**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 = 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 BCTo 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.]

**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.