## 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

- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.1
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.2
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.4
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.5
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.6
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula VSAQS
- RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula MCQS

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)

∴ ∠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 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

∴ ∆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)

⇒ 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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