RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3: Want to have the perfect score in the Class 9 Maths exam? Start studying the RD Sharma Solutions Class 9 Maths. It is a much need guide that will clear your doubts and make you score good marks. All the solutions of RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3 are designed by subject matter experts.
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RD Sharma Class 9 Solutions Chapter 10 Exercise 10.3
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Question 1: In two right triangles, one side and an acute angle of one triangle are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.
Solution:
In two right triangles, one side and an acute angle of one triangle are equal to the corresponding side and angles of the other. (Given)
To prove: Both triangles are congruent.
Consider two right triangles such that
∠ B = ∠ E = 90o …….(i)
AB = DE …….(ii)
∠ C = ∠ F ……(iii)
Here we have two right triangles, △ ABC and △ DEF
From (i), (ii) and (iii),
By the AAS congruence criterion, we have Δ ABC ≅ Δ DEF
Both triangles are congruent. Hence proved.
Question 2: If the bisector of the exterior vertical angle of a triangle is parallel to the base. Show that the triangle is isosceles.
Solution:
Let ABC be a triangle such that AD is the angular bisector of the exterior vertical angle, ∠EAC and AD ∥ BC.
From figure,
∠1 = ∠2 [AD is a bisector of ∠ EAC]
∠1 = ∠3 [Corresponding angles]
and ∠2 = ∠4 [alternative angle]
From above, we have ∠3 = ∠4
This implies, AB = AC
Two sides, AB and AC, are equal.
⇒ Δ ABC is an isosceles triangle.
Question 3: In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.
Solution:
Let Δ ABC be isosceles where AB = AC and ∠ B = ∠ C
Given: Vertex angle A is twice the sum of the base angles B and C. i.e., ∠ A = 2(∠ B + ∠ C)
∠ A = 2(∠ B + ∠ B)
∠ A = 2(2 ∠ B)
∠ A = 4(∠ B)
Now, We know that the sum of angles in a triangle =180°
∠ A + ∠ B + ∠ C =180°
4 ∠ B + ∠ B + ∠ B = 180°
6 ∠ B =180°
∠ B = 30°
Since, ∠ B = ∠ C
∠ B = ∠ C = 30°
And ∠ A = 4 ∠ B
∠ A = 4 x 30° = 120°
Therefore, the angles of the given triangle are 30° and 30° and 120°.
Question 4: PQR is a triangle in which PQ = PR and is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.
Solution: Given that PQR is a triangle such that PQ = PR and S is any point on the side PQ and ST ∥ QR.
To prove: PS = PT
Since, PQ= PR, so △PQR is an isosceles triangle.
∠ PQR = ∠ PRQ
Now, ∠ PST = ∠ PQR and ∠ PTS = ∠ PRQ
[Corresponding angles as ST parallel to QR]
Since, ∠ PQR = ∠ PRQ
∠ PST = ∠ PTS
In Δ PST,
∠ PST = ∠ PTS
Δ PST is an isosceles triangle.
Therefore, PS = PT.
Hence proved.
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