NCERT Solutions for Exercise 7.4 Class 9 Maths Chapter 7 - Triangles

Triangles Class 9 Chapter 7 Exercise: 7.4

Q1 Show that in a right-angled triangle, the hypotenuse is the longest side.

Answer:

Consider a right-angled triangle ABC with right angle at A.

We know that the sum of the interior angles of a triangle is 180.

So, $\angle A\ +\ \angle B\ +\ \angle C\ =\ 180^{\circ}$

or $90^{\circ}\ +\ \angle B\ +\ \angle C\ =\ 180^{\circ}$

or $\angle B\ +\ \angle C\ =\ 90^{\circ}$

Hence $\angle B$ and $\angle C$ are less than $\angle A$ ( $90^{\circ}$ ).

Also, the side opposite to the largest angle is also the largest.

Hence the side BC is largest is the hypotenuse of the $\Delta ABC$ .

Hence it is proved that in a right-angled triangle, the hypotenuse is the longest side.

Q2 In Fig, sides AB and AC of $\small \Delta ABC$ are extended to points P and Q respectively. Also, $\small \angle PBC < \angle QCB$ . Show that $\small AC> AB$ .

Answer:

We are given that,

$\small \angle PBC < \angle QCB$ ......................(i)

Also, $\angle ABC\ +\ \angle PBC\ =\ 180^{\circ}$ (Linear pair of angles) .....................(ii)

and $\angle ACB\ +\ \angle QCB\ =\ 180^{\circ}$ (Linear pair of angles) .....................(iii)

From (i), (ii) and (iii) we can say that :

$\angle ABC\ > \ \angle ACB$

Thus $AC\ > AB$ ( Sides opposite to the larger angle is larger.)

Q3 In Fig., $\small \angle B <\angle A$ and $\small \angle C <\angle D$ . Show that $\small AD <BC$ .

Answer:

In this question, we will use the property that sides opposite to larger angle are larger.

We are given $\small \angle B <\angle A$ and $\small \angle C <\angle D$ .

Thus, $BO\ > AO$ ..............(i)

and $OC\ > OD$ ...............(ii)

Adding (i) and (ii), we get :

$AO\ +\ OD\ <\ BO\ +\ OC$

or $AD\ <\ BC$

Hence proved.

Q4 AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig.). Show that $\small \angle A>\angle C$ and $\small \angle B>\angle D$ .

Answer:

Consider $\Delta ADC$ in the above figure :

$AD\ <\ CD$ (Given)

Thus $\angle CAD\ > \angle ACD$ (as angle opposite to smaller side is smaller)

Now consider $\Delta ABC$ ,

We have : $BC\ > AB$

and $\angle BAC\ > \angle ACB$

Adding the above result we get,

$\angle BAC\ +\ \angle CAD > \angle ACB\ +\ \angle ACD$

or $\small \angle A>\angle C$

Similarly, consider $\Delta ABD$ ,

we have $AB\ <\ AD$

Therefore $\angle ABD\ > \angle ADB$

and in $\Delta BDC$ we have,

$CD\ >\ BC$

and $\angle CBD\ >\ \angle CDB$

from the above result we have,

$\angle ABD\ +\ \angle CBD\ >\ \angle ADB\ +\ \angle CDB$

or $\small \angle B>\angle D$

Hence proved.

Q5 In Fig , $\small PR>PQ$ and PS bisects $\small \angle QPR$ . Prove that $\small \angle PSR>\angle PSQ$ .

Answer:

We are given that $\small PR>PQ$ .

Thus $\angle PQR\ =\ \angle PRQ$

Also, PS bisects $\small \angle QPR$ , thus :

$\angle QPS\ =\ \angle RPS$

Now, consider $\Delta QPS$ ,

$\angle PSR\ =\ \angle PQR\ +\ \angle QPS$ (Exterior angle)

Now, consider $\Delta PSR$ ,

$\angle PSQ\ =\ \angle PRQ\ +\ \angle RPS$

Thus from the above the result we can conclude that :

$\small \angle PSR>\angle PSQ$

Q6 Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Answer:

Consider a right-angled triangle ABC with right angle at B.

Then $\angle B\ >\ \angle A\ or\ \angle C$ (Since $\angle B\ =\ 90^{\circ}$ )

Thus the side opposite to largest angle is also largest. $AC\ >\ BC\ or\ AB$

Hence the given statement is proved that all line segments are drawn from a given point, not on it, the perpendicular line segment is the shortest.