# 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)

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.

**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`