Jharkhand Board JAC Class 9 Maths Important Questions Chapter 7 Triangles Important Questions and Answers.

## JAC Board Class 9th Maths Important Questions Chapter 7 Triangles

Question 1.

If D is the mid-point of the hypotenuse AC of a right triangle ABC, prove that BD = \(\frac {1}{2}\)AC.

Solution :

Let ΔABC is a right triangle such that ∠B = 90° and D is midpoint of AC then we have to prove that BD = \(\frac {1}{2}\)AC, we produce BD to E such that BD = DE and join EC

Now, in ΔADB and ΔCDE we have

AD = DC (Given)

BD = DE (By construction)

And, ∠ADB = ∠CDE

[Vertically opposite angles]

∴ By SAS criterion of congruence, we have

ΔADB ≅ ΔCDE

⇒ EC = AB and ∠CED = ∠ABD ….(i)

(By CPCT)

But ∠CED and ∠ABD are alternate interior angles .

∴ CE || AB ⇒ ∠ABC + ∠ECB = 180°

(interior angles)

⇒ 90° + ∠ECB = 180°

⇒ ∠ECB = 90°

Now, in ΔABC and ΔECB, we have

AB = EC [By (i)]

BC = BC [Common]

And, ∠ABC = ∠ECB = 90°

∴ BY SAS criterion of congruence

ΔABC ≅ ΔECB

⇒ AC = EB = 2BD [By CPCT and also D is a midpoint of AC and BE]

⇒ BD = \(\frac {1}{2}\) AC Hence, proved.

Question 2.

In a right-angled triangle, one acute angle is double the other. Prove that the hypotenuse is double the smallest side.

Solution :

Let ΔABC is a right triangle such that ∠B = 90° and ∠ACB = 2∠CAB, then we have to prove AC = 2BC. We produce CB to D such that BD = CB and join AD. Let

⇒ ∠ACB = 2x and ∠CAB = x

Proof: In ΔABD and ΔABC, we have

BD = BC (By construction)

AB = AB [Common]

∠ABD = ∠ABC = 90°

∴ By SAS criterion of congruence we get

ΔABD ≅ ΔABC

⇒ AD = AC and ∠DAB = ∠CAB

(By CPCT)

⇒ AD = AC and ∠DAB = x[∵ ∠CAB = x]

Now, ∠DAC = ∠DAB + ∠CAB = x + x = 2x

∴ ∠DAC = ∠ACD

⇒ DC = AD

[Sides opposite to equal angles] (∵ DC = 2BC)

⇒ 2BC = AD

⇒ 2BC = AC (AD = AC)

Hence, Proved.

Question 3.

In figure, T is a point on side QR of ΔPQR and S is a point such that RT = ST. Prove that PQ + PR > QS.

Solution :

In ΔPQR we have

PQ + PR > QR

⇒ PQ + PR > QT + TR

⇒ PQ + PR > QT + ST

[∵ RT = ST]

In ΔQST, QT + ST > SQ

∴ PQ + PR > SQ Hence, proved.

Question 4.

In the given figure, PQ = QR and ∠x = ∠y. Prove that AR = PB.

Solution :

Proof: In the figure, ∠QAR + ∠PAR = 180°(Linear pair axiom)

⇒ ∠QAR + ∠x = 180°

⇒ ∠QAR = 180° – ∠x°

Similarly ∠QBP + ∠RBP = 180° (Linear pair axiom)

⇒ ∠QBP + ∠y = 180°

⇒ ∠QBP = 180° – ∠y …(ii)

But given, ∠x = ∠y

∴ ∠QAR = ∠QBP [From (i) and (ii)]

Now, in ΔQAR and ΔQBP, QR = PQ (Given)

∠QAR = ∠QBP (As proved above)

∠Q = ∠Q (Common)

⇒ ΔQAR = ΔQBP

(AAS congruence rule)

⇒ AR = PB (CPCT)

Hence proved.

Question 5.

Diagonal AC and BD of quadrilateral ABCD intersect each other at O. Prove that

(i) AB + BC + CD + DA > AC + BD

(ii) AB + BC + CD + DA < 2 (AC + BD)

Solution :

Given: AC and BD are the diagonals of quadrilateral ABCD. (i) To prove: AB + BC + CD + DA > AC + BD

Proof: We know that the sum of any two sides of a triangle is always greater than the third side. Therefore,

In ΔABC, AB + BC > AC …(i)

In ΔBCD, BC + CD > BD …(ii)

In ΔCDA CD + DA > CA …(iii)

In ΔABD, AB + AD > BD …(iv)

Adding (i), (ii), (iii) and (iv), we get

2 (AB + BC + CD + DA) > 2 (AC + BD)

⇒ AB + BC + CD + DA > AC + BD

Hence proved.

(i) To prove: AB + BC + CD + DA < 2 (AC + BD) Proof : In ΔOAB, OA + OB > AB ……(i)

In ΔBOC, OB + OC > BC …(ii)

In ΔCOD, OC + OD > CD …(iii)

In ΔAOD, OA + OD > DA …(iv)

Adding (i), (ii), (iii) and (iv), we get

2 (OA + OB + OC + OD) > AB + BC + CD + DA

2 [(OA + OC) + (OB + OD)] > AB + BC + CD + DA

2 (AC + BD) > AB + BC + CD + DA

AB + BC + CD + DA < 2 (AC + BD)

Hence proved

Multiple Choice Questions

Question 1.

If the three altitudes of a Δ are equal then triangle is :

(a) isosceles

(b) equilateral

(c) right-angled

(d) none

Solution :

(b) equilateral

Question 2.

ABCD is a square and P, Q, R are points on AB, BC and CD respectively such that AP = BQ = CR and ∠PQR = 90°, then ∠QPR =

(a) 45°

(b) 50°

(c) 60°

(d) 75°

Solution :

(a) 45°

Question 3.

In a ΔXYZ, LM ⊥ YZ and bisectors YN and ZN of ∠Y and ∠Z respectively meet at Non LM then YL + ZM =

(a) YZ

(b) XY

(c) XZ

(d) LM

Solution :

(d) LM

Question 4.

In a ΔPQR, PS is bisector of ∠P and ∠Q = 70°, ∠R = 30°, then

(a) QS > PQ > PR

(b) QS < PQ < PR

(c) PQ > QS > SR

(d) PQ < QS < SR

Solution :

(b) QS < PQ < PR

Question 5.

If D is any point on the side BC of a ΔABC, then:

(a) AB + BC + CA > 2AD

(b) AB + BC + CA < 2AD

(c) AB + BC + CA > 3AD

(d) None of these

Solution :

(a) AB + BC + CA > 2AD

Question 6.

For given figure, which one is correct:

(a) ΔABC ≅ ΔDEP

(b) ΔABC ≅ ΔFED

(c) ΔABC ≅ ΔDFE

(d) ΔABC ≅ ΔEDF

Solution :

(a) ΔABC ≅ ΔDEP

Question 7.

In a right-angled triangle, one acute angle is double the other then the hypotenuse is :

(a) Equal to the smallest side

(b) Double the smallest side

(c) Triple the smallest side

(d) None of these

Solution :

(d) None of these