## RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1

Other Exercises

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.

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)
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 mid points of the sides PQ, QR and PR respectively

To prove : LM = MN
Proof : In ∆LPN and ∆MRH
PN = RN (∵ M is mid point of PR)
LP = MR (Half of 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 triangle and AB = BC = CA

To prove : AD = BE = CF
Proof : In ∆BCE and ∆BCF,
BC = BC (Common side)
CE = BF (Half of 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
From (i) and (ii)
⇒ 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)

⇒ 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)

∴ ∠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°

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 is produced both ways

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.

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 rightangled isosceles triangle.
OR
ABC is a rightangled 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

∵ ∠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 a square and SRT is an equilateral triangle. Prove that
(i) PT = QT
(ii) ∠TQR = 15°

Solution:
Given : PQRS is a square and SRT is an equilateral triangle. PT and QT are joined.

To prove : (i) PT = QT; (ii) ∠TQR = 15°
Proof : In ∆TSP and ∆TQR
ST = RT (Sides of equilateral triangle)
SP = PQ (Sides of 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 the ponits A and B (see figure). Show that the line PQ is perpendicular bisector of AB.

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, PQ are joined
To prove : PQ is perpendicular bisector of AB

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

Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1 are helpful to complete your math homework.

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