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.

congruence triangle 1

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.

congruence triangle 3

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.

congruence triangle 4

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)
in ΔADC
`∠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.

congruence triangle 6

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`



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