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

The dot mark field are mandatory, So please fill them in carefully
To download the complete Syllabus (PDF File), Please fill & submit the form below.

    Q. Prove that measure of each angle an equilateral triangle is 60°.

    Ans. Given : DABC is an equilateral triangle.
    To prove : ∠A = ∠B = ∠C = 60°
    equilateral triangle

    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.

    parallel lines

    Ans. Given : l || m and p || q
    To prove : △ABC △CDA
    Proof : l || m
    ∠ACB = ∠DAC ...(i)
    (Alternative angles)
    p || q
    ∠BAC = ∠ACD ...(ii)
    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.

    triangle bisects

    To prove : ABC is an isosceles triangle.
    Construction : Extend AD such that AD = DE and join
    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.

    isosceles triangle

    Ans. Given : △ABC is an isosceles triangle and BA = AD.

    isosceles triangle 2

    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°.

    Share page on