RD Sharma Class 9 Solutions Chapter 12 Exercise 12.1 (Updated for 2023)

RD Sharma Class 9 Solutions Chapter 12 Exercise 12.1

RD Sharma Class 9 Solutions Chapter 12 Exercise 12.1: You can easily prepare for your upcoming Class 9 Maths exams with the RD Sharma Solutions Class 9 Maths. The download link for the PDF of RD Sharma Class 9 Solutions Chapter 12 Exercise 12.1 will be given in this blog. You can download the PDF for free and access it offline as well. To know more, read the whole blog.

Download RD Sharma Class 9 Solutions Chapter 12 Exercise 12.1 PDF

RD Sharma Class 9 Solutions Chapter 12 Exercise 12.1

 


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Question 1.
In the figure, the sides BA and CA have been produced such that BA = AD and CA = AE. Prove the segment DE || BC.RD Sharma Class 9 Chapter 12 Heron's Formula
Solution:
Given: Sides BA and CA of ∆ABC are produced such that BA = AD are CA = AE. ED is joined.
To prove: DE || BC
Proof: In ∆ABC and ∆DAE AB=AD (Given)
AC = AE (Given)
∠BAC = ∠DAE (Vertically opposite angles)
∴ ∆ABC ≅ ∆DAE (SAS axiom)
∴ ∠ABC = ∠ADE (c.p.c.t.)
But there are alternate angles
∴ DE || BC

Question 2.
In a ∆PQR, if PQ = QR and L, M, and N are the mid-points of the sides PQ, QR, and RP respectively. Prove that LN = MN.
Solution:
Given: In ∆PQR, PQ = QR
L, M, and N are the midpoints of the sides PQ, QR, and PR respectively
RD Sharma Class 9 Solutions Chapter 12 Heron's Formula
To prove: LM = MN
Proof: In ∆LPN and ∆MRH
PN = RN (∵ M is the midpoint of PR)
LP = MR (Half of the equal sides)
∠P = ∠R (Angles opposite to equal sides)
∴ ALPN ≅ AMRH (SAS axiom)
∴ LN = MN (c.p.c.t.)

Question 3.
Prove that the medians of an equilateral triangle are equal.
Solution:
Given: In ∆ABC, AD, BE and CF are the medians of the triangle, and AB = BC = CA
RD Sharma Solutions Class 9 Chapter 12 Heron's Formula
To prove : AD = BE = CF
Proof: In ∆BCE and ∆BCF,
BC = BC (Common side)
CE = BF (Half of the equal sides)
∠C = ∠B (Angles opposite to equal sides)
∴ ABCE ≅ ABCF (SAS axiom)
∴ BE = CF (c.p.c.t.) …(i)
Similarly, we can prove that
∴ ∆CAD ≅ ∆CAF
∴ AD = CF …(ii)
From (i) and (ii)
BE = CF = AD
⇒ AD = BE = CF

Question 4.
In a ∆ABC, if ∠A = 120° and AB = AC. Find ∠B and ∠C.
Solution:
In ∆ABC, ∠A = 120° and AB = AC
∴ ∠B = ∠C (Angles opposite to equal sides)
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
RD Sharma Class 9 PDF Chapter 12 Heron's Formula
⇒ 120° + ∠B + ∠B = 180°
⇒ 2∠B = 180° – 120° = 60°
∴ ∠B = \(\frac { { 60 }^{ \circ } }{ 2 }\) = 30°
and ∠C = ∠B = 30°
Hence ∠B = 30° and ∠C = 30°

Question 5.
In a ∆ABC, if AB = AC and ∠B = 70°, find ∠A.
Solution:
In ∆ABC, ∠B = 70°
AB =AC
∴ ∠B = ∠C (Angles opposite to equal sides)
But ∠B = 70°
∴ ∠C = 70°
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ ∠A + 70° + 70° = 180°
⇒ ∠A + 140°= 180°
∴∠A = 180°- 140° = 40°

Question 6.
The vertical angle of an isosceles triangle is 100°. Find its base angles.
Solution:
In ∆ABC, AB = AC and ∠A = 100°
But AB = AC (In isosceles triangle)
RD Sharma Class 9 Solution Chapter 12 Heron's Formula
∴ ∠C = ∠B (Angles opposite to equal sides)
∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ 100° + ∠B + ∠B = 180° (∵ ∠C = ∠B)
⇒ 2∠B = 180° – 100° = 80°
∴ ∠C = ∠B = 40°
Hence ∠B = 40°, ∠C = 40°

Question 7.
In the figure, AB = AC and ∠ACD = 105°, find ∠BAC.
Solution:
In ∆ABC, AB = AC
∴ ∠B = ∠C (Angles opposite to equal sides)
But ∠ACB + ∠ACD = 180° (Linear pair)
⇒ ∠ACB + 105°= 180°
⇒ ∠ACB = 180°-105° = 75°
∴ ∠ABC = ∠ACB = 75°
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ ∠A + 75° + 75° = 180°
⇒ ∠A + 150°= 180°
⇒ ∠A= 180°- 150° = 30°
∴ ∠BAC = 30°

Question 8.
Find the measure of each exterior angle of an equilateral triangle.
Solution:
In an equilateral triangle, each interior angle is 60°
Class 9 Maths Chapter 12 Heron's Formula RD Sharma Solutions
But interior angle + exterior angle at each vertex = 180°
∴ Each exterior angle = 180° – 60° = 120°

Question 9.
If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.
Solution:
Given : In an isosceles ∆ABC, AB = AC
and base BC has produced both ways
RD Sharma Book Class 9 PDF Free Download Chapter 12 Heron's Formula
To prove: ∠ACD = ∠ABE
Proof: In ∆ABC,
∵ AB = AC
∴∠C = ∠B (Angles opposite to equal sides)
⇒ ∠ACB = ∠ABC
But ∠ACD + ∠ACB = 180° (Linear pair)
and ∠ABE + ∠ABC = 180°
∴ ∠ACD + ∠ACB = ∠ABE + ∠ABC
But ∠ACB = ∠ABC (Proved)
∴ ∠ACD = ∠ABE
Hence proved.

Question 10.
In the figure, AB = AC and DB = DC, find the ratio ∠ABD: ∠ACD.
RD Sharma Class 9 Book Chapter 12 Heron's Formula
Solution:
In the given figure,
In ∆ABC,
AB = AC and DB = DC
In ∆ABC,
∵ AB = AC
∴ ∠ACD = ∠ABE …(i) (Angles opposite to equal sides)
Similarly, in ∆DBC,
DB = DC
∴ ∠DCB = ∠DBC .. (ii)
Subtracting (ii) from (i)
∠ACB – ∠DCB = ∠ABC – ∠DBC
⇒ ∠ACD = ∠ABD
∴ Ratio ∠ABD : ∠ACD = 1 : 1

Question 11.
Determine the measure of each of the equal angles of a right-angled isosceles triangle.
OR
ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Solution:
Given: In a right angled isosceles ∆ABC, ∠A = 90° and AB = AC
To determine, each equal angle of the triangle

Heron's Formula With Solutions PDF RD Sharma Class 9 Solutions
∵ ∠A = 90°
∴ ∠B + ∠C = 90°
But ∠B = ∠C
∴ ∠B + ∠B = 90°
⇒ 2∠B = 90°
90°
⇒ ∠B = \(\frac { { 90 }^{ \circ } }{ 2 }\)  = 45°
and ∠C = ∠B = 45°
Hence ∠B = ∠C = 45°

Question 12.
In the figure, PQRS is square and SRT is an equilateral triangle. Prove that
(i) PT = QT
(ii) ∠TQR = 15°
RD Sharma Class 9 Maths Book Questions Chapter 12 Heron's Formula
Solution:
Given: PQRS is square and SRT is an equilateral triangle. PT and QT are joined.
RD Sharma Mathematics Class 9 Solutions Chapter 12 Heron's Formula
To prove : (i) PT = QT; (ii) ∠TQR = 15°
Proof: In ∆TSP and ∆TQR
ST = RT (Sides of an equilateral triangle)
SP = PQ (Sides of the square)
and ∠TSP = ∠TRQ (Each = 60° + 90°)
∴ ∆TSP ≅ ∆TQR (SAS axiom)
∴ PT = QT (c.p.c.t.)
In ∆TQR,
∵ RT = RQ (Square sides)
∠RTQ = ∠RQT
But ∠TRQ = 60° + 90° = 150°
∴ ∠RTQ + ∠RQT = 180° – 150° = 30°
∵ ∠PTQ = ∠RQT (Proved)
∠RQT = \(\frac { { 30 }^{ \circ } }{ 2 }\)  = 15°
⇒ ∠TQR = 15°

Question 13.
AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from points A and B (see figure). Show that the line PQ is the perpendicular bisector of AB.
Solution Of Rd Sharma Class 9 Chapter 12 Heron's Formula
Solution:
Given: AB is a line segment.
P and Q are points such that they are equidistant from A and B
i.e. PA = PB and QA = QB AP, PB, QA, QB, and PQ are joined
To prove : PQ is perpendicular bisector of AB
RD Sharma Math Solution Class 9 Chapter 12 Heron's Formula
Proof: In ∆PAQ and ∆PBQ,
PA = PB (Given)
QA = QB (Given)
PQ = PQ (Common)
∴ ∆PAQ ≅ ∆PBQ (SSS axiom)
∴ ∠APQ = ∠BPQ (c.p.c.t.)
Now in ∆APC = ∆BPC
PA = PB (Given)
∆APC ≅ ∆BPC (Proved)
PC = PC (Common)
∴ ∆APC = ∆BPC (SAS axiom)
∴ AC = BC (c.p.c.t.)
and ∠PCA = ∠PCB (c.p.c.t.)
But ∠PCA + ∠PCB = 180° (Linear pair)
∴ ∠PCA = ∠PCB = 90°
∴ PC or PQ is perpendicular bisector of AB.

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