NCERT Solutions for Class 9 Maths Chapter 7: Triangles
NCERT Solutions for Class 9 Mathematics Chapter 7 Free PDF Download
Please Click on Free PDF Download link to Download the NCERT Solutions for Class 9 Mathematics Chapter 7 Triangles
Q. Prove that measure of each angle an equilateral triangle is 60°.
To prove : ∠A = ∠B = ∠C = 60°
Proof : In DABC,
AB = BC
∠C = ∠A ...(i)
(Opposite angles of equal sides)
AB = AC
∠C = ∠B ...(ii)
From equations (i) and (ii), ∠A = ∠B = ∠C ...(iii)
In △ABC,
∠A + ∠B + ∠C = 180° (Sum of all angles of △)
⇒ 3∠A = 180°
⇒ ∠A = 60°
∠A = ∠B = ∠C = 60°.
Q. In the given figure, l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that △ABC ≅ △CDA.
Ans. Given : l || m and p || q
To prove : △ABC ≅ △CDA
Proof : l || m
∠ACB = ∠DAC ...(i)
(Alternative angles)
p || q
∠BAC = ∠ACD ...(ii)
(Alternativ
e angles)
In △ABC and △ADC,
∠ACB = ∠DAC (Proved above)
∠BAC = ∠ACD (Proved above)
AC = AC (Common)
△ABC ≅ △CDA (ASA congruency rule)
Q. AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B. Show that the line PQ is perpendicular bisector of AB.
Ans. Give : P and Q are equidistant from the point A and B respectively.
To prove : PQ is perpendicular bisector of AB
Proof : In DAPQ and DBPQ
AP = BP (Given)
AQ = BQ (Given)
PQ = PQ (Common)
△APQ ≅ △BPQ (SSS congruency rule)
∠APQ = ∠BPQ (CPCT)
AP = BP (Given)
OP = OP (Common)
△AOP ≅ △BOP (SAS congruency rule)
AO = BO (CPCT) ∠AOP = ∠BOP (CPCT)
∠AOP + ∠BOP = 180° (Linear pair)
∠AOP + ∠AOP = 180°
∠AOP = 90°
PQ ⊥ AB
Q. If the bisector of the vertical angle of a triangle bisects the base of a triangle then the triangle is isosceles.
Ans. Given : In DABC, AD is the angle bisector of ∠BAC and also bisects the base.
To prove : △ABC is an isosceles triangle.
Construction : Extend AD such that AD = DE and join
CE
Proof : In △ABD and △DCE,
∠ADB = ∠CDE
(Vertically opposite angles)
BD = DC (Given)
AD = DE (By construction)
△ABD ≅ △ECD (SAS congruency rule)
AB = CE (CPCT) ...(i)
∠BAD = ∠CED (CPCT) ...(ii)
∠BAD = ∠CAD (Given) ...(iii)
∠BAD = ∠CAD = ∠CED
[From (ii) and (iii)]
In △ACE, ∠CAE = ∠CEA
(Opposite sides of equal angles)
CE = AB [from (i)]
Form (iv) and (i),
AB = AC
△ABC is an isosceles triangle.
Q. DABC is an isosceles triangle with AB = AC, side BA is produced to D such that AB = AD. Prove that ∠BCD is a right angle.
Ans. Given : △ABC is an isosceles triangle and BA = AD.
To prove : ∠BCD = 90°
Proof : AB = AC
∠ABC = ∠ACB ...(i)
(Opposite angles of equal sides)
AD = AC
∠ACD = ∠ADC ...(ii)
(Opposite angles of equal sides)
∠CAD = ∠ABC + ∠ACB
(Exterior angle property)
⇒ ∠CAD = 2∠ACB (from (i)) ...(iii)
∠BAC = ∠ADC + ∠ACD
(Exterior angle property)
⇒ ∠BAC = 2∠ACD (from (ii)) ...(iv)
From (iii) and (iv)
2∠ACB + 2∠ACD = ∠CAD + ∠BAC
⇒ 2(∠ACB + ∠ACD) = ∠BAD
⇒ 2(∠BCD) = 180°
⇒ ∠BCD = 90°.
