Selina Concise Class 9 Maths Chapter 9 Triangles Exercise 9B Solutions
EXERCISE – 9B
(1) On the sides AB and AC of triangles ABC, equilateral triangle ABD and ACE are drawn. prove that:
(i) ∠CAD = ∠BAE
(ii) CD = BE
Solution :
Given, ABD and ACE are equilateral triangles.
To prove :
(i) ∠CAD = ∠BAE
(ii) CD = BE
Proof:
In △CAD and △BAE,
AD = AB (∵ because ABD is equilateral triangle)
AC = AE (because ACE is equilateral triangle)
∠BAD = ∠CAE (each 60°)
Adding on both side by ∠BAC
∠BAD + ∠BAC = ∠CAE + ∠BAC
∠CAD = ∠BAE ……(i)
△CAD ≅ △BAE (by side angle side (SAS) congruency)
(ii) Then, CD = BE (by CPCTC)
(2) In the following diagrams, ABCD is a square and APB is an equilateral triangle.
In case each case,
(i) Prove that : △APD ≅ △BPC
(ii) Find the angles of △DPC
Solution :
Given, ABCD is a square and DPC is an equilateral triangle.
To prove : (i) △APD ≅ △BPC
(ii) Angles of △DPC
Proof:
In △APD and △BPC,
AD = BC (∵ sides of square)
∠DAP = ∠CBP (each 30°)
AP = BP (sides of equilateral triangle)
∴ △APD ≅ △BPC (by side-angle-side (SAS) congruency)
In △APD,
AD = AP
∠ADP = ∠APD = x
In △APD,
∠APD + ∠ADP + ∠DAP = 180°
x + x + 30 = 180°
2x = 180° – 30°
2x = 150°
x = 150°/2
x = 75°
∠ADC = ∠ADP + ∠PDC
90° = 75° + ∠PDC
90° – 75° = ∠PDC
15° = ∠PDC
Similarly,
∠PCD = 15°
In △DPC,
∠DPC + ∠PDC + ∠PCD = 180°
∠DPC + 15° + 15° = 180°
∠DPC + 30° = 180° – 30°
∠DPC = 150°
(ii) Given, ABCD is a square and APB is an equilateral triangle.
To prove : (i) △APD ≅ △BPC
(ii) Find the angles of △DPC
Proof: In △APD and △BPC,
AP = BP (sides of equilateral triangle)
∠DAP = ∠CBP (each 150°)
AD = BC (sides of square)
(i) ∴ △APD ≅ △BPC (by side-angle-side (SAS) congruency)
(ii) In △ADP,
AD = AP
∠ADP = APD
In △ADP,
∠ADP + ∠APD + ∠DAP = 180°
x + x + 150° = 180°
2x + 150° = 180°
2x = 180° – 150°
2x = 30°
x = 30°/2
x = 15°
Now, ∠ADC = ∠ADP + ∠PDC
90° = 15° + ∠PDC
90° – 15° = ∠PDC
75° = ∠PDC
Similarly,
∠PCD = 75°
In △DPC,
∠PDC + ∠PCD + ∠DPC = 180°
75° + 75° + ∠DPC = 180°
150° + ∠DPC =180°
∠DPC = 180° – 150°
∠DPC = 30°
(3) In the figure, given below, triangle ABC is right angled at B. ABP Q and ACRS are squares.
Prove that:
(i) △ACQ and △ASB are congruent
(ii) CQ = BS
Solution :
Given, △ABC is right angled at B.
ABPQ and ACRS are squares
To prove : (i) △ACQ ≅ △ASB
(ii) CQ = BS
Proof:
∠QAB = ∠CAS (each 90°)
Adding on both side by ∠BAC
∠QAB + ∠BAC = ∠CAS + ∠BAC
∠QAC = ∠SAB …… (i)
In △ACQ and △ASB,
AQ = AB (sides of square)
∠QAC = ∠SAB (from equation (i))
AC = AS (sides of square)
(i) ∴ △ACQ ≅ △ASB (by side angle side (SAS) congruency)
(ii) Then,
CQ = BS (by CPCTC)
(4) In a △ABC, BD is the median to the side AC, BD is produced to E such that BD = DE. Prove that: AE is parallel to BC.
Solution :
Given, BD = DE
To prove: AE || BC (AE parallel BC)
Proof : In △ADE and △CDB
AD = CD (because BD is a median)
∠ADE = ∠BDC (vertically opposite angle)
DE = BD (∵ Given)
△ADE ≅ △CDB (by side angle side (SAS) congruency)
Then,
∠QAE = ∠DCB (by CPCTC)
∠QEA = ∠DBC (by CPCTC)
These are alternate interior angles
Hence, AE || BC [Hence Proved]
(5) In the adjoining figures, QX and RX are the bisectors of the angles Q and R respectively of the triangle PQR. If XS ⊥ QR and XT ⊥ PQ ;
Prove than,
(i) △XTQ = △XSQ
(ii) PX bisects angle ∠P
Solution:
Given, QX and RX are bisectors of ∠Q and ∠R respectively.
XS perpendicular QR and XT perpendicular (XS ⊥ QR) and (XT ⊥ PQ)
To prove: (i) △XTQ ≅ △XSQ
(ii) PX bisects ∠P
Proof:
In △XTQ and △XSQ
∠XTQ = ∠XSQ (each 90°)
∠TQX = ∠SQX (because QX bisects ∠Q)
QX = QX (common side)
∴ △XTQ ≅ △XSQ (by angle angle side (AAS) congruency)
Then, XT = XS ……(i) (by CPCTC)
Now, Draw XW ⊥ PR
In △XSR and △XWR,
∠XSR = ∠XWR (each 90°)
∠XRS = ∠XRW (XR bisect ∠R)
XR = XR (common side)\
∴ △XSR ≅ △XWR (by angle angle side (AAS) congruency)
Then, XS = XW ….. (ii)
from equation (i) and (ii),
XT = XW ….. (iii)
(ii) In △TXP and △WXP
∠PTX = ∠PWX (each 90°)
TX = XW (from equation (iii)
PX = PX (Common side)
∴ △TXP ≅ △WXP (by right angle hypotenuse side (RHS) congruency rule)
Then,
∠TPX = ∠WPX
PX bisects ∠P [Hence Proved]
(6) In the parallelogram ABCD, the angles A and C are obtuse. Points X and Y are taken on the diagonal BD such that the angles XAD and YCB are right angles Prove that : XA = YC
Solution :
Given, ABCD is a parallelogram
To prove: XA = YC
proof : In △XAD and △YCB
∠XAD = ∠YCB (each 90°)
∠ADB = ∠CBY (Alternate interior angle)
AD = BC (opposite sides of parallelogram)
∴ △XAD ≅ △XCB (by angle side angle congruency)
Then, XA = YC (by CPCTC) [Hence proved]
(7) ABCD is a parallelogram. The sides AB and AD are produced to E and F respectively such produced to E and F respectively, such that AB = BE and AD = DF. Prove that : △BEC ≅ △DCF
Solution :
Given, ABCD is a parallelogram
AB = BE
AD = DF
To prove : △BEC ≅ △DCF
Proof : AB = BE …..(i)
AB = DC ……(ii)
From (i) and (ii),
BE = DC ……(iii)
Similarly,
AD = DF …..(iv)
AD = BC ….. (v)
From (iv) and (v)
DF = BC …..(vi)
∠FDC = ∠DAB ….. (vii) (corresponding angles)
∠DAB = ∠CBE ….. (viii)
from equation (vii) and (viii),
∠FDC = ∠CBE …..(ix)
In △BEC and △DCF,
BE = DC (from equation (iii))
∠CBE = ∠FDC (from equation (ix))
BC = DF (from equation (vi))
∴ △BEC ≅ △DCF (by side angle side (SAS) congruency) [Hence proved]
(8) In the following figures, the sides AB and BC and the median AD of triangle ABC are equal to the sides PQ and QR and median PS of the triangle PQR. Prove that △ABC and △PQR are congruent
Solution :
Given,
AB = PQ
BC = QR
AD = PS
To prove :
△ ABC ≅ △PQR
proof: BC = QR
2BD = 2QS
BD = QS …..(i)
In △ABD and △PQS,
AB = PQ (∵ Given)
BD = QS (from equation (i))
AD = PS (∵ Given)
∴ △ABD ≅ △PQS (by side-side-side (SSS) congruency)
Then, ∠B = ∠Q (by CPCTC)
In △ABC and △PQR
AB = PQ (∵ Given)
∠B = ∠Q (proved above)
BC = QR (∵ Given)
∴ △ABC ≅ △PQR (by side-angle-side congruency) [Hence Proved]
(9) In the following diagram, AP and BQ are equal and parallel to each other
Prove that : (i) △ADP = △BOQ
(ii) AB and PQ bisect each other
Solution:
Given, AP = BQ
AP parallel BQ (AP || BQ)
To prove : (i) △NOP ≅ △BOQ
(ii) AB and PQ bisect each other.
proof : In △AOP and △BOQ,
∠OPA = ∠OQB (Alternate interior angle)
AP = BQ (∵ Given)
∠OAP = ∠OBQ (Alternate interior angle)
∴ △AOP ≅ △BOQ (by Angle-side-Angle (ASA) congruency)
(ii) AO = OB (by CPCTC)
PO = OQ
Hence, AB and PQ bisect each other.
(10) In the following figure, OA = OC and AB = BC
(i) ∠P = 90°
(ii) △AOD ≅ △COD
(iii) AD = CD
Solution :
Given : OA = OC
and AB = BC
To prove :
(i) ∠AOB = 90°
(ii) △AOD ≅ △COD
(iii) AD = CD
proof : In △BOA and △BOC
AB = BC (∵ Given)
AO = OC (∵ Given)
BO = BO (common side)
∴ △BOA ≅ △BOC (by side-side-side (SSS) congruency)
Then,
∠BOA = ∠BOC = x (by CPCTC)
Also,
∠BOA + ∠BOC = 180° (Linear pair)
x + x = 180°
2x = 180°
x = 180°/2
x = 90°
i.e. ∠AOB = 90°
(ii) In △AOD and △COD,
AO = OC (∵ Given)
∠AOD = ∠COD (each 90°)
OD = OD (common side)
∴ △AOD ≅ △COD (by side angle side (SAS) congruency)
(iii) AD = CD (by CPCTC)
Here is your solution of Selina Concise Class 9 Maths Chapter 9 Triangles Exercise 9B
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