Practice Set 3.4 Geometry 9th Std Maths Part 2 Answers Chapter 3 Triangles
Question 1
Maharashtra Board Solution
In the adjoining figure, point A is on the bisector of ∠XYZ. If AX = 2 cm, then find AZ.
Solution & Step-by-Step Answer:
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
Maharashtra Board Solution
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
Solution & Step-by-Step 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 = ∠RST = x 56° ∴ ∠RSP = 28°
Question 3
Maharashtra Board Solution
In ∆PQR, PQ = 10 cm, QR = 12 cm, PR triangle. 8 cm. Find out the greatest and the smallest angle of the triangle.
Solution & Step-by-Step Answer:
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
Maharashtra Board Solution
In ∆FAN, ∠F = 80°, ∠A = 40°. Find out the greatest and the smallest side of the triangle. State the reason.
Solution & Step-by-Step Answer:
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
Maharashtra Board Solution
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
Maharashtra Board Solution
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 & Step-by-Step Answer:
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Solution & Step-by-Step Answer:
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
Maharashtra Board Solution
In the adjoining figure, in AABC, seg AD and seg BE are altitudes and AE = BD. Prove that seg AD = seg BE.
Solution & Step-by-Step Answer:
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
Maharashtra Board Solution
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)
Solution & Step-by-Step Answer:
From the given figure, ∠Z = 25° and ∠Y = 51° ∴ ∠Y is greater.
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