Exercise 10.1
Question 1: In figure, the sides BA and CA have been produced such that BA = AD and CA = AE. Prove that segment DE ∥ BC.
Solution:
In ∆ABC and ∆ADE
It is given that,
BA = AD (Given)
CA = AE (Given)
∠BAC = ∠DAE (Vertically opposite angle)
∆ABC ≅ ∆ADE (By SAS congruent criterion)
BC = DE (CPCT)
∠C = ∠E and ∠B = ∠D (alternate interior angle)
BC || ED
Hence Prove.
Question 2: In a PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.
Solution:
Draw a figure based on given instruction,
In △PQR,
It is given that,
PQ = QR and L, M, N are midpoints of the sides PQ, QP and RP.
∠QPR = ∠QRP (two sides of the triangle are equal, so △PQR is an isosceles triangle)
L is the midpoint of PQ and M is midpoints QR.
PL = LQ = QM = MR
Now, consider Δ LPN and Δ MRN,
LP = MR (Proved above)
∠LPN = ∠MRN (∠QPR = ∠LPN and ∠QRP = ∠MRN)
PN = NR (N is midpoint of PR)
Δ LPN ≅ Δ MRN (By SAS congruence criterion)
LN = MN (CPCT)
Hence Proved.
Question 3: In figure, PQRS is a square and SRT is an equilateral triangle. Prove that
(i) PT = QT (ii) ∠ TQR = 15°
Solution:
(i) It is given that, PQRS is a square and SRT is an equilateral triangle.
PQRS is a square
PQ = QR = RS = SP_________(i)
∠SPQ = ∠PQR = ∠QRS = ∠RSP = 90° (Property of a square)
△SRT is an equilateral triangle
SR = RT = TS_________(ii)
∠ TSR = ∠ SRT = ∠ RTS = 60° (Property of equilateral triangle)
From (i) and (ii) we get,
PQ = QR = SP = SR = RT = TS_________(iii)
From figure we get,
∠TSP = ∠TSR + ∠ RSP
∠TSP = 60° + 90°
∠TSP = 150°
∠TRQ = ∠TRS + ∠ SRQ
∠TRQ = 60° + 90°
∠TRQ = 150°
∠TSP = ∠TRQ = 150°__________(iv)
From ΔTSP and ΔTRQ we get,
TS = TR (Proved above)
SP = RQ (Proved above)
∠TSP = ∠TRQ (each 150°)
ΔTSP ≅ ΔTRQ (SAS congruence criterion)
PT = QT (Corresponding parts of congruent triangles)
Hence proved
(ii) In Δ TQR.
QR = TR
ΔTQR is an isosceles triangle.
∠QTR = ∠TQR (angles opposite to equal sides)
∠QTR + ∠ TQR + ∠TRQ = 180° (Sum of angles in a triangle is 180°)
∠TQR + ∠ TQR + ∠TRQ = 180°
2∠TQR + 150° = 180°
2∠TQR = 30°
∠TQR = 15°
Hence proved
Question 4: Prove that the medians of an equilateral triangle are equal.
Solution:
It is given that, △ABC is an equilateral and Let D, E, F are midpoints of BC, CA and AB.
Here, AD, BE and CF are medians of △ABC.
D is midpoint of BC
BD = DC
F is midpoint of AB
CE = EA
E is midpoint of AC
AF = FB
ΔABC is an equilateral triangle
AB = BC = CA (equilateral triangle)
BD = DC = CE = EA = AF = FB
∠ABC = ∠BCA = ∠CAB = 60° (angles in a equilateral triangle are 60°)
In ΔABD and ΔBCE
AB = BC (sides of a equilateral triangle)
BD = CE (sides of a equilateral triangle)
∠ABD = ∠BCE (each 60°)
ΔABD ≅ ΔBCE (By SAS congruence criterion)
AD = BE (Corresponding parts of congruent triangles) __________(1)
In ΔBCE and ΔCAF,
BC = CA (sides of a equilateral triangle)
∠BCE = ∠CAF (each 60°)
CE = AF (Proved above)
ΔBCE ≅ ΔCAF (By SAS congruence criterion)
BE = CF (CPCT) ____________(2)
From equation (1) and (2), we get
AD = BE = CF (Median of equilateral triangle)
The medians of an equilateral triangle are equal.
Hence proved
Question 5: In a Δ ABC, if ∠A = 120° and AB = AC. Find ∠B and ∠C.
Solution:
ΔABC is an isosceles triangle since AB = AC
∠ B = ∠ C (Angles opposite to equal sides are equal)
∠ A + ∠ B + ∠ C = 180° (Sum of angles in a triangle is 180°)
∠ A + ∠ B + ∠ B = 180°
120° + 2∠B = 180°
2∠B = 180° – 120°
2∠B = 60°
∠ B = 30°
Hence, the value of ∠B = ∠C = 30°.
Question 6: In a Δ ABC, if AB = AC and ∠ B = 70°, find ∠ A.
Solution:
It is given that, in ΔABC, AB = AC and ∠B = 70°
∠ B = ∠ C (Angles opposite to equal sides are equal)
∠ B = ∠ C = 70°
∠ A + ∠ B + ∠ C = 180° (Sum of angles in a triangle is 180°)
∠ A + 70° + 70° = 180°
∠ A = 180° – 140°
∠ A = 40°
Hence, the value of ∠A is 40°.
Question 7: The vertical angle of an isosceles triangle is 100°. Find its base angles.
Solution:
It is given that ΔABC is an isosceles triangle.
AB = AC
Given that vertical angle A is 100°
∠B = ∠C (Angles opposite to equal sides are equal)
∠A + ∠B + ∠C = 180° (Sum of interior angles of a triangle is 180°)
100° + ∠B + ∠B = 180°
2∠B = 180° - 100°
∠B = (80°)/2
∠B = 40°
Hence, the value of base angles ∠B and ∠C is 40°.
Question 8: In a figure AB = AC and ∠ACD = 105°. Find ∠BAC.
Solution:
It is given that,
AB = AC and ∠ACD = 105°
BCD is a Straight line.
∠BCA + ∠ACD = 180°
∠BCA + 105° = 180°
∠BCA = 180° - 105°
∠BCA = 75°
ΔABC is an isosceles triangle.
AB = AC
∠ABC = ∠ ACB (Angles opposite to equal sides are equal)
∠ACB = 75°
∠ABC = ∠ACB = 75°
∠ABC + ∠BCA + ∠CAB = 180° (Sum of Interior angles of a triangle is 180°)
75° + 75° + ∠CAB =180°
150° + ∠BAC = 180°
∠BAC = 180° - 150°
∠BAC = 30°
Hence, the value of ∠BAC is 30°.
Question: 9 Find the measure of each exterior angle of an equilateral triangle.
Solution:
It is given that, ∆ABC equilateral triangle.
AB = BC = CA
∠ABC = ∠BCA = CAB
∠ABC = ∠BCA = CAB = 60° (all angles in a equilateral triangle are equal)
Extend side BC to D,
CA to E
AB to F
BCD is a straight line.
∠BCA + ∠ACD = 180° (Angles made in on it are 180°)
60° + ∠ACD = 180°
∠ACD = 120°
The exterior angle of an equilateral triangle is 120°
Hence proved
Question: 10 If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.
Solution:
AB = AC (Sides of a isosceles triangle)
∠B = ∠C (Angles opposite to equal sides are equal)
Exterior angles ∠ACD
∠A + ∠B = ∠ACD _______(1) (Sum of opposite interiors angles)
Exterior angles ∠ACD
∠A + ∠C = ∠ABE _______(2) (Sum of opposite interiors angles)
By equation (1) and (2)
∠A + ∠B = ∠A + ∠C
∠ACD = ∠ABE
Hence proved
Question 11: In Figure AB = AC and DB = DC, find the ratio ∠ABD : ∠ACD.
Solution:
In ∆ABC
AB = AC
∠ABC = ∠ACB (Angles opposite to equal sides are equal)
In ∆DBC
DB = DC
∠DBC = ∠DCB (Angles opposite to equal sides are equal)
∠ABC = ∠ABD + ∠DBC
∠ABC - ∠DBC = ∠ABD
∠ACB = ∠ACD = ∠DCB
∠ACB - ∠DCB = ∠ACD
(∠ABC)/(∠ACD)=(∠ABC-∠DBC)/(∠ACB-∠DCB)
(∠ABC)/(∠ACD)=(∠ABC-∠DBC)/(∠ABC-∠DBC)
(∠ABC)/(∠ACD)=1/1
∠ABC∶ ∠ACD=1:1
Question: 12 Determine the measure of each of the equal angles of a right-angled isosceles triangle.
OR
ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Solution:
ABC is a right angled triangle.
Consider on a right - angled isosceles triangle ABC such that
∠A = 90°
AB = AC
∠C = ∠B _______(i) (Angles opposite to equal sides are equal)
∠A + ∠B + ∠C =180° (Sum of angles in a triangle is 180°)
90° + ∠ B+ ∠ B = 180°
2∠B = 90°
∠B = 45°
∠B = 45°
∠C = 45°
Hence, the angles of a right-angled isosceles triangle is 45°
Question: 13 AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (See Fig. (10.26). Show that the line PQ is perpendicular bisector of AB.
It is given that,
AB is line segment and P, Q are points.
AP = BP _________(i)
AQ = BQ ________(ii)
We have to prove that PQ is perpendicular bisector of AB.
In ΔPAQ and ΔPBQ,
AP = BP (Proved above)
AQ = BQ (Proved above)
PQ - PQ (Common)
Δ PAQ ≅ Δ PBQ (by SSS congruence) ________(iii)
∠APC = ∠BPC (CPCT)
Now, we can observe that ∆APB and ∆ABQ are isosceles triangles. [From (i) and (ii)]
∠PAB = ∠ABQ (Angles opposite to equal sides are equal)
∠QAB = ∠QBA (Angles opposite to equal sides are equal)
In ΔPAC and ΔPBC
AP = BP (Proved above)
∠APC = ∠BPC (Proved above by equation 3)
PC = PC (common side)
ΔPAC ≅ ΔPBC (SAS congruency of triangle)
AC = CB (CPCT)
∠PCA = ∠PBC (CPCT)
ACB is Straight line
∠ACP + ∠ BCP = 180°
∠ACP = ∠PCB
∠ACP = ∠PCB = 90°
AC = CB, C is the midpoint of AB
PC is the perpendicular bisector of AB
Hence, C is a point on the line PQ, PQ is the perpendicular bisector of AB.
Exercise 10.2
Question 1: In figure, it is given that RT = TS, ∠ 1 = 2∠2 and ∠4 = 2∠3. Prove that ΔRBT ≅ ΔSAT.
Solution:
It is given that,
RT = TS
∠1 = 2 ∠2
∠4 = 2∠3
Need to prove: ΔRBT ≅ ΔSAT
RB and SA intersect each other at point O.
∠AOR = ∠BOS (Vertically opposite angles)
∠1 = ∠4
2∠2 = 2∠3 (Given)
∠2 = ∠3
In ΔTRS,
RT = TS
ΔTRS is an isosceles triangle
∠TRS = ∠TSR (Angles opposite to equal sides are equal)
∠TRS = ∠TRB + ∠2_______(1)
∠ TSR = ∠TSA + ∠3_______(2)
From equation (1) and (2) we get
∠TRB + ∠2 = ∠TSA + ∠3
∠TRB = ∠TSA (∠2 = ∠3)
In ΔRBT and ΔSAT
RT = ST (Given)
∠TRB = ∠TSA (Proved above)
∠T = ∠T (common)
Δ RBT ≅ Δ SAT (ASA criterion of congruence)
Hence Prove.
Question 2: Two lines AB and CD intersect at O such that BC is equal and parallel to AD. Prove that the lines AB and CD bisect at O.
Solution:
It is given that, BC ∥ AD and BC = AD
Need to prove: AB and CD bisect at O.
In ΔAOD and ΔBOC
∠OCB =∠ODA (alternate interior angles)
AD = BC (given)
∠OBC = ∠OAD (alternate interior angles)
Δ AOD ≅ Δ BOC (ASA Criterion)
OA = OB (By CPCT)
OD = OC (By CPCT)
AB and CD bisect each other at O.
Hence Proved.
Question 3: BD and CE are bisectors of ∠B and ∠C of an isosceles Δ ABC with AB = AC. Prove that BD = CE.
Solution:
It is given that,
AB = AC
∠ABC = ∠ACB________(i) (Angles opposite to equal sides are equal)
BD is bisector of ∠B
∠ABD = ∠DBC
CE is bisectors of ∠C
∠BCE = ∠ECA
In ΔEBC and ΔDCB we get,
∠EBC = ∠DCB (Proved above)
BC = BC (Common)
∠BCE = ∠CBD (Proved above)
ΔEBC ≅ ΔDCB (By ASA congruence criterion)
CE = BD (CPCT)
BD = CE (CPCT)
Hence proved
Exercise 10.3
Question 1: In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.
Solution:
It is given that, two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other.
In right triangle △ABC and △DEF, we get
∠B = ∠E (each 90°)
AB = DE (give)
∠C = ∠F (give)
ΔABC ≅ ΔDEF (By AAS congruence criterion)
Both the triangles are congruent.
Hence proved
Question 2: If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.
Solution:
Let ABC be a triangle such that AD is the angular bisector of exterior vertical angle, ∠EAC and AD ∥ BC.
From figure,
∠1 = ∠2 (as AD is a bisector of ∠EAC) _______(i)
∠2 = ∠4 (alternative angle) _______(ii)
∠1 = ∠3 (corresponding angles) _______(iii)
From equation (i), (ii) and (iii), we get
∠3 = ∠4
AB = AC
ΔABC is an isosceles triangle.
Question 3: In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.
Solution:
It is given that, ΔABC be isosceles.
AB = AC
∠B = ∠C
Vertex angle A is twice the sum of the base angles B and C.
∠A = 2(∠B + ∠C)
∠A = 2(∠B + ∠B)
∠A = 2(2∠B)
∠A = 4∠B
∠A + ∠B + ∠C = 180° (sum of angles in a triangle is 180°)
4∠B + ∠B + ∠B = 180°
6∠B =180°
∠B = 30°
∠B = ∠C
∠B = ∠C = 30°
∠A = 4∠B
∠A = 4 x 30°
∠A = 120°
Hence, the value of angles of the given triangles are 30° and 30° and 120°.
Question 4: PQR is a triangle in which PQ = PR and is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.
Solution:
It is given that, PQR is a triangle such that PQ = PR and S is any point on the side PQ and ST ∥ QR.
Here,
PQ = PR
So, △PQR is an isosceles triangle.
∠PQR = ∠PRQ
∠PST = ∠PQR (Corresponding angles)
∠PTS = ∠PRQ (Corresponding angles)
As, ∠PQR = ∠PRQ
∠PST = ∠PTS
In Δ PST,
∠ PST = ∠ PTS
Δ PST is an isosceles triangle.
Hence, PS = PT.
Hence proved
Question 1: In figure, It is given that AB = CD and AD = BC. Prove that ΔADC ≅ ΔCBA.
Solution:
It is given that,
AB = CD and AD = BC
In ΔADC and ΔCBA
AB = CD (Given)
BC = AD (Given)
And AC = AC (Common)
ΔADC ≅ ΔCBA (by SSS congruence criterion)
Hence proved
Question 2: In a Δ PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.
Solution:
It is given that, in ΔPQR, PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively.
Join L to M, M to N, N to L
PL = LQ, QM = MR and RN = NP (As L, M and N are mid-points of PQ, QR and RP)
PQ = QR
PL = LQ = QM = MR = PN = LR ________(i) (by mid-point theorem)
MN ∥ PQ
MN = PQ/2
MN = PL = LQ _______(ii)
LN ∥ QR
LN = QR/2
LN = QM = MR _______(iii)
From equation (i), (ii) and (iii), we get
PL = LQ = QM = MR = MN = LN
LN = MN
Hence Proved
Question 1: ABC is a triangle and D is the mid-point of BC. The perpendiculars from D to AB and AC are equal. Prove that the triangle is isosceles.
Solution:
It is given that D is the midpoint of BC.
PD = DQ in a triangle ABC.
In △BDP and △CDQ
PD = QD (Given)
BD = DC (D is mid-point)
∠BPD = ∠CQD (each 90°)
△BDP ≅ △CDQ (By RHS Criterion)
BP = CQ (CPCT) ________(i)
In △APD and △AQD
PD = QD (given)
AD = AD (common)
APD = AQD (each 90°)
△APD ≅ △AQD (By RHS Criterion)
PA = QA (CPCT) ________(ii)
Adding (i) and (ii)
BP + PA = CQ + QA
AB = AC
Hence it is proved that two sides of the triangle are equal, so ABC is an isosceles.
Question 2: ABC is a triangle in which BE and CF are, respectively, the perpendiculars to the sides AC and AB. If BE = CF, prove that Δ ABC is isosceles
Solution:
ABC is a triangle in which BE and CF are perpendicular to the sides AC and AB respectively s.t. BE = CF.
In Δ BCF and Δ CBE,
∠BFC = ∠CEB (each 90°)
BC = CB [Common side]
And CF = BE (Given)
ΔBFC ≅ ΔCEB (By RHS congruence criterion)
∠ FBC = ∠ EBC (CPCT)
∠ABC = ∠ACB (CPCT)
AC = AB (Opposite sides to equal angles are equal in a triangle)
Two sides of triangle ABC are equal.
Hence, ΔABC is isosceles.
Question 3: If perpendiculars from any point within an angle on its arms are congruent. Prove that it lies on the bisector of that angle.
It is given that, an angle ABC and BP is the mid-point of angle B.
Construction: Draw perpendicular PN and PM on the sides BC and BA.
In Δ BPM and Δ BPN
∠ BMP = ∠ BNP (each 90°)
BP = BP (Common)
MP = NP (given)
ΔBPM ≅ ΔBPN (By RHS congruence criterion)
∠MBP = ∠NBP (CPCT)
BP is the angular bisector of ∠ABC.
Hence proved
Question 4: In figure, AD ⊥ CD and CB ⊥ CD. If AQ = BP and DP = CQ, prove that ∠DAQ = ∠CBP.
AD ⊥ CD and CB ⊥ CD.
It is given that, in the figure, AD ⊥ CD and CB ⊥ CD.
AQ = BP
DP = CQ ______(i)
PQ added on both sides of equation (i)
DP + PQ = PQ + QC
DQ = PC ... (i)
In ∆DAQ and ∆CBP
∠ADQ = ∠BCP (each 90°)
AQ = BP (given)
DQ = PC (proved above)
ΔDAQ ≅ ΔCBP ((by RHS congruence criterion)
∠DAQ = ∠CBP (CPCT)
Hence proved
Question 5: ABCD is a square, X and Y are points on sides AD and BC respectively such that AY = BX. Prove that BY = AX and ∠BAY = ∠ABX.
It is given that ABCD is a square, X and Y are points on sides AD and BC respectively and AY = BX.
Join B and X, A and Y.
ABCD is a square,
∠DAB = ∠CBA (each 90°)
∠XAB = ∠YAB (each 90°)
In ∆XAB and ∆YBA
∠XAB = ∠YBA (each 90°)
BX = AY (given)
And AB = BA (Common)
ΔXAB ≅ ΔYBA (by RHS congruence criterion)
BY = AX (CPCT)
∠BAY = ∠ABX (CPCT)
Hence proved
Question 6: Which of the following statements are true (T) and which are false (F):
(i) Sides opposite to equal angles of a triangle may be unequal.
(ii) Angles opposite to equal sides of a triangle are equal
(iii) The measure of each angle of an equilateral triangle is 60
(iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isosceles.
(v) The bisectors of two equal angles of a triangle are equal.
(vi) If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.
(vii) The two altitudes corresponding to two equal sides of a triangle need not be equal.
(viii) If any two sides of a right triangle are respectively equal to two sides of other right triangle, then the two triangles are congruent.
(ix) Two right-angled triangles are congruent if hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle.
Solution:
(i) False
(ii) True
(iii) True
(iv) False
(v) True
(vi) False
(vii) False
(viii) False
(ix) True
Question 7: Fill the blanks.
In the following so that each of the following statements is true.
(i) Sides opposite to equal angles of a triangle are _________
(ii) Angle opposite to equal sides of a triangle are _________
(iii) In an equilateral triangle all angles are _________
(iv) In ΔABC, if ∠A = ∠C, then AB = _________
(v) If altitudes CE and BF of a triangle ABC are equal, then AB _________
(vi) In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is _________ CE.
(vii) In right triangles ABC and DEF, if hypotenuse AB = EF and side AC = DE, then ΔABC ≅ Δ _________
Solution:
(i) Sides opposite to equal angles of a triangle are equal.
(ii) Angles opposite to equal sides of a triangle are equal.
(iii) In an equilateral triangle all angles are equal.
(iv) In a ΔABC, if ∠A = ∠C, then AB = BC.
(v) If altitudes CE and BF of a triangle ABC are equal, then AB = AC
(vi) In an isosceles triangle ΔABC with AB = AC, if BD and CE are its altitudes, then BD is equal to CE
(vii) In right triangles ABC and DEF, if hypotenuse AB = EF and side AC = DE, then, ΔABC = ΔEFD.
Question 1: In Δ ABC, if ∠A = 40° and ∠B = 60°. Determine the longest and shortest sides of the triangle.
Solution:
It is given that, ΔABC, ∠A = 40° and ∠B = 60°
∠A + ∠B + ∠C = 180° (sum of angles in a triangle is 180°)
40° + 60° + ∠C = 180°
∠C = 180° – 100° = 80°
∠C = 80°
40° < 60° < 80°
∠A < ∠B < ∠C
∠C is greater angle and ∠ A is smaller angle.
∠A < ∠B < ∠C (side opposite to greater angle is larger and side opposite to smaller angle is smaller)
BC < AC < AB
Hence we can say that, AB is longest and BC is shortest side.
Question 2: In a Δ ABC, if ∠ B = ∠ C = 45°, which is the longest side?
Solution:
It is given that, ΔABC, ∠B = ∠C = 45°
∠A + ∠B + ∠C = 180° (Sum of angles in a triangle is 180°)
∠A + 90° = 180°
∠A = 180° – 90°
∠A = 90°
∠B = ∠C < ∠A
Hence, BC is the longest side.
Question 3: In Δ ABC, side AB is produced to D so that BD = BC. If ∠ B = 60° and ∠ A = 70°.
Prove that: (i) AD > CD (ii) AD > AC
Solution:
It is given that, Δ ABC, side AB is produced to D so that BD = BC ∠ B = 60°, and ∠ A = 70°
Construction: Join C and D
∠A + ∠B + ∠C = 180° (sum of angles in a triangle is 180°)
70° + 60° + ∠C = 180°
130° + ∠C = 180°
∠C = 180° – 130°
∠C = 50°
∠ACB = 50° ____________(i)
In ΔBDC
∠DBC = 180° – ∠ ABC
∠DBC = 180 – 60°
∠DBC = 120°
DBA is a straight line
BD = BC (given)
∠BCD = ∠BDC (angles opposite to equal sides are equal)
∠DBC + ∠BCD + ∠BDC = 180° (Sum of angles in a triangle is 180°)
120° + ∠ BCD + ∠ BCD = 180°
120° + 2∠BCD = 180°
2∠BCD = 180° – 120°
2∠BCD = 60°
∠BCD = 30°
∠BCD = ∠BDC = 30° ____________(ii)
In Δ ADC
∠DAC = 70° (given)
∠ADC = 30° (Proved above)
∠ACD = ∠ ACB+ ∠ BCD
∠ACD = 50° + 30°
∠ACD = 80°
∠ADC < ∠DAC < ∠ACD
AC < DC < AD (Side opposite to greater angle is longer and smaller angle is smaller)
AD > CD and AD > AC
Hence proved
Question 4: Is it possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?
Solution:
We know that, Sum of 2 sides > third side
Case 1
2 + 3 > 7
2 + 3 ≯ 7
Case 2
2 + 7 > 3
9 > 3
Case 3
3 + 7 > 2
10 > 2
All the 3 case should be true.
Hence, it is not possible to form a triangle.
Question 5: O is any point in the interior of ΔABC. Prove that
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + QB + OC
(iii) OA + OB + OC > (1/2) (AB + BC +CA)
(i) It is given that O is any point in the interior of ΔABC.
We know that, Sum of 2 sides > third side
In Δ ABC
AB + BC > AC
BC + AC > AB
AC + AB > BC
In ΔOBC
OB + OC > BC _______(i)
In ΔOAC
OA + OC > AC _______(ii)
In ΔOAB
OA + OB > AB _______(iii)
In ΔABD, we have
AB + AD > BD
AB + AD > BO + OD _______(iv) (BD = BO + OD)
In ΔODC, we have
OD + DC > OC _______ (v)
(i) By adding equation (iv) and (v)
AB + AD + OD + DC > BO + OD + OC
AB + (AD + DC) > OB + OC
AB + AC > OB + OC
we have
BC + BA > OA + OC ... (vii)
CA+ CB > OA + OB ... (viii)
(ii) Adding equation (vi), (vii) and (viii),
AB + AC + BC + BA + CA + CB > OB + OC + OA + OC + OA + OB
2AB + 2BC + 2CA > 2OA + 2OB + 2OC
2(AB + BC + CA) > 2(OA + OB + OC)
AB + BC + CA > OA + OB + OC
(iii) Adding equations (i), (ii) and (iii)
OB + OC + OA + OC + OA + OB > BC + AC + AB
2OA + 2OB + 2OC > AB + BC + CA
We get = 2(OA + OB + OC) > AB + BC +CA
(OA + OB + OC) > (1/2)(AB + BC +CA)
Question 6: Prove that the perimeter of a triangle is greater than the sum of its altitudes.
Solution:
It is given that, ΔABC in which AD ⊥ BC, BE ⊥ AC and CF ⊥ AB.
Therefore
AD ⊥ BC
AB > AD and AC > AD
AB + AC > 2AD _______(i) (perpendicular line segment is the shortest)
BE ⊥ AC
BA > BE and BC > BE
BA + BC > 2BE _______(ii) (perpendicular line segment is the shortest)
CF ⊥ AB
CA > CF and CB > CF
CA + CB > 2CF _______ (iii) (perpendicular line segment is the shortest)
By Adding equation (i), (ii) and (iii), we get
AB + AC + BA + BC + CA + CB > 2AD + 2BE + 2CF
2AB + 2BC + 2CA > 2(AD + BE + CF)
AB + BC + CA > AD + BE + CF
Hence we can say that the sum of its altitudes are smaller then the perimeter of the triangle.
Question 7: In Fig., prove that:
(i) CD + DA + AB + BC > 2AC
(ii) CD + DA + AB > BC
Solution:
To prove
(i) CD + DA + AB + BC > 2AC
(ii) CD + DA+ AB > BC
From the given figure,
We know that, in a triangle sum of any two sides is greater than the third side
(i) In ΔABC, we have
AB + BC > AC ________(i)
In ΔADC, we have
CD + DA > AC ________ (ii)
By adding (i) and (ii), we get
AB + BC + CD + DA > AC + AC
AB + BC + CD + DA > 2AC
(ii) In Δ ABC, we have,
AB + AC > BC ________ (iii)
And in ΔADC, we have
CD + DA > AC
Add AB on both sides
CD + DA + AB > AC + AB
From equation (iii) and (iv), we get,
CD + DA + AB > AC + AB > BC
CD + DA + AB > BC
Hence proved
Question 8: Which of the following statements are true (T) and which are false (F)?
(i) Sum of the three sides of a triangle is less than the sum of its three altitudes.
(ii) Sum of any two sides of a triangle is greater than twice the median drown to the third side
(iii) Sum of any two sides of a triangle is greater than the third side.
(iv) Difference of any two sides of a triangle is equal to the third side.
(v) If two angles of a triangle are unequal, then the greater angle has the larger side opposite to it
(vi) Of all the line segments that can be drawn from a point to a line not containing it, the perpendicular line segment is the shortest one.
Solution:
(i) False
(ii) True
(iii) True
(iv) False
(v) True
(vi) True
Question 9: Fill in the blanks to make the following statements true.
(i) In a right triangle the hypotenuse is the ___ side.
(ii) The sum of three altitudes of a triangle is ___ than its perimeter.
(iii) The sum of any two sides of a triangle is ___ than the third side.
(iv) If two angles of a triangle are unequal, then the smaller angle has the ___ side opposite to it.
(v) Difference of any two sides of a triangle is ___ than the third side.
(vi) If two sides of a triangle are unequal, then the larger side has ___ angle opposite to it.
Solution:
(i) In a right triangle the hypotenuse is the largest side
(ii) The sum of three altitudes of a triangle is less than its perimeter.
(iii) The sum of any two sides of a triangle is greater than the third side.
(iv) If two angles of a triangle are unequal, then the smaller angle has the smaller side opposite to it.
(v) Difference of any two sides of a triangle is less than the third side.
(vi) If two sides of a triangle are unequal, then the larger side has greater angle opposite to it.
Exercise VSAQs
Question 1: In two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.
Solution:
It is given that, ∆ABC ≅ ∆DEF,
AB = DE and BC = EF
∠A = ∠D
∠B = ∠E
∠C = ∠F
Question 2: In two triangles ABC and DEF, it is given that ∠A = ∠D, ∠B = ∠ E and ∠ C = ∠F. Are the two triangles necessarily congruent?
Solution:
No.
Two triangles are not necessarily congruent, because we know only angle-angle-angle (AAA) criterion. This criterion can produce similar but not congruent triangles.
Question 3: If ABC and DEF are two triangles such that AC = 2.5 cm, BC = 5 cm, C = 75°, DE = 2.5 cm, DF = 5 cm and D = 75°. Are two triangles congruent?
Solution:
Yes
It is given that,
AC = DE (each 2.5cm)
BC = DF (each 5cm
∠D = ∠C (each 75°)
∆ABC ≅ ∆EDF (By SAS theorem triangle)
Question 4: In two triangles ABC and ADC, if AB = AD and BC = CD. Are they congruent?
Solution:
Yes
It is given that,
AB = AD (given)
BC = CD (given)
AC = AC (common)
∆ABC ≅ ∆ADC (By SSS theorem triangle)
Question 5: In triangles ABC and CDE, if AC = CE, BC = CD, ∠A = 60°, ∠C = 30° and ∠D = 90°. Are two triangles congruent?
Solution:
Yes.
AC = CE (Given)
BC = CD (Given)
∠B = ∠D (each 90°)
∆ABC ≅ ∆CDE (By SSA criteria)
Question 6: ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.
Solution:
It is given that AB = AC and BE and CF are two medians.
In △CFB and △BEC
CE = BF (Since, AC = AB = (AC/2) = (AB/2) = CE = BF)
BC = BC (Common)
∠ECB = ∠FBC (Angle opposite to equal sides are equal)
△CFB ≅ △BEC (By SAS theorem)
BE = CF (CPCT)