NCERT Class 9 Maths Solutions Chapter - 7 Triangles, Ex - 7.4

Ex - 7.4

Question 1.  Show that in a right angled triangle, the hypotenuse is the longest side.

Solution


Let us consider a right angled triangle ABC, right angle at B. 
In ABC
A + B + C = 180^o            (angle sum property of a triangle) 
A + 90^o + C = 180^o
A + C = 90^o
Hence, the other two angles have to be acute (i.e. less than 90�).

[In any triangle, the side opposite to the larger (greater) angle is longer]
So, AC is the largest side in ABC.
But AC is the hypotenuse of ABC. Therefore, hypotenuse is the longest side in a right angled triangle.

Question 2.  In the given figure sides AB and AC of ABC are extended to points P and Q respectively. Also, PBC  QCB. Show that AC > AB.


Solution

In the given figure,
ABC + PBC = 180^p            (linear pair)
 ABC = 180^o - PBC             ... (1)
Also,
ACB + QCB = 180^o
ACB = 180^o - QCB                    ... (2)
As PBC < QCB
  180� - PBC > 180^o - QCB.
 ABC > ACB                [From equations (1) and (2)]
 AC > AB                           (side opposite to larger angle is larger)

Question 3.  In the given figure, B < A and C < D. Show that AD < BC.


Solution

In AOB
B < A
 AO < BO     (side opposite to smaller angle is smaller)        ... (1)
Now in COD
C < D
 OD < OC     (side opposite to smaller angle is smaller)        ... (2)
On adding equations (1) and (2), we have
AO + OD < BO + OC
AD < BC

Question 4.  AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that A > C and B > D.


Solution

Let us join AC.
In ABC
AB < BC           (AB is smallest side of quadrilateral ABCD)

(1)

In ADC
AD < CD          (CD is the largest side of quadrilateral ABCD)

(2)

On adding equations (1) and (2), we have
2 + 4 < 1 + 3
 C < A
 A > C
Let us join BD.

In ABD
AB < AD            (AB is smallest side of quadrilateral ABCD)

(3)

In BDC
 BC < CD         (CD is the largest side of quadrilateral ABCD)

On adding equations (3) and (4), we have
8 + 7 < 5 + 6
 D < B
 B > D

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

Solution

  As PR > PQ

  PS is the bisector of QPR

  

   

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

Solution

Let us take a line l and from point P (i.e. not on line l) we have drawn two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.
In PNM
N = 90^o
Now, P + N + M = 180^o    (Angle sum property of a triangle)    
P + M = 90^o             
Clearly, M is an acute angle

Ex-7.3

Ex-7.5