In this chapter, we provide RD Sharma Solutions for Chapter 12 Heron’s Formula Ex 12.1 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 Ex 12.1 Maths pdf, free RD Sharma Solutions for Chapter 12 Heron’s Formula Ex 12.1 Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 9 |

Subject | Maths |

Chapter | Chapter 12 |

Chapter Name | Heron’s Formula |

Exercise | Ex 12.1 |

**RD Sharma Solutions for Class 9 Chapter 12 Heron’s Formula Ex 12.1 Download PDF**

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 = 60∘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 = 90∘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 = 30∘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

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