RD Sharma Solutions for Class 9 Mathematics Chapter 10 Exercise 10.5 Congruent Triangles are provided in this article. Here, students will learn about the congruence criterion of right triangles, mainly using RHS (Right-angle-Hypotenuse-Side) and its related concepts in an interesting and interactive way. Now, this topic can easily be learnt with RD Sharma Solutions for Class 9 with step-by-step illustrations. Download the solutions in PDF for free from the links provided below.
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Access Answers to D Sharma Solutions for Class 9 Maths Chapter 10 Congruent Triangles Exercise 10.5 Page Number 10.51
Question 1: ABC is a triangle, and D is the mid-point of BC. The perpendiculars from D to AB and AC are equal. Prove that the triangle is isosceles.
Solution:
Given: D is the midpoint of BC and PD = DQ in a triangle ABC.
To prove: ABC is isosceles triangle.
In △BDP and △CDQ
PD = QD (Given)
BD = DC (D is mid-point)
∠BPD = ∠CQD = 90o
By RHS Criterion: △BDP ≅ △CDQ
BP = CQ … (i) (By CPCT)
In △APD and △AQD
PD = QD (given)
AD = AD (common)
APD = AQD = 90 o
By RHS Criterion: △APD ≅ △AQD
So, PA = QA … (ii) (By CPCT)
Adding (i) and (ii)
BP + PA = CQ + QA
AB = AC
Two sides of the triangle are equal, so ABC is an isosceles.
Question 2: ABC is a triangle in which BE and CF are, respectively, the perpendiculars to the sides AC and AB. If BE = CF, prove that Δ ABC is isosceles
Solution:
ABC is a triangle in which BE and CF are perpendicular to the sides AC and AB, respectively, s.t. BE = CF.
To prove: Δ ABC is isosceles
In Δ BCF and Δ CBE,
∠ BFC = CEB = 90o [Given]
BC = CB [Common side]
And CF = BE [Given]
By RHS congruence criterion: ΔBFC ≅ ΔCEB
So, ∠ FBC = ∠ EBC [By CPCT]
⇒∠ ABC = ∠ ACB
AC = AB [Opposite sides to equal angles are equal in a triangle]
Two sides of triangle ABC are equal.
Therefore, Δ ABC is isosceles. Hence Proved.
Question 3: If perpendiculars from any point within an angle on its arms are congruent. Prove that it lies on the bisector of that angle.
Solution:
Consider an angle ABC and BP be one of the arms within the angle.
Draw perpendiculars PN and PM on the arms BC and BA.
In Δ BPM and Δ BPN,
∠ BMP = ∠ BNP = 90° [given]
BP = BP [Common side]
MP = NP [given]
By RHS congruence criterion: ΔBPM≅ΔBPN
So, ∠ MBP = ∠ NBP [ By CPCT]
BP is the angular bisector of ∠ABC.
Hence proved
RD Sharma Solutions for Class 9 Maths Chapter 10 Congruent Triangles Exercise 10.5
RD Sharma Solutions Class 9 Maths Chapter 10 Congruent Triangles Exercise 10.5 is based on the topic Congruence Criterion – Right-angle- Hypotenuse-Side. According to the criteria, two triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and one side of the other triangle.
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