent to each other.
Therefore, △ABC ≅ △PQR
(2) The given figure shows a circle with centre O. P is midpoint of chord AB
Show that OP is perpendicular to AB.
Solution :
Given – P is midpoint of AB.
O is centre of given circle.
To Prove : OP ⊥ AB (OP perpendicular AB)
Proof :
In △OPA and △OPB
OA = OB (∵ radii of circle)
OP = OP (Common side)
AP = PB (∵ P is a midpoint)
∴ △OPA ≅ △OPB (∵ by SSS congruency)
Then, ∠OPA = ∠OPB = x (by CPCTC)
Also,
∠OPA + ∠OPB = 180° (by Linear pair angles)
x + x = 180°
2x = 180°
x = 180°/2
x = 90°
Hence, OP perpendicular AB (DP ⊥ AB)
(3) The following figure shows a circle with centre O. If OP is perpendicular to AB, prove that AP = BP
Solution :
Given, OP perpendicular AB (OP ⊥ AB)
To prove: AP = PB
Proof, In △OPA and △OPB,
OP = OP (Common side)
OA = OP (radii of circle)
∠OPA = ∠OAB (each 90°)
∴ △OPA ≅ △OPB (by right angle Hypotenuse side (RHS))
Then, AP = PB (by CPCTC)
(4) In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD prove that :
(i) △ABD and △ECD are congruent
(ii)AB = EC
(iii) AB is parallel to EC
Solution :
Given, D is a mid-point of BC
AD = DE
To prove : (i) △ABD ≅ △ECD
(ii) AB = EC
(iii) AB parallel EC (AB || EC)
Proof, In △ABD and △ECD,
BD = CD (D is a midpoint)
∠BDA = ∠EDC (vertically opposite angle)
AD = DE (∵ Given)
(i) ∴ △ABD ≅ △ECD (By side-angle-side (SAS) congruency)
(ii) AB = EC (by CPCTC)
(iii) ∠BAD = ∠BEC (by CPCTC)
∠ABD = ∠DCE (by CPCTC)
These are alternate interior angle.
Hence, AB parallel EC
AB || EC
(5) A triangle ABC has ∠B = ∠C. Prove that :
(i) The perpendiculars from the mid-point of BC to AB and AC are equal.
(ii) The perpendiculars form B and C to the opposite sides are equal.
Solution :
Given, ∠B = ∠C
To Prove :
(i) PD = QD
Proof : In △DBP and △DCQ
∠DPB = ∠DQC (each 90°)
∠PBD = ∠QCD (∵ Given)
BD = DC (D is a mid-point)
∴ △DBP ≅ △DCQ (by angle-angle-side (AAS) congruency)
PD = QD (by CPCTC)
(ii) In △BPC and △CQB
∠BPC = ∠CQB (each 90°)
∠PBC = ∠QCB (∵ Given)
BC = BC (∵ Common)
∴ △BPC ≅ △CQB (by angle-angle-side (AAS) congruency)
BQ = CP (by CPCTC)
(6) The perpendicular bisectors of the sides of a triangle AB meet at I. Prove that :
IA = IB = IC
Solution :
Given, IP, IQ and IR are perpendicular bisector of side AB, BC and DC respectively
To prove : IA = IB = IC
Proof : In △API and △BPI,
AP = BP
∠API = ∠BPI
PI = PI
∴ △API ≅ △BPI (by side angle side (SAS) congruency)
=> IA = IB ……. (i) (by CPCTC)
In △BQI and △CQI
BQ = CQ (∵ Given)
∠BQI = ∠CQI (each 90°)
IQ = IQ (Common side)
△BQI ≅ △CQI (by side angle side (SAS) congruency)
IB = IC ….. (ii)
IA = IB …..(i)
From equation (i) and (ii)
IA = IB = IC [Hence proved]
(7) A line segment AB is bisected at point P and through point P another line segment PQ, which is perpendicular to AB, is drawn show that : QA = QB
Solution :
Given, PQ is perpendicular bisector of AB
To prove : AQ = QB
Proof: In △APQ and △BPQ
AP = BP (∵ Given)
∠APQ = ∠BPQ (each 90°)
QP = QP (common side)
∴ △APQ ≅ △BPQ (by side angle side (SAS) congruency)
Then, AQ = QB (by CPCTC) [Hence proved]
(8) If AP bisects angle BAC and M is any point of AP, prove that the perpendiculars drawn from M to AB and AC are equal.
Solution :
Given, ∠PAM = ∠QAM
MP perpendicular AB (M ⊥ AB)
MQ perpendicular AC (MQ ⊥ AC)
To prove: MP = MQ
Proof:- In △PAM and △QAM
∠MPA = ∠MQA (each 90°)
∠PAM = ∠QAM (∵ Given)
NM = AM (common side)
∴ △PAM ≅ △QAM (by angle-angle-side (AAS) congruency).
Then, MP = MQ (by CPCTC) [Hence proved]
(9) From the given diagram, in which ABCD is a parallelogram, ABL is AL line segment and E is mid-point of BC.
Prove that:
(i) △DCE ≅ △LBE
(ii) AB = BL
(iii) AL = 2DC
Solution :
Given, ABCD is a parallelogram
E is a mid-point of BC
To prove :
(i) DCE ≅ △LBE
(ii) AB = BL
(iii) AL = 2DC
Proof : In △DCE and △LBE
∠DCE = ∠LBE (Alternate interior angle)
CE = BE (E is a mid-point)
∠CED = ∠BEL (vertically opposite angle)
(i) ∴ △DCE ≅ △LBE (by angle side angle (ASA) congruency)
(ii) DC = BL …… (i) (by CPCTC)
DC = AB …… (ii) (Opposite sides of parallelogram)
From equation (i) and (ii),
AB = BL
(iii) AL = AB + BL
AL = DC + DC (∵From equation (i) and (ii)
AL = 2DC
(10) In the given figure, AB = DB and AC = DC. If ∠ABD = 58°,
∠DBC = (2x – 4) °
∠ACB = y + 15° and
∠DCB = 63°, find the values of x and y.
Solution :
Given, AB = BD
AC = DC
∠ABD = 58°
∠DBC = 2x – 4
∠ACB = y + 15
∠DCB = 63°
find : x and y
Solution :
In △ABC and △DBC
AB = BD (∵ Given)
AC = DC (∵ Given)
BC = BC (common side)
∴ △ABC ≅ △DBC (by side-side-side (SSS) congruency)
=> ∠ABC = ∠DBC …..(i) (by CPCTC)
=> ∠ACB = ∠DCB ……(ii)
Now,
∠ABC + ∠DBC = ∠ABD
(2x – 4) + (2x – 4) = 58°
2x – 4 + 2x – 4 = 58°
4x – 8 = 58°
4x = 58 + 8
4x = 66
x = 66/4
x = 16.5
Now, we have to find the value of y.
Also, ∠ACB = ∠DCB
y + 15 = 63
y = 63 – 15
y = 48°
Here is your solution of Selina Concise Class 9 Maths Chapter 9 Triangles Exercise 9A
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