RD Chapter 12 Heron s Formula Ex VSAQS Solutions
Question - 1 : - In two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.
Answer - 1 : - 
In AABC and ADEF,
∆ABC ≅ ∆DEF
and AB = DE, BC = EF
∴ ∠A = ∠D, ∠B = ∠E and ∠C = ∠F
Question - 2 : - In two triangles ABC and DEF, it is given that ∠A = ∠D, ∠B = ∠E and ∠C = ∠F. Are the two triangles necessarily congruent?
Answer - 2 : -
No, as the triangles are equiangular, so similar.
Answer - 3 : - Yes, triangles are congruent (SAS axiom)
Question - 4 : - In two triangles ABC and ADC, if AB = AD and BC = CD. Are they congruent?
Answer - 4 : -
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 : - In triangles ABC and CDE, if AC = CE, BC = CD, ∠A = 60°, ∠C – 30° and ∠D = 90°. Are two triangles congruent?
Answer - 5 : -
Yes, triangles are congruent because,
In ∆ABC, and ∆CDE,
AC = CE
BC = CD ∠C = 30°
∴ ∆ABC ≅ ∆CDE (SAS axiom)
Question - 6 : - ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.
Answer - 6 : -
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 : - Find the measure of each angle of an equilateral triangle.
Answer - 7 : -
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 =
= 60°
Question - 8 : - CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that ∆ADE ≅ ∆BCE.
Answer - 8 : -
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 : - Prove that the sum of three altitude of a triangle is less than the sum of its sides.
Answer - 9 : -
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 : - In the figure, if AB = AC and ∠B = ∠C. Prove that BQ = CP.
Answer - 10 : -
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.)