NCERT Class 9 Solutions: Triangles (Chapter 7) Exercise 7.4 – Part 3

Interior Angle and Exterior Angle

Give Interior Angle and Exterior Angle ,interior angle is and exterior angle is 150

Give Interior Angle and Exterior Angle

Give Interior Angle and Exterior Angle ,interior angle is and exterior angle is 150

Interior Angle: interior angle (or internal angle) if a point within the angle is in the interior of the polygon. A polygon has exactly one internal angle per vertex.

Exterior Angle: exterior angle (or external angle) is an angle formed by one side of a simple polygon and a line extended an adjacent side

Q-5 In the figure ,PR>PQ and PS bisects QPR . Prove that PSR>PSQ.

Give triangle of PQR and PS is bisects also PR > PQ

Give Triangle of PQR and PS Is Bisects

Give triangle of PQR and PS is bisects also PR > PQ

Solution:

Given, PR>PQ and PS bisects QPR

  • To prove,

    PSR>PSQ

  • Proof,

    PQR>PRQ ……equation (1) ( PR>PQ as angle opposite to larger side is larger.) QPS=RPS ……equation (2) (PS bisects QPR )

    PSR=PQR+QPS …..equation (3) (exterior angle of a triangle equals to the sum of opposite interior angles) PSQ=PRQ+RPS ….equation(4) (exterior angle of a triangle equals to the sum of opposite interior angles)

  • Adding (1) and (2) PQR+QPS>PRQ+RPS PSR>PSQ [from (1), (2), (3) and (4)]

Q-6 Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Solution:

Give triangle of ABC ,let l is a line segment and B is a point lying.Line AB perpendicular to l.

Give Triangle of ABC

Give triangle of ABC ,let l is a line segment and B is a point lying.Line AB perpendicular to l.

  • Let l is a line segment and B is a point lying .We drew a line AB perpendicular to l. Let C be any other point on l.

  • To prove,

    AB<AC

  • Proof,In ΔABC , B=90°

  • Now, A+B+C=180° A+C=90° ( B=90) C must be acute angle. (Angle sum property of a triangle)

    or C<B

    AB<AC (Side opposite to the larger angle is larger.)

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