Question

Can anybody send the solution of algebra practice set 3.1 Q.1 and 2 of 9th Maths Ssc board​

Answer 1

Hope it helps you..........

Answer 2

Answer:

Question 1. Can you give an alternative proof of the above theorem by drawing a line through point R and parallel to seg PQ in the above figure? (Textbook pg. no. 25)

Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.1 11

Solution:

Yes.

Construction: Draw line RM parallel to seg PQ through a point R.

Proof:

seg PQ || line RM and seg PR is their transversal. [Construction]

∴ ∠PRM = ∠QPR ……..(i) [Alternate angles]

seg PQ || line RM and seg QR is their transversal. [Construction]

∴ ∠SRM = ∠PQR ……..(ii) [Corresponding angles]

∴ ∠PRM + ∠SRM = ∠QPR + ∠PQR [Adding (i) and (ii)]

∴ ∠PRS = ∠PQR + ∠QPR [Angle addition property]

3 Triangles Question 2. Observe the given figure and find the measures of ∠PRS and ∠RTS. (Textbook pg. no.25)

Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.1 12

Solution:

∠PRS is an exterior angle of ∆PQR.

So from the theorem of remote interior angles,

∠PRS = ∠PQR + ∠QPR

= 40° + 30°

∴ ∠PRS = 70°

∴ ∠TRS=70° …[P – T – R]

In ∆RTS,

∠TRS + ∠RTS + ∠TSR = 180° …[Sum of the measures of the angles of a triangle is 180°]

∴ 70° + ∠RTS + 20° = 180°

∴ ∠RTS + 90° = 180°

∴ ∠RTS = 180°

∴ ∠RTS = 90