RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS

In this chapter, we provide RD Sharma Solutions for Chapter 12 Heron’s Formula VSAQS for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 12 Heron’s Formula VSAQS Maths pdf, free RD Sharma Solutions for Chapter 12 Heron’s Formula VSAQS Maths book pdf download. Now you will get step by step solution to each question.

RD Sharma Solutions for Class 9 Chapter 12 Heron’s Formula VSAQS Download PDF

Question 1.
Solution:
In two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.

In AABC and ADEF,
∆ABC ≅ ∆DEF
and AB = DE, BC = EF
∴ ∠A = ∠D, ∠B = ∠E and ∠C = ∠F
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS – 1

Question 2.Solution:
In two triangles ABC and DEF, it is given that ∠A = ∠D, ∠B = ∠E and ∠C = ∠F. Are the two triangles necessarily congruent?

No, as the triangles are equiangular, so similar.

Question 3.Solution:
If ABC and DEF are two triangles such that AC = 2.5 cm, BC = 5 cm, ∠C = 75°, DE = 2.5 cm, DF = 5 cm and ∠D = 75°. Are two triangles congruent?

Yes, triangles are congruent (SAS axiom)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS – 3

Question 4.Solution:
In two triangles ABC and ADC, if AB = AD and BC = CD. Are they congruent?

Yes, these are congruent
In two triangles ABC are ADC,
AB = AD (Given)
BC = CD (Given)
and AC = AC (Common)
∴ ∆sABC ≅ AADC (SSS axiom)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS – 4

Question 5.Solution:
In triangles ABC and CDE, if AC = CE, BC = CD, ∠A = 60°, ∠C – 30° and ∠D = 90°. Are two triangles congruent?

Yes, triangles are congruent because,
In ∆ABC, and ∆CDE,
AC = CE
BC = CD ∠C = 30°
∴ ∆ABC ≅ ∆CDE (SAS axiom)
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS – 5

Question 6.Solution:
ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.

Given : In ∆ABC, AB = AC
BE and CF are two medians
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS – 6
To prove : BE = CF
Proof: In ∆ABE and ∆ACF.
AB = AC (Given)
∠A = ∠A (Common)
AE = AF (Half of equal sides)
∴ ∆ABE ≅ ∆ACF (SAS axiom)
∴ BE = CF (c.p.c.t.)

Question 7.Solution:
Find the measure of each angle of an equilateral triangle.

In ∆ABC,
AB = AC = BC
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS – 7
∵ AB = AC
∴ ∠C = ∠B …(i)
(Angles opposite to equal sides)
Similarly,
AC = BC
∴ ∠B = ∠A …(ii)
From (i) and (ii),
∠A = ∠B = ∠C
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
∴ ∠A + ∠B + ∠C = 180∘3 = 60°

Question 8.Solution:
CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that ∆ADE ≅ ∆BCE.

Given : An equilateral ACDE is formed on the side of square ABCD. AE and BE are joined
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS – 8
To prove : ∆ADE ≅ ∆BCE
Proof : In ∆ADE and ∆BCE,
AD = BC (Sides of a square)
DE = CE (Sides of equilateral triangle)
∠ADE = ∠BCE(Each = 90° + 60° = 150°)
∴ AADE ≅ ABCE (SAS axiom)

Question 9.Solution:
Prove that the sum of three altitude of a triangle is less than the sum of its sides.

Given : In ∆ABC, AD, BE and CF are the altitude of ∆ABC
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS – 9
To prove : AD + BE + CF < AB + BC + CA
Proof : In right ∆ABD, ∠D = 90°
Then other two angles are acute
∵ ∠B < ∠D
∴ AD < AB …(i)
Similarly, in ∆BEC and ∆ABE we can prove thatBE and CF < CA …(iii)
Adding (i), (ii), (iii)
AD + BE -t CF < AB + BC + CA

Question 10.Solution:
In the figure, if AB = AC and ∠B = ∠C. Prove that BQ = CP.
RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS – 10

Given : In the figure, AB = AC, ∠B = ∠C
To prove : BQ = CP
Proof : In ∆ABQ and ∆ACP
AB = AC (Given)
∠A = ∠A (Common)
∠B = ∠C (Given)
∴ ∆ABQ ≅ ∆ACP (ASA axiom)
∴ BQ = CP (c.p.c.t.)

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