Triangles

Exercise 7.4

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

Answer: In a right angled triangle, the angle opposite To the hypotenuse is 90°, while other two angles are Always less than 90°. As you know that the side opposite to the largest angle is always the largest in a triangle.

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

Answer: ∠ABC=180°-∠PBC
∠ACB=180°-∠OCB
Since ∠PBC <∠OCB
So, ∠ABC >∠ACB

As you know side opposite to the larger angle is larger than the side opposite to the smaller angle.
Hence, AC > AB

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

Answer: AO < BO (Side opposite to smaller angle)
DO < CO (Side opposite to smaller angle)
So, AO+DO < BO+CO
Or, AD < BC

Question 4: AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that
angle A > angle C and angle B > angle D.

Answer: Let us draw two diagonals BD and AC as shown in the figure.
In ΔABD Sides AB < AD < BD

So, ∠ADB<∠ABD --------(1)
(angle opposite to smaller side is smaller)
In ΔBCD sides BC<DC<BD
So, ∠BDC<∠CBD ---------(2)

Adding equations (1) and (2)
∠ADB+∠BDC<∠ABD+∠CBD
Or, ∠ADC<∠ABC

Similarly, in ΔABC
∠BAC>∠ACD ---------(3)
∠DAC>∠DCA --------(4)

Adding equations (3) and (4)
∠BAC+∠DAC&gat;∠ACB+∠DCA
Or, ∠BAD>∠BCD

Question 5: In following figure, PR > PQ and PS bisects angle QPR. Prove that angle PSR > angle PSQ.

Answer: for convenience let us name these angles as follows:
∠PQR=1 ∠PRQ=2 ∠QPR=3 ∠QPS=4
∠RPS=5 ∠PSQ=6 ∠PSR=7
Since PR>PQ, so ∠1>∠2
In ΔPQS
∠1+∠4+∠6=180°
In ΔPRS
∠2+∠5+∠7=180°
In both these triangles
∠4=∠5
∠1>∠2
So, for making the sum total equal to 180° the following will always be true:
∠6<∠7