** 9th Class Maths Solutions Chapter 7 Triangles **

Exercise:- 7.1

In quadrilateral ACBD, AC = AD and AB bisects ∠A. Show that Î”ABC ≅ Î”ABD What can you say about BC and BD ?

Solution:

Given that AC = AD

∠BAC = ∠BAD (∵ AB bisects∠A)

Now in Î”ABC and Î”ABD

AC = AD (∵ given)

∠BAC = ∠BAD (Y given)

AB = AB (common side)

∴ Î”ABC ≅ Î”ABD

(∵ SAS congruence rule)

ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA, prove that i) Î”ABD ≅Î”BAC ii) BD = AC

iii) ∠ABD = ∠BAC.

Solution :

i) Given that AD = BC and

∠DAB = ∠CBA

Now in Î”ABD and Î”BAC

AB = AB (∵ Common side)

AD = BC (∵ given)

∠DAB = ∠CBA (∵ given)

∴ Î”ABD ≅ Î”BAC

(∵ SAS congruence)

ii) From (i) AC = BD (∵ CPCT)

iii) ∠ABD = ∠BAC [ ∵ CPCT from (i)]

AD and BC are equal and perpendi-culars to a line segment AB. Show that CD bisects AB.

Solution:

Given that AD = BC; AD ⊥ AB; BC ⊥ AB

In Î”BOC and Î”AOD

∠BOC = ∠AOD (∵ vertically opposite angles)

∴ Î”OBC = Î”OAD (∵ right angle)

BC = AD

Î”OBC ≅ Î”OAD (∵ AAS congruence)

∴ OB = OA (∵ CPCT)

∴ ‘O’ bisects AB

Also OD = OC

∴ ‘O’ bisects CD

⇒ AB bisects CD

l and m are two parallel lines inter-sected by another pair of parallel lines p and q. Show that Î”ABC ≅ Î”CDA.

Solution:

Given that l // m; p // q.

In Î”ABC and Î”CDA

∠BAC = ∠DCA (∵ alternate interior angles)

∠ACB = ∠CAD

AC = AC

∴ Î”ABC ≅ Î”CDA (∵ ASA congruence)

In the figure given below AC = AE; AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Solution:

Given that AC = AE, AB = AD and

∠BAD = ∠EAC

In Î”ABC and Î”ADE

AB = AD

AC = AE

∠BAD = ∠EAC

∴ Î”ABC ≅ Î”ADE (∵ SAS congruence)

⇒ BC = DE (CPCT)

In right triangle ABC, right angle is 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 fig.). Show that

i) Î”AMC = Î”BMD

ii) ∠DBC is a right angle

iii) Î”DBC = Î”ACB

iv) CM =

Solution:

Given that ∠C = 90°

M is mid point of AB;

DM = CM (i.e., M is mid point of DC)

i) In Î”AMC and Î”BMD

AM = BM (∵ M is mid point of AB)

CM = DM ( ∵ M is mid point of CD)

∠AMC = ∠BMD ( ∵ Vertically opposite angles)

∴ Î”AMC ≅ Î”BMD

(∵ SAS congruence)

(CPCT of Î”AMC and Î”BMD)

But these are alternate interior angles for the lines DB and AC and DC as transversal.

∴DB || AC

As AC ⊥ BC; DB is also perpendicular to BC.

∴ ∠DBC is a right angle.

DB = AC (CPCT of Î”BMD and Î”AMC)

∠DBC = ∠ACB = 90°(already proved)

BC = BC (Common side)

∴ Î”DBC ≅ Î”ACB (SAS congruence rule)

DC =

CM =

In the given figure Î”BCD is a square and Î”APB is an equilateral triangle.

Prove that Î”APD ≅ Î”BPC.

[Hint: In Î”APD and Î”BPC;

Solution:

Given that □ABCD is a square.

Î”APB is an equilateral triangle.

Now in Î”APD and Î”BPC

AP = BP ( ∵ sides of an equilateral triangle)

AD = BC (∵ sides of a square)

∠PAD = ∠PBC [ ∵ 90° – 60°]

∴ Î”APD ≅ Î”BPC (by SAS congruence)

In the figure given below Î”ABC is isosceles as

(Hint: Compare Î”APB and Î”ACQ)

Solution:

Given that Î”ABC is isosceles and

AP = AQ

Now in Î”APB and Î”AQC

AP = AQ (given)

AB = AC (given)

∠PAB = ∠QAC (∵ Vertically opposite angles)

∴ Î”APB ≅ Î”AQC (SAS congruence)

∴

In the figure given below AABC, D is the midpoint of BC. DE ⊥ AB, DF ⊥ AC and DE = DF. Show that Î”BED ≅ AÎ”CFD.

Solution:

Given that D is the mid point of BC of Î”ABC.

DF ⊥ AC; DE = DF

DE ⊥ AB

In Î”BED and Î”CFD

∠BED = ∠CFD (given as 90°)

BD = CD (∵D is mid point of BC)

ED = FD (given)

∴ Î”BED ≅ Î”CFD (RHS congruence)

If the bisector of an angle of a triangle also bisects the opposite side, prove that the triangle is isosceles.

Solution:

Let Î”ABC be a triangle.

The bisector of ∠A bisects BC

To prove: Î”ABC is isosceles

(i.e., AB = AC)

We know that bisector of vertical angle divides the base of the triangle in the ratio of other two sides.

∴

Thus

⇒ AB = AC

Hence the Triangle is isosceless.

In the given figure Î”ABC is a right triangle and right angled at B such that ∠BCA = 2 ∠BAC. Show that the hypotenuse AC = 2BC.

[Hint : Produce CB to a point D that BC = BD]

Solution:

Given that ∠B = 90°; ∠BCA = 2∠BAC

To prove : AC = 2BC

Produce CB to a point D such that

BC = BD

Now in Î”ABC and Î”ABD

AB = AB (common)

BC = BD (construction)

∠ABC =∠ABD (∵ each 90°)

∴ Î”ABC ≅ Î”ABD

Thus AC = AD and ∠BAC = ∠BAD = 30° [CPCT]

[ ∵ If ∠BAC = x then

∠BCA = 2x

x + 2x = 90°

3x = 90°

⇒ x = 30°

∴ ∠ACB = 60°]

Now in Î”ACD,

∠ACD = ∠ADC = ∠CAD = 60°

∴∠ACD is equilateral ⇒ AC = CD = AD

⇒ AC = 2BC (∵ C is mid point)

Exercise 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 (i) OB = OC (ii) AO bisects ∠A.

Solution:

Given that in Î”ABC

AB = AC

Bisectors of ∠B and ∠C meet at ‘O’.

To prove

i) OB = OC

∠B = ∠C (Angles opposite to equal, sides)

∠OBC = ∠OCB

⇒ OB = OC (∵ Sides opposite to equal angles in Î”OBC)

ii) AO bisects ∠A.

In Î”AOB and Î”AOC

AB = AC (given)

BO = CO (already proved)

∠ABO = ∠ACO (∵ ∠B =∠C)

∴ Î”AOB ≅ Î”AOC

⇒ ∠BAO = ∠CAO [ ∵ CPCT of Î”AOB and Î”AOC]

∴ AO is bisector of ∠A.

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

ii) ∠MDB = ∠MCA

iii) In Î”DBC and Î”ACB

iv) DC = AB (CPCT of Î”DBC and Î”ACB)

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Question 2.

In Î”ABC, AD is the perpendicular bisector of BC (see given figure). Show that Î”ABC is an isosceles triangle in which AB = AC

Solution:

Given that AD ⊥ BC; AD = DC

In Î”ABD and Î”ACD

AD = AD (common)

BD = DC (given)

∠ADB = ∠ADC (given)

∴ Î”ABD ≅ Î”ACD (∵ SAS congruence)

⇒ AB = AC (CPCT of Î”ABD and Î”ACD)

Question 3.

ABC is an isosceles triangle in which altitudes BD and CE are drawn to equal sides AC and AB respectively (see figure). Show that these altitudes are equal.

Solution:

Given that AC = AB; BD ⊥ AC; CE ⊥ AB

In Î”BCD and Î”CBE

∠BDC = ∠CEB (90° each)

∠BCD = ∠CBE (∵ angles opp. to equal sides of a triangle)

BC = BC

∴ Î”BCD ≅ Î”CBE (∵ AAScongruence)

⇒ BD = CE (CPCT)

Question 4.

ABC is a triangle in which altitudes BD and CE to sides AC and AB are equal (see figure). Show that

i) Î”ABD ≅ Î”ACE

ii) AB = AC i.e., ABC is an isosceles triangle.

Solution:

Given that BD ⊥ AC; CE ⊥ AC

BD = CE

Now in Î”ABD and Î”ACE

∠ADB = ∠AEC (∵ given 90°)

∠A = ∠A (commori angle)

BD = CE

∴ Î”ABD = Î”ACE (∵ AAS congruence)

⇒ AB = AC (∵ C.P.C.T)

Question 5.

Î”ABC and Î”DBC are two isosceles triangles on the same base BC (see figure). Show that ∠ABD = ∠ACD.

Solution:

Given that Î”ABC and Î”DBC are isosceles.

To prove ∠ABD = ∠ACD

Join A and D.

Now in Î”ABD and ACD

AB = AC (∵ equal sides of isosceles triangles)

BD = CD (∵ equal sides of isosceles triangles)

AD = AD (∵ common side)

∴ Î”ABD ≅ Î”ACD (∵ SSS congruence)

⇒ ∠ABD = ∠ACD (CPCT)

## Exercise 7.3

Question 1.

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that, (i) AD bisects BC (ii) AD bisects ∠A.

Solution:

Given that in Î”ABC, AB = AC

and AD ⊥ BC

i) Now in Î”ABD and Î”ACD

AB = AC (given)

∠ADB = ADC (given AD ⊥ BC)

AD = AD (common)

∴ Î”ABD ≅ Î”ACD (∵ RHS congruence)

⇒ BD = CD (CPCT)

⇒ AD, bisects BC.

ii) Also ∠BAD = ∠CAD

(CPCT of Î”ABD ≅ Î”ACD )

∴ AD bisects ∠A.

Question 2.

Two sides AB, BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Î”PQR (see figure). Show that:

(i) Î”ABM ≅ Î”PQN

ii) Î”ABC ≅ Î”PQR

Solution:

Given that

AB = PQ

AM = PN

i) Now in Î”ABM and Î”PQN

AB = PQ (given)

AM = PN (given)

BM = QN (∵ BC = QR ⇒

∴ Î”ABM ≅ Î”PQN

(∵ SSS congruence)

ii) In Î”ABC and Î”PQR

AB = PQ (given)

BC = QR (given)

∠ABC = ∠PQN [∵ CPCT of Î”ABM and Î”PQN from (i)]

∴ Î”ABC ≅ Î”PQR

(∵ SAS congruence)

Question 3.

BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Solution:

In Î”ABC altitude BE and CF are equal.

Now in Î”BCE and Î”CBF

∠BEC = ∠CFB (∵ given 90°)

BC = BC (common; hypotenuse)

CF = BE (given)

∴ Î”BEC ≅ Î”CBF

⇒ ∠EBC = ∠FCB (∵ CPCT)

But these are also the interior angles opposite to sides AC and AB of Î”ABC.

⇒ AC = AB

Hence proved.

Question 4.

Î”ABC is an isosceles triangle in which AB = AC. Show that ∠B = ∠C.

(Hint : Draw AP ⊥ BQ (Using RHS congruence rule)

Solution:

Given the Î”ABC is an isosceles triangle and AB = AC

Let D be the mid point of BC; Join A, D.

Now in Î”ABD and Î”ACD

AB = AC (given)

BD = DC (construction)

AD = AD (common)

∴ Î”ABD ≅ Î”ACD (∵ SSS congruence)

⇒ ∠B = ∠C [∵ CPCT]

Question 5.

Î”ABC 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.

Solution:

Given that in Î”DBC; AB = AC; AD = AB

In Î”ABC

∠ABC + ∠ACB = ∠DAC …………… (1)

[∵ exterior angle]

In Î”ACD

∠ADC + ∠ACD = ∠BAC ………………(2)

Adding (1) & (2)

∠DAC + ∠BAC = 2 ∠ACB + 2∠ACD

[∵ ∠ABC = ∠ACB

∠ADC = ∠ACD]

180° = 2 [∠ACB + ∠ACD]

180° = 2[∠BCD]

∴ ∠BCD =

(or)

From the figure

∠2 = x + x = 2x

∠1 = y + y = 2y

∠1 + ∠2 = 2x + 2y

180° = 2 = (x + y)

∴ x + y =

Hence proved.

Question 6.

ABC is a right angled triangle in which ∠A = 90° and AB = AC, Show that ∠B = ∠C.

Solution:

Given Î”ABC; AB – AC

Join the mid point D of BC to A.

Now in Î”ADC and Î”ADB

AD = AD (common)

AC = AB (giyen)

DC = DB (construction)

⇒ Î”ADC ≅ Î”ADB

⇒ ∠C = ∠B (CPCT)

Question 7.

Show that the angles of an equilateral triangle are 60° each.

Solution:

Given Î”ABC is an equilateral triangle

AB = BC = CA

∠A = ∠B (∵ angles opposite to equal sides)

∠B = ∠C (∵ angles opposite to equal sides)

⇒ ∠A = ∠B = ∠C = x say

Also ∠A+∠B + ∠C =180°

⇒ x + x + x = 180°

3x = 180°

⇒ x =

Hence proved.

Triangles Exercise 7.4

Question 1.

Show that in a right angled triangle, the hypotenuse is the longest side.

Solution:

Let a Î”ABC be right angled at ∠B.

Then ∠A + ∠C = 90°

(i.e.,) ∠A and ∠C are both acute.

Now, ∠A < ∠B ⇒ BC < AC

Also ∠C < ∠B ⇒ AB < AC

∴ AC, the hypotenuse is the longest side.

Question 2.

In the given figure, sides AB and AC of Î”ABC are extended “to points P and Q respectively. Also ∠PBC < ∠QCB. Show that AC > AB.

Solution:

From the figure,

∠PBC = ∠A + ∠ACB

∠QCB = ∠A + ∠ABC

Given that ∠PBC < ∠QCB

⇒∠A + ∠ACB < ∠A + ∠ABC

⇒ ∠ACB < ∠ABC

⇒ AB < AC

⇒ AC > AB

Hence proved.

Question 3.

In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution:

Given that ∠B < ∠A; ∠C < ∠D

∠B < ∠A ⇒ AO < OB [in Î”AOB] ……………… (1)

∠C < ∠D ⇒ OD < OC [in Î”COD]…… (2)

Adding (1) & (2)

AO + OD < OB + OC

AD < BC

Hence proved.

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:

Given that AB and CD are the smallest and longest sides of quadrilateral ABCD.

From the figure,

In Î”BCD

∠1 > ∠2 [∵ DC > BC] ………………(1)

In Î”BDA

∠4 > ∠3 [∵ AD > AB] ………….(2)

Adding (1) & (2)

∠1 + ∠4 > ∠2 + ∠3

∠B > ∠D

Similarly,

In Î”ABC, ∠6 < ∠7 [ ∵AB < BC] ……………….(3)

In Î”ACD

∠5 < ∠8 …………. (4)

Adding (3) & (4)

∠6 + ∠5 < ∠7 + ∠8

∠C < ∠A ⇒ ∠A > ∠C

Hence proved.

Question 5.

In the given figure, PR > PQ and PS bisects ∠QPR. Prove that

∠PSR > ∠PSQ.

Solution:

Given that PR > PQ;

∠QPS =∠RPS

PR> PQ

∠Q > ∠R

Now ∠Q +∠QPS > ∠R + ∠RPS

⇒ 180° – (∠Q + ∠QPS) < 180° – (∠R + ∠RPS)

⇒ ∠PSQ < ∠PSR ⇒ ∠PSR > ∠PSQ

Hence proved.

Question 6.

If two sides of a triangle measure 4 cm and 6 cm find all possible measurements (positive integers) of the third side. How many distinct triangles can be obtained ?

Solution:

Given that two sides of a triangle are 4 cm and 6 cm.

∴ The measure of third side > Differ-ence between other two sides.

third side > 6 – 4

third side > 2

Also the measure of third side < sum of other two sides

third side <6 + 4 < 10

∴ 2 < third side <10

∴ The measure of third side may be 3 cm, 4 cm, 5 cm, 6 cm, 7 cm, 8 cm, 9 cm

∴ Seven distinct triangles can be obtained.

Question 7.

Try to construct a triangle with 5 cm, 8 cm and 1 cm. Is it possible or not ? Why ? Give your justification.

Solution:

As the sum (6 cm) of two sides 5 cm and 1 cm is less than third side. It is not possible to construct a triangle with the given measures.