Gujarat Board Textbook Solutions Class 9 Maths Chapter 7 Triangles
GSEB Solutions Class 9 Maths Chapter 7 Triangles
Ex 7.1
Question 1.
In quadrilateral ABCD, AC = AD, and AB bisect ∠A (see in figure). Show that ΔABC ≅ ΔABD. What can you say about BC and BD?
Solution:
Given: Quadrilateral ABCD in which AB is the bisector of ∠A and AC = AD.
To show:
ΔABC ≅ ΔABD
Proof: In ΔABC and ΔABD,
AC = AD (Given)
∠BAC = ∠BAD
(AB is the bisector of ∠A)
AB = AB (Common)
∴ ΔABC ≅ ΔABD (by SAS Congruency)
Hence BC = BD (by CPCT)
Question 2.
ABCD is a quadrilateral in which AD = BC and
∠DAB = ∠CBA (see figure). Prove that
1. ΔABD ≅ ΔBAC
2. BD = AC
3. ∠ABD = ∠BAC
Solution:
Given: ABCD is a quadrilateral in which
AD = BC and ∠DAB = ∠CBA.
1. In ΔABD and ΔBAC,
AD = BC (given)
∠DAB = ∠CBA (given)
AB = AB (common)
∴ ΔABD ≅ ΔBAC (by SAS congruency)
2. ΔABD ≅ ΔBAC (proved above)
∴ BD = AC (byCPCT)
3. ΔABD ≅ ΔBAC (proved above)
∴ ∠ABD = ∠BAC (by CPCT)
Question 3.
AD and BC are equal perpendiculars to a line segment AB (see fig). Show that CD bisects AB.
Solution:
Given: AD and BC are perpendiculars on AB
such that AD = BC.
To Prove: CD bisects AB.
i.e., OA = OD
Proof: In ΔAOD and ΔBOC,
AD = BC (Given)
∠OAD = ∠OBC (each 900)
AD || BC (If alternate interior angles are equal then lines are parallel)
∴ ∠ADO = ∠OCB (Alternate interior angle)
Hence ΔAOD ≅ ΔBOC (by ASA congruency)
Therefore, OA = OB (by CPCT)
∴ CD bisects AB.
Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see in figure). Show that
ΔABC ≅ ΔCDA
Solution:
Given: l || m and both are intersected by another pair of parallel lines p and q
To Prove: ΔABC ≅ ΔCDA
Proof: Since AB || CD
and AD||BC
Now in ΔABC and ΔADC,
∠1 = ∠2 (Alternate interior angles
∠3 = ∠4 (Alternate interior angles)
and AC = AC (common)
ΔABC ≅ ΔADC (by ASA congruency)
Question 5.
Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see figure). Show that
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
Solution:
Given: l is the bisector of ∠A, i.e., ∠BAP = ∠BAQ.
BP and BQ are perpendiculars from B to arms of ∠A.
(i) In ΔAPB and ΔAQB
∠BPA = ∠BQA (each 90°)
∠BAP = ∠BAQ (l is the angle bisector of ∠A)
and AB = AB (Common)
Hence ΔAPB ≅ ΔAQB (By AAS congruency)
(ii) Since ΔAPB ≅ ΔAQB
∴ BP = BP (By CPCT)
Question 6.
In figure,AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Solution:
Given: In figure AC = AE.
and AB = AD
and ∠BAD = ∠EAC
To show: BC = DE
Proof: In ΔABC and ΔADE,
AB = AD (given)
∠BAD = ∠EAC (given)
Adding ∠DAC on both sides, we get
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠DAE
AC = AE (given)
∴ ΔBAC ≅ ΔDAE (By SAS congruency)
Hence BC = DE (By CPCT)
Question 7.
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see figure). Show that.
(i) ADAP AEBP
(ii) AD = BE
Solution:
Proof: (i) In ΔDAP and ΔEBP,
∠BAD = ∠ABE (given)
or ∠DAP = ∠PBE
AP = PB (P is the mid≅points)
∠EPA = ∠DPB (given)
Adding ∠DPE on both sides, we get
∠EPA + ∠DPE = ∠DPB + ∠DPE
∠DPA = ∠EPB
∴ ΔDAP ≅ ΔEBP (By ASA congruency)
(ii) ΔDAP ≅ ΔEBP
∴ AD = BE (by CPCT)
Question 8.
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that:
1. ∠AMC ≅ ΔBMD
2. ∠DBC is a right angle.
3. ΔDBC ≅ ΔACB
4. CM = 1/2AB
Solution:
Proof:
1. In ΔAMC and ΔBMD,
AM = BM (M is the mid-point of AB)
∠AMC = ∠BMD (Vertically opposite angle)
CM = DM (given)
Hence, ΔAMC ≅ ΔBMD (by SAS congruency)
AC = BD (by CPCT)
2. ΔAMC ≅ ΔBMD (Proved in part i)
∴ ∠ACM = ∠BDM (by CPCT)
(Alternate interior angles)
Hence AC || BD (If alternate interior angles are equal then lines are parallel)
∴ ∠ACB + ∠DBC = 180°
(Sum of consecutive interior angles is equal to 180°
∴ ∠DBC = 180°≅ 90°
⇒ ∠DBC = 90°
Hence ∠DBC is a right angle.
3. Now in ΔACB and ΔDBC,
∠ACB = ∠DBC (each 900)
BC = CB (common)
AC = BD (Proved in part i)
Hence ΔACB ≅ ΔDBC (by SAS congruency)
4. ΔDBC ≅ ΔACB (Proved in part iii)
DC = AB (by CPCT)
⇒ 2CM = AB (DM = CM)
CM = 1/2 AB
Ex 7.2
Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that.
1. OB = OC
2. AO bisects ∠A
Solution:
1. AB = AC (given)
∠B = ∠C
(Angle opposite to equal sides are equal)
∠B = ∠C (Dividing by 2)
(OB and OC angle bisectors of ∠B and ∠C)
⇒ ∠1 = ∠2
Hence OB = OC
(Sides opposite to equal angles are equal)
2. Now in ΔAOB and ΔAOC
AB = AC (given)
∠B = ∠C (Angle opposite to equal sides are equal)
⇒1/2∠B = 1/2∠C (Dividing by 2)
⇒∠3 = ∠4
and OB = OC (Proved above)
Hence ∠AOB ≅ ∠AOC (by SAS congruency)
∴ ∠OAB = ∠OAC (By CPCT)
Hence AO bisects ∠A.
Question 2.
In ΔABC, AD is the perpendicular bisector of BC (see in figure). Show that ΔABC is an isosceles triangle in which AB = AC.
Solution:
Given: ΔABC in which AD ⊥ BC
and BD = CD
To Prove: ΔABC is an isosceles triangle in which
AB = AC
Proof: In ΔADB and ΔADC,
BD = CD (AD is lar bisector of BC)
∠ADB = ∠ADC (each 90°)
AD = AD (Common)
∴ ΔADB ≅ ΔADC (By SAS congruency)
∴ AB = AC (by CPCT)
Therefore ABC is an isosceles triangle in which AB = AC.
Question 3.
ABC is an isosceles triangle in which altitudes BF and CF are drawn to equal sides AC and AB respectively (See figure). Show that these altitudes are equal.
Solution:
Given: ABC is an isosceles triangle in which BE and CF are altitudes on AC and AB respectively.
To Prove:
BE = CF
Proof: In ΔABE and ΔACF,
AB = AC (Equal sides of an isosceles triangle)
∠AEB = ∠AFC (each 90°)
∠BAE = ∠CAF (Common)
∴ ΔAEB ≅ ΔACF (By AAS congruency)
Hence BE = CF (by CPCT)
Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that
(i) ΔABE ≅ ΔACF
(ii) AB = AC (i.e. ABC is an isosceles triangle
Solution:
Given: ΔABC in which BE = CF.
(i) In ΔABE and ΔACF,
∠BAE = ∠CAF (Common)
∠AEB = ∠AFC (Each 900)
BE = CF (given)
∴ ΔABE ≅ ΔACF (By AAS congruency)
(ii) ΔABE ≅ ΔACF (Proved above in part i)
∴ AB = AC (by CPCT)
ΔABC is an isosceles triangle.
Question 5.
ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that ∠ABD = ∠ACD.
Solution:
ABC is an isosceles triangle.
AB = AC (given)
∠ABC = ∠ACB ……..(1) (Angles opposite to equal sides are equal)
Now DBC is an isosceles triangle.
BD = CD (given)
∴ ∠DBC = ∠DCB (Angles opposite to equal sides are equal)
Adding equation (1) and (2), we get
∠ABC + ∠DBC = ∠ACB + ∠DCB
∠ABD = ∠ACD
Question 6.
ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. (see figure). Show that ∠BCD is a right angle.
Or
In the given figure AB = AC and AB = AD, Prove that ∠BCD = 90°.
Solution:
ΔABC is an isosceles triangle.
AB = AC (given)
∴ ∠ABC = ∠ACB
(Angles opposite to equal sides are equal)
In ΔACD,
AB = AD (by construction)
But AB = AC
∴ AD = AC
∠ADC = ∠ACD …….(2)
(Angles opposite to equal sides are equal)
Adding eqn. (1) and (2), we get
∠ABC + ∠ADC = ∠ACB + ∠ACD
∠ABC + ∠ADC = ∠BCD
∠ABC + ∠BDC = ∠BCD ………(3)
(∠ADC = ∠BDC)
Now in ∠ABC,
∠ABC + ∠BDC + ∠BCD = 180° (By Angle Sum Property)
∠BCD + ∠BCD = 180°
[From eqn. (3)]
2 ∠BCD = 180°
∠BCD = 180°/2
∠BCD = 90°
Question 7.
ABC is a right-angled triangle in which ∠A = 90°, and AB = AC. Find ∠B and ∠C.
Solution:
In ΔABC,
AB = AC
∠B = ∠C
(Angles opposite to equal sides are equal)
Now in ∠ABC,
∠A + ∠B + ∠C = 180° (by angle sum property)
90°+ ∠C + ∠C = 180° (∠B = ∠C)
2∠C = 180° – 90°
2∠C = 90°
∠C = 45°
∠B = ∠C = 45°
Question 8.
Show that the angles of an equilateral triangle are 60° each.
Solution:
Given: ABC is an equilateral triangle.
To Show:
∠A = ∠B = ∠C = 60°
Proof: ABC is an equilateral triangle.
∴ AB = AC
∴ ∠C = ∠B …….(1) (Angles opposite to equal sides are equal)
BC = AC
∠A = ∠B ………(2) (Angles opposite to equal sides are equal)
From eqn. (1) and (2), we get
∠A = ∠B = ∠C = x (Say)
Now in ΔABC,
∠A + ∠B + ∠C = 180°
x + x + x = 180°
3x = 180°
x = 60°
Hence ∠A + ∠B = ∠C = 60°
Ex 7.3
Question 1.
ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that
1. ΔABD ≅ ΔACD
2. ΔABP ≅ ΔACP
3. AP bisects ∠A as well as ∠D
4. AP is the perpendicular bisector of BC.
Solution:
1. In ΔABD and ΔACD,
AB = AC (given) (ΔABC is an isosceles triangle)
BD = CD (given) (ΔDBC is an isosceles triangle)
AD = AD (Common)
∴ ΔABD ≅ ΔACD (by SSS congruency)
2. In ΔABP and ΔACP,
AB = AC (given)
∠BAP = ∠CAP (CPCT as AABD ≅ ∠ACD)
AP = AP (common)
∴ ΔABP ≅ ΔACP (by SAS congruency)
3. ΔABP ≅ ΔACP (Proved in part (ii))
BP = CP ………(1) (byCPCT)
and ∠BAP = ∠CAP (by CPCT)
∴ AP bisect ∠A.
In ΔBDP and ΔCDP,
BD = CD (given ΔBDC is an isosceles triangle)
BP = CP (Proved above in part (ii))
DP = DF (Common)
∠BDP = ∠CDP (by SSS congruency)
Hence ∠BDP = ∠CDP (by CPCT)
∴ AP bisect ∠D.
4. Since ΔBDP≅ΔCDP (Proved above)
BP = CP (by CPCT)
∠BPD = ∠CPD (by CPCT)
∠BPD + ∠CPD = 180° (linear pair)
∴ ∠BPD + ∠BPD = 180°
2∠BPD = 180°
∠BPD = 90°
∴ AP is the perpendicular bisector of BC.
Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A.
solution:
(i) In ΔADB and ΔADC,
∠ADB = ∠ADC (each 900)
AB = AC (ΔABC is an isosceles triangle)
AD = AD (common)
∴ ΔABD ≅ ΔACD (by RHS congruency)
BD = CD (by CPCT)
AD bisects BC.
(ii)∴ ΔADB ≅ ΔADC (Proved in part (i))
∴ ∠BAD = ∠CAD (CPCT)
Hence AD bisects ∠A.
Question 3.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of åPQR (see figure). Show that
(i) ΔABM ≅ ΔPQN
(ii) ΔABC ≅ ΔPQR.
Solution:
Given In ΔABC and ΔPQR,
AB = PQ
BC = QR
and AM = PN
(i) In ΔABM and ΔPQN,
AB = PQ (given)
BC = QR (given)
∴ 1/2BC = 1/2QR (Dividing by 2)
Hence BM = QN (median bisect opposite sides)
AM = PN
∴ ΔABM ≅ ΔPQN (by SSS congruency)
∠ABM = ∠PQN (by CPCT)
⇒ ∠ABC = ∠PQR
(ii) In ΔABC and ΔPQR,
AB = PQ (given)
∠ABC = ∠PQR (proved above in part (i))
BC = QR (Given)
∴ ΔABC ≅ ΔPQR (by SAS congruency)
Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Solution:
Given: ΔABC in which BE = CF.
To Prove: ΔABC is an isosceles triangle
Proof: In ABEC and ΔCFB,
∠BEC = ∠CFB (each 900)
BC = CB (Common)
BE = CF (given)
∴ ∠BEC = ∠CFB (by RHS congruency)
∴ ∠CBF = ∠BCE (by CPCT)
Hence AB = AC (Sides opposite to equal angle are equal)
Therefore, ABC is an isosceles triangle.
Question 5.
ABC is an isoscele8 triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Solution:
Given: ABC is an isosceles triangle in which
AB = AC.
To Prove: ∠B = ∠C
Construction: Draw AP ⊥ BC
Proof: In right triangles APB and APC,
∠APB = ∠APC (each 90°)
AB = AC (given)
AP = AP (common)
∴ ΔAPB ≅ ΔAPC (By RHS congruency)
∠ABP = ∠ACP (by CPCT)
⇒ ∠B = ∠C
Ex 7.4
Question 1.
Show that in a right-angled triangle, the hypotenuse is the longest side.
Solution:
Given: Right ΔABC which is right-angled at
B, i.e., ∠B 900,
Then ∠A + ∠B + ∠C = 180°
(by angle sum property)
∠A + 90° + ∠C = 180°
∠A + ∠C = 90°
∴ AC > BC (Side opposite to greater angle is longer)
and ∠B>∠C
∴ AC > AB (Side opposite to greater angle is longer)
Hence, AC, the hypotenuse is the longest side.
Question 2.
Infiguresides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
Solution:
∠PBC < ∠QCB (given)
⇒ 180°- ∠PBC>180°- ∠QCB
⇒∠ABC > ∠ACB
∴ AC > AB (Side opposite to greater angle is longer)
Question 3.
In the figure,∠B<∠A and ∠C<∠D. Show that AD∠BC
Solution:
∠B <∠A (given)
∠A >∠B
OB > OA ……….(1)
(Side opposite to greater angle is longer)
Now ∠C <∠D (given)
∴ ∠D > ∠C
∴ OC > OD ……….(2)
(Side opposite to greater angle is longer)
Adding eqn. (1) and (2), we get
OB + OC > OA + OD
⇒ BC >AD
⇒ AD > ∠C
Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠A> ∠C and ∠B> ∠D.
Solution:
Construction: Join AC.
Proof: In ΔABC,
AB AB
∠BAC > ∠BCA …….. (1) (Angle opposite to longer side is greater)
In ΔACD
CD>AD ( CDisthelongest side of quadrilateral)
∴ ∠CMD > ∠ACD ……(2) (Angle
opposite to longer side is greater)
Adding eqn. (1) and (2)
∠BAC + ∠CAD> ∠BCA + ∠ACD
⇒ ∠A > ∠C
Similarly, join B to D and we can prove
∠B > ∠D
Question 5.
In figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR> ∠PSQ.
Given:
In figure PR>PQ
and PS bisects ∠QPR.
To Prove:
∠PSR > ∠PSQ
Proof In ΔPQR,
PR > PQ (given)
∠PQR > ∠PRQ
∠PQS > ∠PRS ……..(1)
∠QPS = ∠RPS …….(2) (PS is the angle bisector of ∠QPR)
Adding ∠QPS on both sides in eqn. (1)
∠PQS + ∠QPS > ∠PRS + ∠QPS ……..(3)
From eqn. (2) and (3)
⇒ ∠PQS + ∠QPS> ∠PRS + ∠RPS …….(4)
⇒ ∠PSQ = ∠PRS + ∠RPS ……….(5) (By exterior angle theorem)
and ∠PSR = ∠QPS + ∠PQS …….(6) (by exterior angle theorem)
From eqn. (4), (5) and (6), we have
∠PSR > ∠PSQ
Question 6.
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Solution:
Given: A line I and P is a point not lying online l and PQ ⊥ l and R is any point on l other than Q.
To Prove: PQ < PR
Proof: in Δ PQR
∠Q = 90º
∠P + ∠Q + ∠R = 180° m(by angle sum property)
⇒ ∠P + 90° + ∠R = 180°
⇒ ∠P + ∠R = 90°
Hence ∠R is an acute angle.
∴ ∠Q > ∠R
(Side opposite to greater angle is longer)
∴ PR >PQ
⇒ PQ<PR
Hence perpendicular line segment is the shortest.
Hence the perpendicular line segment is the shortest.
Ex 7.5
Question 1.
ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.
Solution:
Draw perpendicular bisector of any two sides of ΔABC. The point of intersection of these perpendicular bisectors is equidistant from all the vertices of Hence point O will be the centre of the circle which passes through all vertices.
Question 2.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Solution:
The draw angle bisector of any two angles of a triangle. Where the point of intersection of these angle bisectors would be equidistant from all the three sides of the triangle. Thus, point O is the in centre of the triangle.
Question 3.
In a huge park, people are concentrated at three points (See figure)
A: where there are different slides and swings for children.
B: near which a man-made lake is situated.
C: which is near to a large parking and exit. Where should an ice-cream parlour be set up so that the maximum number of persons can approach it?
(Hint: the parlour should be equidistant from A, B and C)
Solution:
If ice-cream parlour is situated on the circumcentre of the circle, then the maximum number of people will reach. It will be equidistant from each vertex A, B and C.
Question 4.
Complete the hexagonal and star shaped Rangolies (See figure (i) and (ii)) by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?
No. of equilateral triangles of side 1 cm formed in hexagonal Rangoli
No. of equilateral triangles of side 1 cm formed in stare shaped Rangoli
= 75 x 4 = 300
Therefore, star shaped Rangoli has more triangles.