<\/span><\/h2>\nQuestion 1.
Solution:
In two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.<\/p>\n
In AABC and ADEF,
\u2206ABC \u2245 \u2206DEF
and AB = DE, BC = EF
\u2234 \u2220A = \u2220D, \u2220B = \u2220E and \u2220C = \u2220F
RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula VSAQS \u2013 1<\/p>\n
Question 2.Solution:
In two triangles ABC and DEF, it is given that \u2220A = \u2220D, \u2220B = \u2220E and \u2220C = \u2220F. Are the two triangles necessarily congruent?<\/p>\n
No, as the triangles are equiangular, so similar.<\/p>\n
Question 3.Solution:
If ABC and DEF are two triangles such that AC = 2.5 cm, BC = 5 cm, \u2220C = 75\u00b0, DE = 2.5 cm, DF = 5 cm and \u2220D = 75\u00b0. Are two triangles congruent?<\/p>\n
Yes, triangles are congruent (SAS axiom)
RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula VSAQS \u2013 3<\/p>\n
Question 4.Solution:
In two triangles ABC and ADC, if AB = AD and BC = CD. Are they congruent?<\/p>\n
Yes, these are congruent
In two triangles ABC are ADC,
AB = AD (Given)
BC = CD (Given)
and AC = AC (Common)
\u2234 \u2206sABC \u2245 AADC (SSS axiom)
RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula VSAQS \u2013 4<\/p>\n
Question 5.Solution:
In triangles ABC and CDE, if AC = CE, BC = CD, \u2220A = 60\u00b0, \u2220C \u2013 30\u00b0 and \u2220D = 90\u00b0. Are two triangles congruent?<\/p>\n
Yes, triangles are congruent because,
In \u2206ABC, and \u2206CDE,
AC = CE
BC = CD \u2220C = 30\u00b0
\u2234 \u2206ABC \u2245 \u2206CDE (SAS axiom)
RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula VSAQS \u2013 5<\/p>\n
Question 6.Solution:
ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.<\/p>\n
Given : In \u2206ABC, AB = AC
BE and CF are two medians
RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula VSAQS \u2013 6
To prove : BE = CF
Proof: In \u2206ABE and \u2206ACF.
AB = AC (Given)
\u2220A = \u2220A (Common)
AE = AF (Half of equal sides)
\u2234 \u2206ABE \u2245 \u2206ACF (SAS axiom)
\u2234 BE = CF (c.p.c.t.)<\/p>\n
Question 7.Solution:
Find the measure of each angle of an equilateral triangle.<\/p>\n
In \u2206ABC,
AB = AC = BC
RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula VSAQS \u2013 7
\u2235 AB = AC
\u2234 \u2220C = \u2220B \u2026(i)
(Angles opposite to equal sides)
Similarly,
AC = BC
\u2234 \u2220B = \u2220A \u2026(ii)
From (i) and (ii),
\u2220A = \u2220B = \u2220C
But \u2220A + \u2220B + \u2220C = 180\u00b0
(Sum of angles of a triangle)
\u2234 \u2220A + \u2220B + \u2220C =\u00a0180<\/span><\/span><\/span>\u2218<\/span><\/span><\/span><\/span>3<\/span><\/span><\/span><\/span><\/span>\u00a0= 60\u00b0<\/p>\nQuestion 8.Solution:
CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that \u2206ADE \u2245 \u2206BCE.<\/p>\n
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\u2019s Formula VSAQS \u2013 8
To prove : \u2206ADE \u2245 \u2206BCE
Proof : In \u2206ADE and \u2206BCE,
AD = BC (Sides of a square)
DE = CE (Sides of equilateral triangle)
\u2220ADE = \u2220BCE(Each = 90\u00b0 + 60\u00b0 = 150\u00b0)
\u2234 AADE \u2245 ABCE (SAS axiom)<\/p>\n
Question 9.Solution:
Prove that the sum of three altitude of a triangle is less than the sum of its sides.<\/p>\n
Given : In \u2206ABC, AD, BE and CF are the altitude of \u2206ABC
RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula VSAQS \u2013 9
To prove : AD + BE + CF < AB + BC + CA
Proof : In right \u2206ABD, \u2220D = 90\u00b0
Then other two angles are acute
\u2235 \u2220B < \u2220D
\u2234 AD < AB \u2026(i)
Similarly, in \u2206BEC and \u2206ABE we can prove thatBE and CF < CA \u2026(iii)
Adding (i), (ii), (iii)
AD + BE -t CF < AB + BC + CA<\/p>\n
Question 10.Solution:
In the figure, if AB = AC and \u2220B = \u2220C. Prove that BQ = CP.
RD Sharma Class 9 Solutions Chapter 12 Heron\u2019s Formula VSAQS \u2013 10<\/p>\n
Given : In the figure, AB = AC, \u2220B = \u2220C
To prove : BQ = CP
Proof : In \u2206ABQ and \u2206ACP
AB = AC (Given)
\u2220A = \u2220A (Common)
\u2220B = \u2220C (Given)
\u2234 \u2206ABQ \u2245 \u2206ACP (ASA axiom)
\u2234 BQ = CP (c.p.c.t.)<\/p>\n
This is the complete blog on RD Sharma Class 9 Solutions Chapter 12 VSAQs. To know more about the CBSE<\/a> Class 9 Maths exam, ask in the comments.<\/p>\n
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-12-VSAQS.jpg\")
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