Practice Set 3.3 Geometry 9th Std Maths Part 2 Answers Chapter 3 Triangles
Solution & Step-by-Step Answer:
i. ∠ACB = 50° [Given] In ∆ABC, seg AC ≅ seg AB [Given] ∴ ∠ABC ≅ ∠ACB [Isosceles triangle theorem] ∴ x = 50°
ii. ∠DBC = 60° [Given]
In ABDC, seg BD ≅ seg DC [Given]
∴ ∠DCB ≅ ∠DBC [Isosceles triangle theorem]
∴ y = 60°
iii. ∠ABD = ∠ABC + ∠DBC [Angle addition property]
= 50° + 60°
∴ ∠ABD = 110°
iv. ∠ACD = ∠ACB + ∠DCB [Angle addition property]
= 50° + 60°
∴ ∠ACD = 110°
∴ x = 50°, y = 60°,
∠ABD = 110°, ∠ACD = 110°
Solution & Step-by-Step Answer:
Length of hypotenuse = 15 [Given] Length of median on the hypotenuse = x length of hypotenuse [In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse] = x 15 = 7.5 ∴ The length of the median on the hypotenuse is 7.5 units.
Solution & Step-by-Step Answer:
i. PQ = 12, QR = 5 [Given] In APQR, ∠Q = 90° [Given] ∴ PR2 = QR2 + PQ2 [Pythagoras theorem] = 25 + 144 ∴ PR2 =169 ∴ PR = 13 units [Taking square root of both sides]
ii. In right angled APQR, seg QS is the median on hypotenuse PR.
∴ QS = PR [In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse]
= x 13
∴ l(QS) = 6.5 units
Solution & Step-by-Step Answer:
i. In ∆PQR, G is the point of concurrence of the medians. [Given] The centroid divides each median in the ratio 2 : 1. PG : GT = 2 : 1 ∴ PG = 2 x 2.5 ∴ PG = 5 units
ii. Now, PT = PG + GT [P – G – T]
= 5 + 2.5
∴ l(PG) = 5 units, l(PT) = 7.5 units
Maharashtra Board Class 9 Maths Chapter 3 Triangles Practice Set 3.3 Intext Questions and Activities
Solution & Step-by-Step Answer:
Yes Construction: Draw seg AD ⊥ seg BC. Proof: In ∆ABD and ∆ACD, seg AB≅ seg AC [Given] ∠ADB ≅ ∠ADC [Each angle is of measure 90°] seg AD ≅ seg AD [Common side] ∴ ∆ABD ≅ ∆ACD [Hypotenuse side test] ∴ ∠ABD ≅ ∠ACD [c.a.c.t.] ∴ ∠ABC ≅ ∠ACB [B-D-C]
Solution & Step-by-Step Answer:
Yes Proof: In ∆ABC and ∆ACB, seg AB ≅ seg AC [Given] ∠BAC ≅ ∠CAB [Common angle] seg AC ≅ seg AB [Given] ∴ ∆ABC ≅ ∆ACB [SAS test] ∴ ∠ABC ≅ ∠ACB [c. a. c. t.]
Solution & Step-by-Step Answer:
AD = DC = BD= 1.9 cm AC = AD + DC [A – D – C] = 1.9 + 1.9 = 2 x 1.9 cm ∴ AC = 2 x BD ∴ BD = AC