NCERT Solutions for Class 9 Maths Chapter 7 Triangles

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

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

Answer:

In the given triangles we are given that:-

(i) $AC=AD$

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

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

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

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 $AD=BC$ and $\mathrm{\angle }DAB=\mathrm{\angle }CBA$ (see Fig. ). Prove that

$\mathrm{\Delta }ABD\cong \mathrm{\Delta }BAC$ .

Answer:

It is given that :-

(i) AD = BC

(ii) $\mathrm{\angle }DAB=\mathrm{\angle }CBA$

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

So, by SAS congruence, we can write :

$\mathrm{\Delta }ABD\cong \mathrm{\Delta }BAC$

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

Answer:

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

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

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

Answer:

In the first part we have proved that $\mathrm{\Delta }ABD\cong \mathrm{\Delta }BAC$ .

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

$\mathrm{\angle }ABD=\mathrm{\angle }BAC$

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

Answer:

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

(i) AD = BC (given)

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

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

Thus by AAS Postulate, we have

$\mathrm{\Delta }AOD\cong \mathrm{\Delta }BOC$

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

And thus CD bisects AB.

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

Answer:

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

(i) $\mathrm{\angle }BCA=\mathrm{\angle }DAC$

(ii) $\mathrm{\angle }BAC=\mathrm{\angle }DCA$

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

Thus by ASA congruence, we have :

$\mathrm{\Delta }ABC\cong \mathrm{\Delta }CDA$

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

Answer:

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

(i) $\mathrm{\angle }P=\mathrm{\angle }Q$ (Right angle)

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

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

Thus AAS congruence, we can write :

$\mathrm{\Delta }APB\cong \mathrm{\Delta }AQB$

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

Answer:

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

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

$BP=BQ$

Thus B is equidistant from arms of angle A.

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

Answer:

From the given figure following result can be drawn:-

$\mathrm{\angle }BAD=\mathrm{\angle }EAC$

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

$\mathrm{\angle }BAD+\mathrm{\angle }DAC=\mathrm{\angle }EAC+\mathrm{\angle }DAC$

$\mathrm{\angle }BAC=\mathrm{\angle }EAD$

Now consider $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }ADE$ , :-

(i) $AC=AE$ (Given)

(ii) $\mathrm{\angle }BAC=\mathrm{\angle }EAD$ (proved above)

(iii) $AB=AD$ (Given)

Thus by SAS congruence we can say that :

$\mathrm{\Delta }ABC\cong \mathrm{\Delta }ADE$

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

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

Answer:

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

Adding $\mathrm{\angle }DPE$ both sides, we get :

$\mathrm{\angle }EPA+\mathrm{\angle }DPE=\mathrm{\angle }DPB+\mathrm{\angle }DPE$

or $\mathrm{\angle }DPA=\mathrm{\angle }EPB$

Now, consider $\mathrm{\Delta }DAP$ and $\mathrm{\Delta }EBP$ :

(i) $\mathrm{\angle }DPA=\mathrm{\angle }EPB$ (Proved above)

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

(iii) $\mathrm{\angle }BAD=\mathrm{\angle }ABE$ (Given)

Hence by ASA congruence, we can say that :

$\mathrm{\Delta }DAP\cong \mathrm{\Delta }EBP$

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 $\mathrm{\angle }BAD=\mathrm{\angle }ABE$ and $\mathrm{\angle }EPA=\mathrm{\angle }DPA$ (see Fig). Show that $AD=BE$

Answer:

In the previous part we have proved that $\mathrm{\Delta }DAP\cong \mathrm{\Delta }EBP$ .

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

$AD=BE$

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 $DM=CM$ . Point D is joined to point B (see Fig.). Show that: $\mathrm{\Delta }AMC\cong \mathrm{\Delta }BMD$

Answer:

Consider $\mathrm{\Delta }AMC$ and $\mathrm{\Delta }BMD$ ,

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

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

(iii) $CM=DM$ (Given)

Thus by SAS congruency, we can conclude that :

$\mathrm{\Delta }AMC\cong \mathrm{\Delta }BMD$

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 $DM=CM$ . Point D is joined to point B (see Fig.). Show that: $\mathrm{\angle }DBC$ is a right angle.

Answer:

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

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

This implies side AC is parallel to BD.

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

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

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

Hence $\mathrm{\angle }DBC$ 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 $DM=CM$ . Point D is joined to point B (see Fig.). Show that: $\mathrm{\Delta }DBC\cong \mathrm{\Delta }ACB$

Answer:

Consider $\mathrm{\Delta }DBC$ and $\mathrm{\Delta }ACB$ ,

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

(ii) $\mathrm{\angle }ACB=\mathrm{\angle }DBC$ (Right angle)

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

Thus SAS congruence we can conclude that :

$\mathrm{\Delta }DBC\cong \mathrm{\Delta }ACB$

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 $DM=CM$ . Point D is joined to point B (see Fig.). Show that: $CM=\frac{1}{2}AB$

Answer:

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

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

$DM+CM=AM+BM$

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

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

Hence proved.

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

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

Answer:

In the triangle ABC,

Since AB = AC, thus $\mathrm{\angle }B=\mathrm{\angle }C$

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

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

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

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

Answer:

Consider $\mathrm{\Delta }AOB$ and $\mathrm{\Delta }AOC$ ,

(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 :

$\mathrm{\Delta }AOB\cong \mathrm{\Delta }AOC$

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

$\mathrm{\angle }BAO=\mathrm{\angle }CAO$

Hence AO bisects $\mathrm{\angle }A$ .

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

Answer:

Consider $\mathrm{\Delta }$ ABD and $\mathrm{\Delta }$ ADC,

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

(ii) $\mathrm{\angle }ADB=\mathrm{\angle }ADC$ (Right angle)

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

Thus by SAS congruence axiom, we can state :

$\mathrm{\Delta }ADB\cong \mathrm{\Delta }ADC$

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

Thus $\mathrm{\Delta }ABC$ 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 $\mathrm{\Delta }AEB$ and $\mathrm{\Delta }AFC$ ,

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

(ii) $\mathrm{\angle }AEB=\mathrm{\angle }AFC$ (Right angles)

(iii) $AB=AC$ (Given)

Thus by AAS congruence axiom, we can conclude that :

$\mathrm{\Delta }AEB\cong \mathrm{\Delta }AFC$

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 $\mathrm{\Delta }ABE\cong \mathrm{\Delta }ACF$

Answer:

Consider $\mathrm{\Delta }ABE$ and $\mathrm{\Delta }ACF$ ,

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

(ii) $\mathrm{\angle }AEB=\mathrm{\angle }AFC$ (Right angles)

(iii) $BE=CF$ (Given)

Thus by AAS congruence, we can say that :

$\mathrm{\Delta }ABE\cong \mathrm{\Delta }ACF$

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

Answer:

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

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

Hence $\mathrm{\Delta }ABC$ is an isosceles triangle.

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

Answer:

Consider $\mathrm{\Delta }ABD$ and $\mathrm{\Delta }ACD$ ,

(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 :

$\mathrm{\angle }ABD\cong \mathrm{\angle }ACD$

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

Answer:

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

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

Similarly in $\mathrm{\Delta }$ ACD,

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

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

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

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

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

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

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

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

Answer:

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

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

Thus, $\mathrm{\angle }B=\mathrm{\angle }C$

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

So, we have :

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

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

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

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

Q8 Show that the angles of an equilateral triangle are ${60}^{\circ }$ 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 : $\mathrm{\angle }A=\mathrm{\angle }B=\mathrm{\angle }C$

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

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

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

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

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

Class 9 Triangles NCERT Solutions - Excercise: 7.3

Q1 (i) $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DBC$ 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 $\mathrm{\Delta }ABD\cong \mathrm{\Delta }ACD$

Answer:

Consider $\mathrm{\Delta }ABD$ and $\mathrm{\Delta }ACD$ ,

(i) $AD=AD$ (Common)

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

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

Thus by SSS congruency we can conclude that :

$\mathrm{\Delta }ABD\cong \mathrm{\Delta }ACD$

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 $\mathrm{\Delta }ABP\cong \mathrm{\Delta }ACP$

Answer:

Consider $\mathrm{\Delta }ABP$ and $\mathrm{\Delta }ACP$ ,

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

(ii) $\mathrm{\angle }PAB=\mathrm{\angle }PAC$ (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 :

$\mathrm{\Delta }ABP\cong \mathrm{\Delta }ACP$

Q1 (iii) $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DBC$ 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 $\mathrm{\angle }A$ as well as $\mathrm{\angle }D$ .

Answer:

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

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

Hence AP bisects $\mathrm{\angle }A$ .

Now consider $\mathrm{\Delta }BPD$ and $\mathrm{\Delta }CPD$ ,

(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 :

$\mathrm{\Delta }BPD\cong \mathrm{\Delta }CPD$

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

or AP bisects $\mathrm{\angle }D$ .

Q1 (iv) $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DBC$ 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 $\mathrm{\Delta }BPD\cong \mathrm{\Delta }CPD$ .

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

Also, $BP=CP$

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

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

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

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 $AB=AC$ . Show that AD bisects BC

Answer:

Consider $\mathrm{\Delta }ABD$ and $\mathrm{\Delta }ACD$ ,

(i) $AB=AC$ (Given)

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

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

Thus by RHS axiom we can conclude that :

$\mathrm{\Delta }ABD\cong \mathrm{\Delta }ACD$

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 $AB=AC$ . Show that AD bisects $\mathrm{\angle }A$ .

Answer:

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

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

$\mathrm{\angle }BAD=\mathrm{\angle }CAD$

Hence $AD$ bisects $\mathrm{\angle }A$ .

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 $\mathrm{\Delta }PQR$ (see Fig). Show that:

(i) $\mathrm{\Delta }ABM\cong \mathrm{\Delta }PQN$

(ii) $\mathrm{\Delta }ABC\cong \mathrm{\Delta }PQR$

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 $\mathrm{\Delta }ABM$ and $\mathrm{\Delta }PQN$ ,

(a) $AM=PN$ (Given)

(b) $AB=PQ$ (Given)

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

Thus by SSS congruence rule, we can conclude that :

$\mathrm{\Delta }ABM\cong \mathrm{\Delta }PQN$

(ii) Consider $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }PQR$ :

(a) $AB=PQ$ (Given)

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

(c) $BC=QR$ (Given)

Thus by SAS congruence rule,

$\mathrm{\Delta }ABC\cong \mathrm{\Delta }PQR$

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 $\mathrm{\Delta }BEC$ and $\mathrm{\Delta }CFB$ ,

(i) $\mathrm{\angle }BEC=\mathrm{\angle }CFB$ (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 : $\mathrm{\Delta }BEC\cong \mathrm{\Delta }CFB$

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

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 $AB=AC$ . Draw $AP\perp BC$ to show that $\mathrm{\angle }B=\mathrm{\angle }C$ .

Answer:

Consider $\mathrm{\Delta }ABP$ and $\mathrm{\Delta }ACP$ ,

(i) $\mathrm{\angle }APB=\mathrm{\angle }APC={90}^{\circ }$ (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 :

$\mathrm{\Delta }ABP\cong \mathrm{\Delta }ACP$

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

$\mathrm{\angle }B=\mathrm{\angle }C$

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

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

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

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

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

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

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

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

Answer:

We are given that,

$\mathrm{\angle }PBC<\mathrm{\angle }QCB$ ......................(i)

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

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

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

$\mathrm{\angle }ABC>\mathrm{\angle }ACB$

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

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

Answer:

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

We are given $\mathrm{\angle }B<\mathrm{\angle }A$ and $\mathrm{\angle }C<\mathrm{\angle }D$ .

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 $\mathrm{\angle }A>\mathrm{\angle }C$ and $\mathrm{\angle }B>\mathrm{\angle }D$ .

Answer:

Consider $\mathrm{\Delta }ADC$ in the above figure :

$AD<CD$ (Given)

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

Now consider $\mathrm{\Delta }ABC$ ,

We have : $BC>AB$

and $\mathrm{\angle }BAC>\mathrm{\angle }ACB$

Adding the above result we get,

$\mathrm{\angle }BAC+\mathrm{\angle }CAD>\mathrm{\angle }ACB+\mathrm{\angle }ACD$

or $\mathrm{\angle }A>\mathrm{\angle }C$

Similarly, consider $\mathrm{\Delta }ABD$ ,

we have $AB<AD$

Therefore $\mathrm{\angle }ABD>\mathrm{\angle }ADB$

and in $\mathrm{\Delta }BDC$ we have,

$CD>BC$

and $\mathrm{\angle }CBD>\mathrm{\angle }CDB$

from the above result we have,

$\mathrm{\angle }ABD+\mathrm{\angle }CBD>\mathrm{\angle }ADB+\mathrm{\angle }CDB$

or $\mathrm{\angle }B>\mathrm{\angle }D$

Hence proved.

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

Answer:

We are given that $PR>PQ$ .

Thus $\mathrm{\angle }PQR=\mathrm{\angle }PRQ$

Also, PS bisects $\mathrm{\angle }QPR$ , thus :

$\mathrm{\angle }QPS=\mathrm{\angle }RPS$

Now, consider $\mathrm{\Delta }QPS$ ,

$\mathrm{\angle }PSR=\mathrm{\angle }PQR+\mathrm{\angle }QPS$ (Exterior angle)

Now, consider $\mathrm{\Delta }PSR$ ,

$\mathrm{\angle }PSQ=\mathrm{\angle }PRQ+\mathrm{\angle }RPS$

Thus from the above the result we can conclude that :

$\mathrm{\angle }PSR>\mathrm{\angle }PSQ$

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 $\mathrm{\angle }B>\mathrm{\angle }Aor\mathrm{\angle }C$ (Since $\mathrm{\angle }B={90}^{\circ }$ )

Thus the side opposite to largest angle is also largest. $AC>BCorAB$

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 $\mathrm{\Delta }ABC$ which is equidistant from all the vertices of $\mathrm{\Delta }ABC$ .

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 $\mathrm{\Delta }ABC$ .

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 $\mathrm{\Delta }ABC$ .

Q4 Complete the hexagonal and star shaped Rangolies [see Fig. (i) and (ii)] by filling them with as many equilateral triangles of side $1$ 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 $=\frac{25\sqrt{3}}{4}c{m}^2$

So, the area of the hexagon is :

$=\text{6}\times \frac{25\sqrt{3}}{4}=\frac{75\sqrt{3}}{2}c{m}^2$

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

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

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

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 :

$=\text{1}2\times \frac{\sqrt{3}}{4}\times 5^2=75\sqrt{3}c{m}^2$

Thus the number of equilateral triangles are :

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

Hence star-shaped rangoli has more equilateral triangles.