## Maharashtra State Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.4

Question 1.

In the adjoining figure, point A is on the bisector of ∠XYZ. If AX = 2 cm, then find AZ.

Solution:

AX = 2 cm [Given]

Point A lies on the bisector of ∠XYZ. [Given]

Point A is equidistant from the sides of ∠XYZ. [Every point on the bisector of an angle is equidistant from the sides of the angle]

∴ A Z = AX

∴ AZ = 2 cm

Question 2.

In the adjoining figure, ∠RST = 56°, seg PT ⊥ ray ST, seg PR ⊥ ray SR and seg PR ≅ seg PT. Find the measure of ∠RSP.

State the reason for your answer.

Solution:

seg PT ⊥ ray ST, seg PR ⊥ ray SR [Given]

seg PR ≅ seg PT

∴ Point P lies on the bisector of ∠TSR [Any point equidistant from the sides of an angle is on the bisector of the angle]

∴ Ray SP is the bisector of ∠RST.

∠RSP = 56° [Given]

∴ ∠RSP = \(\frac { 1 }{ 2 }\)∠RST

= \(\frac { 1 }{ 2 }\) x 56°

∴ ∠RSP = 28°

Question 3.

In ∆PQR, PQ = 10 cm, QR = 12 cm, PR triangle. 8 cm. Find out the greatest and the smallest angle of the triangle.

Solution:

In ∆PQR,

PQ = 10 cm, QR = 12 cm, PR = 8 cm [Given]

Since, 12 > 10 > 8

∴ QR > PQ > PR

∴ ∠QPR > ∠PRQ > PQR [Angle opposite to greater side is greater]

∴ In ∆PQR, ∠QPR is the greatest angle and ∠PQR is the smallest angle.

Question 4.

In ∆FAN, ∠F = 80°, ∠A = 40°. Find out the greatest and the smallest side of the triangle. State the reason.

Solution:

In ∆FAN,

∠F + ∠A + ∠N = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ 80° + 40° + ∠N = 180°

∴ ∠N = 180° – 80° – 40°

∴∠N = 60°

Since, 80° > 60° > 40°

∴ ∠F > ∠N > ∠A

∴ AN > FA > FN [Side opposite to greater angle is greater]

∴ In ∆FAN, AN is the greatest side and FN is the smallest side.

Question 5.

Prove that an equilateral triangle is equiangular.

Given: ∆ABC is an equilateral triangle.

To prove: ∆ABC is equiangular

i.e. ∠A ≅ ∠B ≅ ∠C …(i) [Sides of an equilateral triangle]

In ∆ABC,

seg AB ≅ seg BC [From (i)]

∴ ∠C = ∠A (ii) [Isosceles triangle theorem]

In ∆ABC,

seg BC ≅ seg AC [From (i)]

∴ ∠A ≅ ∠B (iii) [Isosceles triangle theorem]

∴ ∠A ≅ ∠B ≅ ∠C [From (ii) and (iii)]

∴ ∆ABC is equiangular.

Question 6.

Prove that, if the bisector of ∠BAC of ∆ABC is perpendicular to side BC, then AABC is an isosceles triangle.

Given: Seg AD is the bisector of ∠BAC.

seg AD ⊥ seg BC

To prove: AABC is an isosceles triangle.

Proof.

In ∆ABD and ∆ACD,

∠BAD ≅ ∠CAD [seg AD is the bisector of ∠BAC]

seg AD ≅ seg AD [Common side]

∠ADB ≅ ∠ADC [Each angle is of measure 90°]

∴ ∆ABD ≅ ∆ACD [ASA test]

∴ seg AB ≅ seg AC [c. s. c. t.]

∴ ∆ABC is an isosceles triangle.

Question 7.

In the adjoining figure, if seg PR ≅ seg PQ, show that seg PS > seg PQ.

Solution:

Proof.

In ∆PQR,

seg PR ≅ seg PQ [Given]

∴ ∠PQR ≅ ∠PRQ ….(i) [Isosceles triangle theorem]

∠PRQ is the exterior angle of ∆PRS.

∴ ∠PRQ > ∠PSR ….(ii) [Property of exterior angle]

∴ ∠PQR > ∠PSR [From (i) and (ii)]

i.e. ∠Q > ∠S ….(iii)

In APQS,

∠Q > ∠S [From (iii)]

∴ PS > PQ [Side opposite to greater angle is greater]

∴ seg PS > seg PQ

Question 8.

In the adjoining figure, in AABC, seg AD and seg BE are altitudes and AE = BD. Prove that seg AD = seg BE.

Solution:

Proof:

In ∆ADB and ∆BEA,

seg BD ≅ seg AE [Given]

∠ADB ≅ ∠BEA = 90° [Given]

seg AB ≅ seg BA [Common side]

∴ ∆ADB ≅ ∆BEA [Hypotenuse-side test]

∴ seg AD ≅ seg BE [c. s. c. t.]

**Maharashtra Board Class 9 Maths Chapter 3 Triangles Practice Set 3.4 Intext Questions and Activities**

Question 1.

As shown in the given figure, draw ∆XYZ such that side XZ > side XY. Find which of ∠Z and ∠Y is greater. (Textbook pg. no. 41)

Answer:

From the given figure, ∠Z = 25° and ∠Y = 51°

∴ ∠Y is greater.