7. TRIANGLES

Assam board Class 9 Chapter 7. Triangles

Chapter 7. TRIANGLES

Important Note :

1. If ∆ PQR is congruent to ∆ ABC, we write ∆ PQR ≅ ∆ ABC.

2. CPCT : corresponding parts of congruent triangles.

3. SAS Congruence Rule : If two sides and the included angle of one triangle are equal to two sides and the included angle of the other triangle, then the two triangles are congruent).
4. ASA Congruence Rule : If two angles and the included side of one triangle are equal to two angles and the included side of the other triangle, then the two triangles are congruent
5. AAS Congruence Rule : If two angles and one side of one triangle are equal to two angles and the corresponding side of the other triangle, then the two triangles are congruent .
6. Angles opposite to equal sides of a triangle are equal.
7. Sides opposite to equal angles of a triangle are equal.
8. Each angle of an equilateral triangle is of 60°.
9. SSS Congruence Rule : If three sides of one triangle are equal to three sides of the other triangle, then the two triangles are congruent
10. RHS (Right angle-Hypotenuse-Side Congruence rule) : If in two right triangles, hypotenuse and one side of a triangle are equal to the hypotenuse and one side of other triangle, then the two triangles are congruent .
11. In a triangle, angle opposite to the longer side is larger (greater).
12. In a triangle, side opposite to the larger (greater) angle is longer.
13. Sum of any two sides of a triangle is greater than the third side.

EXERCISE 7.1

1. In quadrilateral ABCD, and AB bisects (see Fig. 7.16) .Show that . What can you say about BC and BD ?

Solution : Given, ABCD is a quadrilateral, and AB bisects . Then show we that

Proof: Since , AB bisects .

So,

In and , we have

[Given ]

[Given]

[Common side]

[SAS rule]

[CPCT]

2. ABCD is a quadrilateral in which and (see Fig. 7.17) . Prove that (i) (ii) (iii)

Solution : Given, ABCD is a quadrilateral , and .

To prove (i) (ii) (iii)

Proof : (i) In and , we have

[ Common side]

[Given]

[SAS rule] Proved

(ii) In and , we have

[ Common side]

[Given]

[SAS rule]

[CPCT] Proved

(iii) In and , we have

[ Common side]

[Given]

[SAS rule]

[CPCT] Proved

3. AD and BC are equal perpendicular to a line segment AB (see Fig. 7.18) . Show that CD bisects AB .

Solution: Given AD and BC are equal perpendicular to a line segment AB .Then we show that CD bisects AB .

Proof : In figure ,

In and , we have

[Given]

[ given]

[Vertically opposite angle]

[ASA rule]

[CPCT]

So, CD bisects AB . Proved

4. and are two parallel lines intersected by another pair of parallel lines and (see Fig. 7.18) . Show that .

Solution : Given, and are two parallel lines intersected by another pair of parallel lines and.Then we show that .

Proof : In and , we have

[Alternative interior angle]

[Common side]

[Alternative interior angle]

[ASA rule]

5. Line is the bisector of an angle and B is any point on . BP and BQ are perpendiculars from B to the arms of (see Fig. 7.20) . Show that : (i) (ii) or B is equidistant from the arms of .

Solution: Given , the line is the bisector of an angle and B is any point on . BP and BQ are perpendiculars from B to the arms of .

Then we show that : (i) (ii) or B is equidistant from the arms of .

Proof : (i) Since, is the bisector of an angle .

So,

In and we have

[Given]

[Common side]

[Given]

[ASA rule]

(ii) Since, is the bisector of an angle .

So,

In and we have

[Given]

[Common side]

[Given]

[ASA rule]

[CPCT]

Or B is equidistant from the arms of .

Proved .

6. In Fig. 7.21, , and . Show that .

Solution: Given, ,and .Then we show that .

Proof : We have,

[Add both side ]

In and , we have

[SAS rule]

[CPCT] Proved .

7. AB is a line segment and P is its mid-point D and E are points on the same side of AB such that and (see Fig. 7.22) . Show that (i) (ii) .

Solution : Given, AB is a line segment and P is its mid-point D and E are points on the same side of AB such that and .

Then we show that (i) (ii) .

Proof : Since, P is the mid-point of AB .

Again,

[ Add both side ]

And

In and , we have

[ Given]

[Given]

[ASA rule]

(ii) Proof : Since, P is the mid-point of AB .

Again,

[ Add both side ]

And

In and , we have

[ Given]

[Given]

[ASA rule]

[CPCT] Proved.

(ii) Since, [ASA rule]

[CPCT] Proved .

8. 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 . Point D is joined to point B (see Fig. 7.23) . Show that :

(i) (ii) is a right angle . (iii) (iv)

Solution : Given , ABC is a right triangle angled at C , M is the mid-point of hypotenuse AB . C is joined to M and produced to a point D such that . Point D is joined to point B .

Then we show that : (i) (ii) is a right angle (iii) (iv)

Proof : (i) Since, M is the mid-point of hypotenuse AB .

In and , we have

[Given]

[Vertically opposite angle]

[Given]

[SAS rule]

(ii) Proof : Since , [SAS rule]

[CPCT]

So, and BC is a transversal .

[ ]

is a right angle .

(iii) Proof : Since, we have

[CPCT]

In and , we have

[Given ]

[Given]

[Common side]

[SAS rule]

(iv) Since, [SAS rule]

[ CPCT]

Now,

[ ]

Proved .

EXERCISE 7.2

1. In an isosceles triangle ABC , with , the bisectors of and intersect each other at O . join A to O . Show that : (i) (ii) bisects .

Solution: Given, an isosceles triangle ABC , with , the bisectors of and intersect each other at O . Join A to O . Then we show that : (i) (ii) bisects

Proof: Since, OB and OC are the bisectors of and respectively .

and

So, and

In figure:

(i) In and , we have

[given]

[common side]

[SAS]

[CPCT]

(ii) In and , we have

[given]

[common side]

[SAS]

[CPCT]

bisects .

2. In , AD is the perpendicular bisectors of BC (see Fig. 7.30) . Show that is an isosceles triangle in which .

Solution: Given, ABC is a triangle and AD is the perpendicular bisectors of BC . Then we show that is an isosceles triangle in which .

Proof: Since, AD is the perpendicular bisectors of BC .

and

In and , we have

[given]

[common side]

[SAS]

[CPCT]

So, ABC is an isosceles triangle . Proved

3. ABC is an isosceles triangle in which altitudes BE and CF are draw to equal sides AC and AB respectively (see Fig. 7.31) . Show that thee altitudes are equal .

Solution: Given, ABC is an isosceles triangle in which altitudes BE and CF are draw to equal sides AC and AB respectively . Then we show that these altitudes are equal .

Proof: In figure:

In and , we have

[given]

[= 90° ]

[Common angle]

[SAS]

[CPCT]

Proved .

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32) . Show that :

(i)

(ii) i.e, is an isosceles triangle .

Solution: Given, ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal . Then we show that: (i) (ii) i.e, is an isosceles triangle .

Proof: (i) In and , we have

[given]

[= 90° ]

[Common angle]

[ASA]

Proved .

(ii) In and , we have

[given]

[= 90° ]

[Common angle]

[ASA]

[CPCT]

So, ABC is an isosceles triangle .

Proved .

5. ABC and BDC are two isosceles triangle on the same base BC (see Fig. 7.33) .Show that .

Solution: Given, ABC and BDC are two isosceles triangle on the same base BC . Then we show that .

Proof: In , we have

and

In , we have

and

Adding (i) and (ii) , we have

Proved .

6. is an isosceles triangle in which .Side BA is produced to D such that AD = AB (see Fig. 7.34 ). Show that is a right triangle .

Solution: Given, is an isosceles triangle in which and the side BA is produced to D such that AD = AB Then we show that is a right triangle .

Proof: In given figure :

In , we have

Adding (i) and (ii) , we get

[Adding both side ]

Proved.

7. ABC is right angled triangle in which and . Find and .

Solution: Given, ABC is right angled triangle in which and .

In , we have

So,

We have,

8. Show that the triangles of an equilateral triangles are60° each .

Solution: let ABC is an equilateral triangles and .

Then we show that .

Proof: Since, ABC is an equilateral triangles .

and .

In , we have

Proved.

EXERCISE 7.3

1. and are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39).If AD is extended to intersect BC at P, show that

(i)

(ii)

(iii) bisects as well as .

(iv) is the perpendicular bisector of .

Solution: Given, and are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC and AD is extended to intersect BC at P .

Then we show that :

(i)

(ii)

(iii) bisects as well as .

(iv) is the perpendicular bisector of .

Proof : Since, and are two isosceles triangles .

So, and

Also, and

(i) In and , we have

[Given]

[Common side]

[SSS] Proved.

(ii) In and , we have

[SAS]

(iii) In and , we have

[Given]

[Common side]

[SSS]

[CPCT]

is the bisector of .

Again, and , we have

[Given]

[Common side]

[SSS]

[CPCT]

is the bisector of .

So, bisects as well as .

(iv) In and , we have

[SAS]

and

is the perpendicular bisector of .

2. AD is an altitude of an isosceles triangle ABC in which . Show that : (i) is bisects (ii) bisects .

Solution: Given, AD is an altitude of an isosceles triangle ABC in which . Then we show that : (i) is bisects (ii) bisects .

Proof: (i) In figure ,

In and ,we have

[given]

[ABC is an isosceles triangle]

[ Common side]

[SAS]

[CPCT]

So, AD bisects BC . Proved.

(ii) In and,we have

[given]

[ABC is an isosceles triangle]

[ Common side]

[SAS]

[CPCT]

So, AD is bisects .

3. Two sides AB and BC and median AM of the triangle ABC are respectively equal to sides PQ and QR and median PN of (see Fig. 7.40). Show that :

(i)

(ii)

Solution: Given, Two sides AB and BC and median AM of the triangle ABC are respectively equal to sides PQ and QR and median PN of . Then we show that :

(i)

(ii)

Proof: Since, AM and PN are the median of the two triangles ABC and PQR respectively .

and

(i) In and , we have

[Given]

[SSS]

Proved.

(ii) In and , we have

[ ]

[SAS]

Proved.

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

Solution: Given, BE and CF are two equal altitudes of a triangle ABC . Then we prove that the triangle ABC is isosceles .

Proof: In figure ,

In and we have,

[Common side]

[Given]

[RHS]

[CPCT]

So, ABC is an isosceles triangle .

Proved .

5. ABC is an isosceles triangle with . Draw to show that .

Solution: Given, ABC is an isosceles triangle with and . Then we show that .

Proof: In figure,

In andwe have,

[Given]

[ ]

[Common side]

[RHS]

[CPCT]

Proved .

EXERCISE 7.4

1. Show that in aright angled triangle, the hypotenuse is the longest side .

Solution: let ABC is a right angled triangle such that .

Then we show that : or

Proof: Since, .

So, and are acute angle .

i.e., and

We know that ,

In a triangle, angle opposite to the longer side is larger (greater) .

So, and

or Proved.

2. In Fig. 7.48, sides AB and AC of are extended to points P and Q respectively. Also , .Show that .

Solution: Given, sides AB and AC of are extended to points P and Q respectively and .Then we show that .

Proof: We have,

[Linear pair of angles]

And [Linear pair of angles]

But

[from(i) and (ii)]

Proved.

3. In Fig .7.49 , and . Show that

Solution: Given, and . Then we show that.

Proof:In given figure :

Since,

and

Adding (i) and (ii) , we get

Proved.

4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig .7.50). Show that and .

Solution: Given, AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD . Then we show that ∠A >∠C and ∠B>∠D.

Construction: We join AC and BD .

Proof: In given figure:

In , we have

Adding (i) and (ii), we get

In , we have

Adding (iii) and (iv) , we get

Proved .

5. In Fig .7.51, and PS bisects .Prove that .

Solution: Given, and PS bisects .Prove that.

Proof: In given figure:

Since, PS bisects .

And

In , we have

From(i) ,(ii) and (iii) we get

Proved .

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

Solution: let PQ is a straight line and a point A is not on it . and C is any point on the line PQ other than B .

Then we prove that

Proof: Since,

So,

In , we have

and are acute angle of the triangle ABC .

We know that , the side opposite to the larger (greater) angle is longer of a triangle.

Proved.