NCERT Solutions for Class 9 Maths Chapter 7 Triangles

NCERT solutions for class 10 maths chapter 7 Triangles - Excercise: 7.1

Q1 In quadrilateral $\small ABCD$ , $\small AC=AD$ and $\small AB$ bisects $\small \angle A$ (see Fig.). Show that $\small \Delta ABC\cong \Delta ABD$ . What can you say about $\small BC$ and $\small BD$ ?

Answer:

In the given triangles we are given that:-

(i) $\small AC=AD$

(ii) Further, it is given that AB bisects angle A. Thus $\angle$ BAC $=\ \angle$ BAD.

(iii) Side AB is common in both the triangles. $AB=AB$

Hence by SAS congruence, we can say that : $\small \Delta ABC\cong \Delta ABD$

By c.p.c.t. (corresponding parts of congruent triangles are equal) we can say that $BC\ =\ BD$

Q2 (i) $ABCD$ is a quadrilateral in which $\small AD=BC$ and $\small \angle DAB= \angle CBA$ (see Fig. ). Prove that

$\small \Delta ABD\cong \Delta BAC$ .

Answer:

It is given that :-

(i) AD = BC

(ii) $\small \angle DAB= \angle CBA$

(iii) Side AB is common in both the triangles.

So, by SAS congruence, we can write :

$\small \Delta ABD\cong \Delta BAC$

Q2 (ii) $\small ABCD$ is a quadrilateral in which $\small AD=BC$ and $\small \angle DAB=\angle CBA$ (see Fig.). Prove that $\small BD=AC$

Answer:

In the previous part, we have proved that $\small \Delta ABD\cong \Delta BAC$ .

Thus by c.p.c.t. , we can write : $\small BD=AC$

Q2 (iii) $\small ABCD$ is a quadrilateral in which $\small AD=BC$ and $\small \angle DAB= \angle BAC$ (see Fig.). Prove that $\small \angle ABD= \angle BAC$ .

Answer:

In the first part we have proved that $\small \Delta ABD\cong \Delta BAC$ .

Thus by c.p.c.t. , we can conclude :

$\small \angle ABD= \angle BAC$

Q3 $\small AD$ and $\small BC$ are equal perpendiculars to a line segment $\small AB$ (see Fig.). Show that $\small CD$ bisects $\small AB$ .

Answer:

In the given figure consider $\Delta$ AOD and $\Delta$ BOC.

(i) AD = BC (given)

(ii) $\angle$ A = $\angle$ B (given that the line AB is perpendicular to AD and BC)

(iii) $\angle$ AOD = $\angle$ BOC (vertically opposite angles).

Thus by AAS Postulate, we have

$\Delta AOD\ \cong \ \Delta BOC$

Hence by c.p.c.t. we can write : $AO\ =\ OB$

And thus CD bisects AB.

Q4 $\small l$ and $\small m$ are two parallel lines intersected by another pair of parallel lines $\small p$ and $\small q$ (see Fig. ). Show that $\small \Delta ABC\cong \Delta CDA$ .

Answer:

In the given figure, consider $\Delta$ ABC and $\Delta$ CDA :

(i) $\angle\ BCA\ =\ \angle DAC$

(ii) $\angle\ BAC\ =\ \angle DCA$

(iii) Side AC is common in both the triangles.

Thus by ASA congruence, we have :

$\Delta ABC\ \cong \ \Delta CDA$

Q5 (i) Line $\small l$ is the bisector of an angle $\small \angle A$ and B is any point on $\small l$ . $\small BP$ and $\small BQ$ are perpendiculars from $\small B$ to the arms of $\small \angle A$ (see Fig.). Show that: $\small \Delta APB\cong \Delta AQB$

Answer:

In the given figure consider $\small \Delta APB$ and _{$\small \Delta AQB$} ,

(i) $\angle P\ =\ \angle Q$ (Right angle)

(ii) $\angle BAP\ =\ \angle BAQ$ (Since it is given that I is bisector)

(iii) Side AB is common in both the triangle.

Thus AAS congruence, we can write :

$\small \Delta APB\cong \Delta AQB$

Q5 (ii) Line $\small l$ is the bisector of an angle $\small \angle A$ and $\small B$ is any point on $\small l$ . $\small BP$ and $\small BQ$ are perpendiculars from $\small B$ to the arms of $\small \angle A$ (see Fig. ). Show that: $\small BP=BQ$ or $\small B$ is equidistant from the arms of $\small \angle A$ .

Answer:

In the previous part we have proved that _{$\small \Delta APB\cong \Delta AQB$} .

Thus by c.p.c.t. we can write :

$BP\ =\ BQ$

Thus B is equidistant from arms of angle A.

Q6 In Fig, _{$\small AC=AE,AB=AD$} and $\small \angle BAD= \angle EAC$ . Show that $\small BC=DE$ .

Answer:

From the given figure following result can be drawn:-

$\angle BAD\ =\ \angle EAC$

Adding $\angle DAC$ to the both sides, we get :

$\angle BAD\ +\ \angle DAC\ =\ \angle EAC\ +\ \angle DAC$

$\angle BAC\ =\ \angle EAD$

Now consider $\Delta ABC$ and $\Delta ADE$ , :-

(i) $AC\ =\ AE$ (Given)

(ii) $\angle BAC\ =\ \angle EAD$ (proved above)

(iii) $AB\ =\ AD$ (Given)

Thus by SAS congruence we can say that :

$\Delta ABC\ \cong \ \Delta ADE$

Hence by c.p.c.t., we can say that : $BC\ =\ DE$

Q7 (i) $\small AB$ is a line segment and $\small P$ is its mid-point. $\small D$ and $\small E$ are points on the same side of $\small AB$ such that $\small \angle BAD=\angle ABE$ and $\small \angle EPA=\angle DPB$ (see Fig). Show that $\small \Delta DAP\cong \Delta EBP$

Answer:

From the figure, it is clear that :
$\angle EPA\ =\ \angle DPB$

Adding $\angle DPE$ both sides, we get :

$\angle EPA\ +\ \angle DPE =\ \angle DPB\ +\ \angle DPE$

or $\angle DPA =\ \angle EPB$

Now, consider $\Delta DAP$ and $\Delta EBP$ :

(i) $\angle DPA =\ \angle EPB$ (Proved above)

(ii) $AP\ =\ BP$ (Since P is the midpoint of line AB)

(iii) $\small \angle BAD=\angle ABE$ (Given)

Hence by ASA congruence, we can say that :

$\small \Delta DAP\cong \Delta EBP$

Q7 (ii) AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that $\small \angle BAD=\angle ABE$ and $\small \angle EPA=\angle DPA$ (see Fig). Show that $\small AD=BE$

Answer:

In the previous part we have proved that $\small \Delta DAP\cong \Delta EBP$ .

Thus by c.p.c.t., we can say that :

$\small AD=BE$

Q8 (i) In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that $\small DM=CM$ . Point D is joined to point B (see Fig.). Show that: $\small \Delta AMC\cong \Delta BMD$

Answer:

Consider $\Delta AMC$ and $\Delta BMD$ ,

(i) $AM\ =\ BM$ (Since M is the mid-point)

(ii) $\angle CMA\ =\ \angle DMB$ (Vertically opposite angles are equal)

(iii) $CM\ =\ DM$ (Given)

Thus by SAS congruency, we can conclude that :

$\small \Delta AMC\cong \Delta BMD$

Q8 (ii) In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that $\small DM=CM$ . Point D is joined to point B (see Fig.). Show that: $\small \angle DBC$ is a right angle.

Answer:

In the previous part, we have proved that $\small \Delta AMC\cong \Delta BMD$ .

By c.p.c.t. we can say that : $\angle ACM\ =\ \angle BDM$

This implies side AC is parallel to BD.

Thus we can write : $\angle ACB\ +\ \angle DBC\ =\ 180^{\circ}$ (Co-interior angles)

and, $90^{\circ}\ +\ \angle DBC\ =\ 180^{\circ}$

or $\angle DBC\ =\ 90^{\circ}$

Hence $\small \angle DBC$ is a right angle.

Q8 (iii) In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that $\small DM=CM$ . Point D is joined to point B (see Fig.). Show that: $\small \Delta DBC\cong \Delta ACB$

Answer:

Consider $\Delta DBC$ and $\Delta ACB$ ,

(i) $BC\ =\ BC$ (Common in both the triangles)

(ii) $\angle ACB\ =\ \angle DBC$ (Right angle)

(iii) $DB\ =\ AC$ (By c.p.c.t. from the part (a) of the question.)

Thus SAS congruence we can conclude that :

$\small \Delta DBC\cong \Delta ACB$

Q8 (iv) In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that $\small DM=CM$ . Point D is joined to point B (see Fig.). Show that: $\small CM=\frac{1}{2}AB$

Answer:

In the previous part we have proved that $\Delta DBC\ \cong \ \Delta ACB$ .

Thus by c.p.c.t., we can write : $DC\ =\ AB$

$DM\ +\ CM\ =\ AM\ +\ BM$

or $CM\ +\ CM\ =\ AB$ (Since M is midpoint.)

or $\small CM=\frac{1}{2}AB$ .

Hence proved.

NCERT Class 9 Maths Chapter 7 Question Answer - Excercise: 7.2

Q1 (i) In an isosceles triangle ABC, with $\small AB=AC$ , the bisectors of $\small \angle B$ and $\small \angle C$ intersect each other at O. Join A to O. Show that : $\small OB=OC$

Answer:

In the triangle ABC,

Since AB = AC, thus $\angle B\ =\ \angle C$

or $\frac{1}{2}\angle B\ =\ \frac{1}{2}\angle C$

or $\angle OBC\ =\ \angle OCB$ (Angles bisectors are equal)

Thus $\small OB=OC$ as sides opposite to equal are angles are also equal.

Q1 (ii) In an isosceles triangle ABC, with $\small AB=AC$ , the bisectors of $\small \angle B$ and $\small \angle C$ intersect each other at O. Join A to O. Show that : AO bisects $\small \angle A$

Answer:

Consider $\Delta AOB$ and $\Delta AOC$ ,

(i) $AB\ =\ AC$ (Given)

(ii) $AO\ =\ AO$ (Common in both the triangles)

(iii) $OB\ =\ OC$ (Proved in previous part)

Thus by SSS congruence rule, we can conclude that :

$\Delta AOB\ \cong \ \Delta AOC$

Now, by c.p.c.t.,

$\angle BAO\ =\ \angle CAO$

Hence AO bisects $\angle A$ .

Q2 In $\Delta ABC$ , AD is the perpendicular bisector of BC (see Fig). Show that $\small \Delta ABC$ is an isosceles triangle in which $\small AB=AC$ .

Answer:

Consider $\Delta$ ABD and $\Delta$ ADC,

(i) $AD\ =\ AD$ (Common in both the triangles)

(ii) $\angle ADB\ =\ \angle ADC$ (Right angle)

(iii) $BD\ =\ CD$ (Since AD is the bisector of BC)

Thus by SAS congruence axiom, we can state :

$\Delta ADB\ \cong \ \Delta ADC$

Hence by c.p.c.t., we can say that : $\small AB=AC$

Thus $\Delta ABC$ is an isosceles triangle with AB and AC as equal sides.

Q3 ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig.). Show that these altitudes are equal.

Answer:

Consider $\Delta AEB$ and $\Delta AFC$ ,

(i) $\angle A$ is common in both the triangles.

(ii) $\angle AEB\ =\ \angle AFC$ (Right angles)

(iii) $AB\ =\ AC$ (Given)

Thus by AAS congruence axiom, we can conclude that :

$\Delta AEB\ \cong \Delta AFC$

Now, by c.p.c.t. we can say : $BE\ =\ CF$

Hence these altitudes are equal.

Q4 (i) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig). Show that $\small \Delta ABE \cong \Delta ACF$

Answer:

Consider $\Delta ABE$ and $\Delta ACF$ ,

(i) $\angle A$ is common in both the triangles.

(ii) $\angle AEB\ =\ \angle AFC$ (Right angles)

(iii) $BE\ =\ CF$ (Given)

Thus by AAS congruence, we can say that :

$\small \Delta ABE \cong \Delta ACF$

Q4 (ii) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig.). Show that $\small AB=AC$ , i.e., ABC is an isosceles triangle.

Answer:

From the prevoius part of the question we found out that : $\Delta ABE\ \cong \Delta ACF$

Now, by c.p.c.t. we can say that : $AB\ =\ AC$

Hence $\Delta \ ABC$ is an isosceles triangle.

Q5 ABC and DBC are two isosceles triangles on the same base BC (see Fig.). Show that $\small \angle ABD\ \cong \ \angle ACD$ .

Answer:

Consider $\Delta ABD$ and $\Delta ACD$ ,

(i) $AD\ =\ AD$ (Common in both the triangles)

(ii) $AB\ =\ AC$ (Sides of isosceles triangle)

(iii) $BD\ =\ CD$ (Sides of isosceles triangle)

Thus by SSS congruency, we can conclude that :

$\small \angle ABD\ \cong \ \angle ACD$

Q6 $\Delta ABC$ is an isosceles triangle in which $AB=AC$ . Side BA is produced to D such that $AD=AB$ (see Fig.). Show that $\angle BCD$ is a right angle.

Answer:

Consider $\Delta$ ABC,
It is given that AB = AC

So, $\angle ACB = \angle ABC$ (Since angles opposite to the equal sides are equal.)

Similarly in $\Delta$ ACD,

We have AD = AB
and $\angle ADC = \angle ACD$
So,

$\angle CAB + \angle ACB + \angle ABC = 180^{\circ}$

$\angle CAB\ +\ 2\angle ACB = 180^{\circ}$
or $\angle CAB\ = 180^{\circ}\ -\ 2\angle ACB$ ...........................(i)

And in $\Delta$ ADC,
$\angle CAD\ = 180^{\circ}\ -\ 2\angle ACD$ ..............................(ii)

Adding (i) and (ii), we get :
$\angle CAB\ +\ \angle CAD\ = 360^{\circ}\ -\ 2\angle ACD\ -\ 2\angle ACB$

or $180^{\circ}\ = 360^{\circ}\ -\ 2\angle ACD\ -\ 2\angle ACB$

and $\angle BCD\ =\ 90^{\circ}$

Q7 ABC is a right angled triangle in which $\small \angle A =90^{\circ}$ and $\small AB=AC$ . Find $\small \angle B$ and $\small \angle C$ .

Answer:

In the triangle ABC, sides AB and AC are equal.

We know that angles opposite to equal sides are also equal.

Thus, $\angle B\ =\ \angle C$

Also, the sum of the interior angles of a triangle is $180^{\circ}$ .

So, we have :

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

or $90^{\circ} +\ 2\angle B\ =\ 180^{\circ}$

or $\angle B\ =\ 45^{\circ}$

Hence $\angle B\ =\ \angle C\ =\ 45^{\circ}$

Q8 Show that the angles of an equilateral triangle are $\small 60^{\circ}$ each.

Answer:

Consider a triangle ABC which has all sides equal.

We know that angles opposite to equal sides are equal.

Thus we can write : $\angle A\ =\ \angle B\ =\ \angle C$

Also, the sum of the interior angles of a triangle is $180 ^{\circ}$ .

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

or $3\angle A\ =\ 180^{\circ}$

or $\angle A\ =\ 60^{\circ}$

So, all the angles of the equilateral triangle are equal ( $60^{\circ}$ ).

Class 9 Triangles NCERT Solutions - Excercise: 7.3

Q1 (i) $\small \Delta ABC$ and $\small \Delta DBC$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that $\small \Delta ABD\cong \Delta ACD$

Answer:

Consider $\Delta ABD$ and $\Delta ACD$ ,

(i) $AD\ =\ AD$ (Common)

(ii) $AB\ =\ AC$ (Isosceles triangle)

(iii) $BD\ =\ CD$ (Isosceles triangle)

Thus by SSS congruency we can conclude that :

$\small \Delta ABD\cong \Delta ACD$

Q1 (ii) Triangles ABC and Triangle DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig). If AD is extended to intersect BC at P, show that $\small \Delta ABP \cong \Delta ACP$

Answer:

Consider $\Delta ABP$ and $\Delta ACP$ ,

(i) $AP$ is common side in both the triangles.

(ii) $\angle PAB\ =\ \angle PAC$ (This is obtained from the c.p.c.t. as proved in the previous part.)

(iii) $AB\ =\ AC$ (Isosceles triangles)

Thus by SAS axiom, we can conclude that :

$\small \Delta ABP \cong \Delta ACP$

Q1 (iii) $\small \Delta ABC$ and $\small \Delta DBC$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that AP bisects $\small \angle A$ as well as $\small \angle D$ .

Answer:

In the first part, we have proved that $\small \Delta ABD\cong \Delta ACD$ .

So, by c.p.c.t. $\angle PAB\ =\ \angle PAC$ .

Hence AP bisects $\angle A$ .

Now consider $\Delta BPD$ and $\Delta CPD$ ,

(i) $PD\ =\ PD$ (Common)

(ii) $BD\ =\ CD$ (Isosceles triangle)

(iii) $BP\ =\ CP$ (by c.p.c.t. from the part (b))

Thus by SSS congruency we have :

$\Delta BPD\ \cong \ \Delta CPD$

Hence by c.p.c.t. we have : $\angle BDP\ =\ \angle CDP$

or AP bisects $\angle D$ .

Q1 (iv) $\small \Delta ABC$ and $\small \Delta DBC$ are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that AP is the perpendicular bisector of BC.

Answer:

In the previous part we have proved that $\Delta BPD\ \cong \ \Delta CPD$ .

Thus by c.p.c.t. we can say that : $\angle BPD\ =\ \angle CPD$

Also, $BP\ =\ CP$

SInce BC is a straight line, thus : $\angle BPD\ +\ \angle CPD\ =\ 180^{\circ}$

or $2\angle BPD\ =\ 180^{\circ}$

or $\angle BPD\ =\ 90^{\circ}$

Hence it is clear that AP is a perpendicular bisector of line BC.

Q2 (i) AD is an altitude of an isosceles triangle ABC in which $\small AB=AC$ . Show that AD bisects BC

Answer:

Consider $\Delta ABD$ and $\Delta ACD$ ,

(i) $AB\ =\ AC$ (Given)

(ii) $AD\ =\ AD$ (Common in both triangles)

(iii) $\angle ADB\ =\ \angle ADC\ =\ 90^{\circ}$

Thus by RHS axiom we can conclude that :

$\Delta ABD\ \cong \ \Delta ACD$

Hence by c.p.c.t. we can say that : $BD\ =\ CD$ or AD bisects BC.

Q2 (ii) AD is an altitude of an isosceles triangle ABC in which $\small AB=AC$ . Show that AD bisects $\small \angle A$ .

Answer:

In the previous part of the question we have proved that $\Delta ABD\ \cong \ \Delta ACD$

Thus by c.p.c.t., we can write :

$\angle BAD\ =\ \angle CAD$

Hence $AD$ bisects $\angle A$ .

Q3 Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of $\small \Delta PQR$ (see Fig). Show that:

(i) $\small \Delta ABM \cong \Delta PQN$

(ii) $\small \Delta ABC \cong \Delta PQR$

Answer:

(i) From the figure we can say that :

$BC\ =\ QR$

or $\frac{1}{2}BC\ =\ \frac{1}{2}QR$

or $BM\ =\ QN$

Now, consider $\Delta ABM$ and $\Delta PQN$ ,

(a) $AM\ =\ PN$ (Given)

(b) $AB\ =\ PQ$ (Given)

(c) $BM\ =\ QN$ (Prove above)

Thus by SSS congruence rule, we can conclude that :

$\small \Delta ABM \cong \Delta PQN$

(ii) Consider $\Delta ABC$ and $\Delta PQR$ :

(a) $AB\ =\ PQ$ (Given)

(b) $\angle ABC\ =\ \angle PQR$ (by c.p.c.t. from the above proof)

(c) $BC\ =\ QR$ (Given)

Thus by SAS congruence rule,

$\small \Delta ABC \cong \Delta PQR$

Q4 BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Answer:

Using the given conditions, consider $\Delta BEC$ and $\Delta CFB$ ,

(i) $\angle BEC\ =\ \angle CFB$ (Right angle)

(ii) $BC\ =\ BC$ (Common in both the triangles)

(iii) $BE\ =\ CF$ (Given that altitudes are of the same length. )

Thus by RHS axiom, we can say that : $\Delta BEC\ \cong \Delta CFB$

Hence by c.p.c.t., $\angle B\ =\ \angle C$

And thus $AB\ =\ AC$ (sides opposite to equal angles are also equal). Thus ABC is an isosceles triangle.

Q5 ABC is an isosceles triangle with $\small AB=AC$ . Draw $\small AP\perp BC$ to show that $\small \angle B=\angle C$ .

Answer:

Consider $\Delta ABP$ and $\Delta ACP$ ,

(i) $\angle APB\ =\ \angle APC\ =\ 90^{\circ}$ (Since it is given that AP is altitude.)

(ii) $AB\ =\ AC$ (Isosceles triangle)

(iii) $AP\ =\ AP$ (Common in both triangles)

Thus by RHS axiom we can conclude that :

$\Delta ABP\ \cong \Delta ACP$

Now, by c.p.c.t.we can say that :

$\angle B\ =\ \angle C$

Class 9 maths chapter 7 NCERT solutions - Excercise: 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.

Triangles class 9 NCERT solutions - Excercise: 7.5

Q1 ABC is a triangle. Locate a point in the interior of $\small \Delta ABC$ which is equidistant from all the vertices of $\small \Delta ABC$ .

Answer:

We know that circumcenter of a triangle is equidistant from all the vertices. Also, circumcenter is the point of intersection of the perpendicular bisectors of the sides of a triangle.

Thus, draw perpendicular bisectors of each side of the triangle ABC. And let them meet at a point, say O.

Hence O is the required point which is equidistant from all the vertices.

Q2 In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Answer:

The required point is called in-centre of the triangle. This point is the intersection of the angle bisectors of the interior angles of a triangle.

Hence the point can be found out in this case just by drawing angle bisectors of all the angles of the triangle.

Q3 In a huge park, people are concentrated at three points (see Fig.):

A : where there are different slides and swings for children,

B : near which a man-made lake is situated,

C : which is near to a large parking and exit.

Where should an icecream parlour be set up so that maximum number of persons can approach it? ( Hint : The parlour should be equidistant from A, B and C)

Answer:

The three main points form a triangle ABC. Now we have to find a point which is equidistant from all the three points.

Thus we need to find the circumcenter of the $\Delta ABC$ .

We know that circumcenter is defined as the point as the intersection point of the perpendicular bisectors of the sides of the triangle.

Hence the required point can be found out by drawing perpendicular bisectors of $\Delta ABC$ .

Q4 Complete the hexagonal and star shaped Rangolies [see Fig. (i) and (ii)] by filling them with as many equilateral triangles of side $\small 1$ cm as you can. Count the number of triangles in each case. Which has more triangles?

Answer:

For finding the number of triangles we need to find the area of the figure.

Consider the hexagonal structure :

Area of hexagon = 6 $\times$ Area of 1 equilateral

Thus area of the equilateral triangle :

$=\ \frac{\sqrt{3}}{4}\times a^2$

or $=\ \frac{\sqrt{3}}{4}\times 5^2$

or $=\ \frac{25\sqrt{3}}{4}\ cm^2$

So, the area of the hexagon is :

$=\6\times \frac{25\sqrt{3}}{4}\ =\ \frac{75\sqrt{3}}{2}\ cm^2$

And the area of an equilateral triangle having 1cm as its side is :

$=\ \frac{\sqrt{3}}{4}\times 1^2$

or $=\ \frac{\sqrt{3}}{4}\ cm^2$

Hence a number of equilateral triangles that can be filled in hexagon are :

$=\ \frac{\frac{75\sqrt{3}}{2}}{\frac{\sqrt{3}}{4}}\ =\ 150$

Similarly for star-shaped rangoli :

Area :

$=\12\times \frac{\sqrt{3}}{4}\times 5^2 \ =\ 75\sqrt{3}\ cm^2$

Thus the number of equilateral triangles are :

$=\ \frac{75\sqrt{3}}{\frac{\sqrt{3}}{4}}\ =\ 300$

Hence star-shaped rangoli has more equilateral triangles.