AP State Syllabus AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.3 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 7th Lesson Triangles 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 ΔABC, AB = AC

and AD ⊥ BC

i) Now in ΔABD and ΔACD

AB = AC (given)

∠ADB = ADC (given AD ⊥ BC)

AD = AD (common)

∴ ΔABD ≅ ΔACD (∵ RHS congruence)

⇒ BD = CD (CPCT)

⇒ AD, bisects BC.

ii) Also ∠BAD = ∠CAD

(CPCT of ΔABD ≅ ΔACD )

∴ 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 ΔPQR (see figure). Show that:

(i) ΔABM ≅ ΔPQN

ii) ΔABC ≅ ΔPQR

Solution:

Given that

AB = PQ

AM = PN

i) Now in ΔABM and ΔPQN

AB = PQ (given)

AM = PN (given)

BM = QN (∵ BC = QR ⇒ \(\frac { 1 }{ 2 }\)BC = \(\frac { 1 }{ 2 }\)QR ⇒ BM = QN)

∴ ΔABM ≅ ΔPQN

(∵ SSS congruence)

ii) In ΔABC and ΔPQR

AB = PQ (given)

BC = QR (given)

∠ABC = ∠PQN [∵ CPCT of ΔABM and ΔPQN from (i)]

∴ ΔABC ≅ ΔPQR

(∵ 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 ΔABC altitude BE and CF are equal.

Now in ΔBCE and ΔCBF

∠BEC = ∠CFB (∵ given 90°)

BC = BC (common; hypotenuse)

CF = BE (given)

∴ ΔBEC ≅ ΔCBF

⇒ ∠EBC = ∠FCB (∵ CPCT)

But these are also the interior angles opposite to sides AC and AB of ΔABC.

⇒ AC = AB

Hence proved.

Question 4.

ΔABC is an isosceles triangle in which AB = AC. Show that ∠B = ∠C.

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

Solution:

Given the ΔABC is an isosceles triangle and AB = AC

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

Now in ΔABD and ΔACD

AB = AC (given)

BD = DC (construction)

AD = AD (common)

∴ ΔABD ≅ ΔACD (∵ SSS congruence)

⇒ ∠B = ∠C [∵ CPCT]

Question 5.

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.

Solution:

Given that in ΔDBC; AB = AC; AD = AB

In ΔABC

∠ABC + ∠ACB = ∠DAC …………… (1)

[∵ exterior angle]

In ΔACD

∠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 = \(\frac{180^{\circ}}{2}\) = 90°

(or)

From the figure

∠2 = x + x = 2x

∠1 = y + y = 2y

∠1 + ∠2 = 2x + 2y

180° = 2 = (x + y)

∴ x + y = \(\frac{180^{\circ}}{2}\) = 90°

Hence proved.

Question 6.

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

Solution:

Given ΔABC; AB – AC

Join the mid point D of BC to A.

Now in ΔADC and ΔADB

AD = AD (common)

AC = AB (giyen)

DC = DB (construction)

⇒ ΔADC ≅ ΔADB

⇒ ∠C = ∠B (CPCT)

Question 7.

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

Solution:

Given ΔABC 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 = \(\frac{180}{3}\) = 60°

Hence proved.