RD Chapter 12- Heron-s Formula Ex-12.1 |
RD Chapter 12- Heron-s Formula Ex-12.2 |
RD Chapter 12- Heron-s Formula Ex-12.3 |
RD Chapter 12- Heron-s Formula Ex-12.4 |
RD Chapter 12- Heron-s Formula Ex-12.6 |
RD Chapter 12- Heron-s Formula Ex-VSAQS |
RD Chapter 12- Heron-s Formula Ex-MCQS |

**Answer
1** :

Given : In ∆ABC, D is mid-point of BC and DE ⊥ AB, DF ⊥ AC and DE = DF

To Prove : ∆ABC is an isosceles triangle

Proof : In right ∆BDE and ∆CDF,

Side DE = DF

Hyp. BD = CD

∴ ∆BDE ≅ ∆CDF (RHS axiom)

∴ ∠B = ∠C (c.p.c.t.)

Now in ∆ABC,

∠B = ∠C (Prove)

∴ AC = AB (Sides opposite to equal angles)

∴ AABC is an isosceles triangle

**Answer
2** :

Given : In ∆ABC,

BE ⊥ AC and CF ⊥ AB

BE = CF

To prove : AABC is an isosceles triangle

Proof : In right ABCE and ABCF Side

BE = CF (Given)

Hyp. BC = BC (Common)

∴ ∆BCE ≅ ∆BCF (RHS axiom)

∴ ∠BCE = ∠CBF (c.p.c.t.)

∴ AB = AC (Sides opposite to equal angles)

∴ ∆ABC is an isosceles triangle

**Answer
3** :

Given : A point P lies in the angle ABC and PL ⊥ BA and PM ⊥ BC and PL = PM. PB is joined

To prove : PB is the bisector ∠ABC,

Proof : In right ∆PLB and ∆PMB

Side PL = PM (Given)

Hyp. PB = PB (Common)

∴ ∆PLB ≅ ∆PMB (RHS axiom)

∴ ∆PBL = ∆PBM (c.p.c.t.)

∴ PB is the bisector of ∠ABC

**Answer
4** :

Given : In the figure,

AD ⊥ CD and CB ⊥ CD, AQ = BP and DP = CQ

To prove : ∠DAQ = ∠CBP

Proof : ∵ DP = CQ

∴ DP + PQ = PQ + QC

⇒ DQ = PC

Now in right ∆ADQ and ∆BCP

Side DQ = PC (Proved)

Hyp. AQ = BP

∴ ∆ADQ ≅ ∆BCP (RHS axiom)

∴ ∠DAQ = ∠CBP (c.p.c.t.)

Which of the following statements are true (T) and which are false (F):

(i) Sides opposite to equal angles of a triangle may be unequal.

(ii) Angles opposite to equal sides of a triangle are equal.

(iii) The measure of each angle of an equilateral triangle is 60°.

(iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isosceles.

(v) The bisectors of two equal angles of a triangle are equal.

(vi) If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.

(vii) The two altitudes corresponding to two equal sides of a triangle need not be equal.

(viii)If any two sides of a right triangle are respectively equal to two sides of other right triangle, then the two triangles are congruent.

(ix) Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle.

**Answer
5** :

(i) False : Sides opposite to equal angles of a triangle are equal.

(ii) True.

(iii) True.

(iv) False : The triangle is an isosceles triangle.

(v) True.

(vi) False : The triangle is an isosceles.

(vii) False : The altitude an equal.

(viii) False : If one side and hypotenuse of one right triangle on one side and hypotenuse of the other right triangle are equal, then triangles are congruent.

(ix) True.

Fill in the blanks in the following so that each of the following statements is true.

(i) Sides opposite to equal angles of a triangle are …….

(ii) Angle opposite to equal sides of a triangle are …….

(iii) In an equilateral triangle all angles are …….

(iv) In a ∆ABC if ∠A = ∠C, then AB = …….

(v) If altitudes CE and BF of a triangle ABC are equal, then AB = ……..

(vi) In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is ……… CE.

(vii) In right triangles ABC and DEF, if hypotenuse AB = EF and side AC = DE, then ∆ABC ≅ ∆……

**Answer
6** :

(i) Sides opposite to equal angles of a triangle are equal.

(ii) Angle opposite to equal sides of a triangle are equal.

(iii) In an equilateral triangle all angles are equal.

(iv) In a ∆ABC, if ∠A = ∠C, then AB = BC.

(v) If altitudes CE and BF of a triangle ABC are equal, then AB = AC.

(vi) In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is equal to CE.

(vii) In right triangles ABC and DEF, it hypotenuse AB = EF and side AC = DE, then ∆ABC ≅ ∆EFD.

**Answer
7** :

Given : In square ABCD, X and Y are points on side AD and BC respectively and AY = BX

To prove : BY = AX

∠BAY = ∠ABX

Proof: In right ∆BAX and ∆ABY

AB =AB (Common)

Hyp. BX = AY (Given)

∴ ∆BAX ≅ ∆ABY (RHS axiom)

∴ AX = BY (c.p.c.t.)

∠ABX = ∠BAY (c.p.c.t.)

Hence, BY = AX and ∠BAY = ∠ABX.

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