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

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

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.

Solution:

In two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.

In AABC and ADEF,

∆ABC ≅ ∆DEF

and AB = DE, BC = EF

∴ ∠A = ∠D, ∠B = ∠E and ∠C = ∠F

Question 2.

Solution:

In two triangles ABC and DEF, it is given that ∠A = ∠D, ∠B = ∠E and ∠C = ∠F. Are the two triangles necessarily congruent?

No, as the triangles are equiangular, so similar.

Question 3.

Solution:

If ABC and DEF are two triangles such that AC = 2.5 cm, BC = 5 cm, ∠C = 75°, DE = 2.5 cm, DF = 5 cm and ∠D = 75°. Are two triangles congruent?

Yes, triangles are congruent (SAS axiom)

Question 4.

Solution:

In two triangles ABC and ADC, if AB = AD and BC = CD. Are they congruent?

Yes, these are congruent

In two triangles ABC are ADC,

AB = AD (Given)

BC = CD (Given)

and AC = AC (Common)

∴ ∆sABC ≅ AADC (SSS axiom)

Question 5.

Solution:

In triangles ABC and CDE, if AC = CE, BC = CD, ∠A = 60°, ∠C – 30° and ∠D = 90°. Are two triangles congruent?

Yes, triangles are congruent because,

In ∆ABC, and ∆CDE,

AC = CE

BC = CD ∠C = 30°

∴ ∆ABC ≅ ∆CDE (SAS axiom)

Question 6.

Solution:

ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.

Given : In ∆ABC, AB = AC

BE and CF are two medians

To prove : BE = CF

Proof: In ∆ABE and ∆ACF.

AB = AC (Given)

∠A = ∠A (Common)

AE = AF (Half of equal sides)

∴ ∆ABE ≅ ∆ACF (SAS axiom)

∴ BE = CF (c.p.c.t.)

Question 7.

Solution:

Find the measure of each angle of an equilateral triangle.

In ∆ABC,

AB = AC = BC

∵ AB = AC

∴ ∠C = ∠B …(i)

(Angles opposite to equal sides)

Similarly,

AC = BC

∴ ∠B = ∠A …(ii)

From (i) and (ii),

∠A = ∠B = ∠C

But ∠A + ∠B + ∠C = 180°

(Sum of angles of a triangle)

∴ ∠A + ∠B + ∠C = \(\frac { { 180 }^{ \circ } }{ 3 }\) = 60°

Question 8.

Solution:

CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that ∆ADE ≅ ∆BCE.

Given : An equilateral ACDE is formed on the side of square ABCD. AE and BE are joined

To prove : ∆ADE ≅ ∆BCE

Proof : In ∆ADE and ∆BCE,

AD = BC (Sides of a square)

DE = CE (Sides of equilateral triangle)

∠ADE = ∠BCE(Each = 90° + 60° = 150°)

∴ AADE ≅ ABCE (SAS axiom)

Question 9.

Solution:

Prove that the sum of three altitude of a triangle is less than the sum of its sides.

Given : In ∆ABC, AD, BE and CF are the altitude of ∆ABC

To prove : AD + BE + CF < AB + BC + CA

Proof : In right ∆ABD, ∠D = 90°

Then other two angles are acute

∵ ∠B < ∠D

∴ AD < AB …(i)

Similarly, in ∆BEC and ∆ABE we can prove thatBE and CF < CA …(iii)

Adding (i), (ii), (iii)

AD + BE -t CF < AB + BC + CA

Question 10.

Solution:

In the figure, if AB = AC and ∠B = ∠C. Prove that BQ = CP.

Given : In the figure, AB = AC, ∠B = ∠C

To prove : BQ = CP

Proof : In ∆ABQ and ∆ACP

AB = AC (Given)

∠A = ∠A (Common)

∠B = ∠C (Given)

∴ ∆ABQ ≅ ∆ACP (ASA axiom)

∴ BQ = CP (c.p.c.t.)

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

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