RD Sharma 2020 solution class 9 chapter 12 Congruent Triangles Exercise 12.6

Exercise 12.6

Page-12.81

Question 1:

In ΔABC, if ∠A = 40° and ∠B = 60°. Determine the longest and shortest sides of the triangle.

Answer 1:

In the triangle ABC it is given that 

We have to find the longest and shortest side.

Here

Now is the largest angle of the triangle.

So the side in front of the largest angle will be the longest side.

Hence will be the longest

Since is the shortest angle so that side in front of it will be the shortest.

And is shortest side

Hence Is longest and is shortest.

Question 2:

In a ΔABC, if ∠B = ∠C = 45°, which is the longest side?

Answer 2:

In the triangle ABC it is given that 

We have to find the longest side.

Here

(Since)

Now is the largest angle of the triangle.

So the side in front of the largest angle will be the longest side.

Hence BC will be the longest side.

Question 3:

In Δ ABC, side AB is produced to D so that BD = BC. If ∠B = 60° and ∠A = 70°, prove that :
(i) AD > CD                           (ii) AD>AC

Answer 3:

It is given that

, and

We have to prove that 

(1)

(2)

(1) 

Now

And since BD=BC, so, and 

 

That is,

Now

And, so

 

Hence (1) (Side in front of greater angle will be longer)

And (2) Proved.

Question 4:

Is a possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?

Answer 4:

As we know that a triangle can only be formed if

The sum of two sides is greater than the third side.

Here we have 2 cm, 3 cm and 7 cm as sides.

If we add

(Since 5 is less than 7)

Hence the sum of two sides is less than the third sides 

So, the triangle will not exist.

Question 5:

In Δ ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.

Answer 5:

It is given that

AP is the bisector of

We have to arrange, and in descending order.

In we have

(As AP is the bisector of)

So (Sides in front or greater angle will be greater)              ........(1)

In we have

(As AP is the bisector of)

Since, 

So                   ..........(2)

Hence 

From (1) & (2) we have

Question 6:

Prove that the perimeter of a triangle is greater than the sum of its altitudes.

Answer 6:

We have to prove that the perimeter of a triangle is greater than the sum of its altitude.

In

, ,

We have to prove

Since

So and

By adding, we have

           ........(1)

Now consider then

, and

Now by adding     .......(2)

Again consider

, and

By adding   ...........(3)

Adding (1), (2) and (3), we get

Hence the perimeter of a triangle is greater than the sum of all its altitude.

Question 7:

In the given figure, prove that:

(i) CD + DA + AB + BC > 2AC

(ii) CD + DA + AB > BC
 

Answer 7:

(1) We have to prove that

In we have

(As sum of two sides of triangle is greater than third one)   ........(1)

In we have

(As sum of two sides of triangle is greater than third one)   .........(2)

Hence 

Adding (1) & (2) we get

Proved.

(2) We have to prove that

In we have

(As sum of two sides of triangle is greater than third one)

(Adding both sides)

Proved.

Question 8:

Which of the following statements are true (T) and which are false (F)?

(i) Sum of the three sides of a triangle is less than the sum of its three altitudes.

(ii) Sum of any two sides of a triangle is greater than twice the median drawn to the third side.

(iii) Sum of any two sides of a triangle is greater than the third side.

(iv) Difference of any two sides of a triangle is equal to the third side.

(v) If two angles of a triangle are unequal, then the greater angle has the larger side opposite to it.

(vi) Of all the line segments that can be drawn from a point to a line not containing it, the perpendicular line segment is the shortest one.

Answer 8:

(1)

(2)

(3)

(4)

(5)

(6)

Page-12.82

Question 9:

Fill in the blanks to make the following statements true.

(i) In a right triangle the hypotenuse is the ...... side.

(ii) The sum of three altitudes of a triangle is ....... than its perimeter.

(iii) The sum of any two sides of a triangle is ..... than the third side.

(iv) If two angles of a triangle are unequal, then the smaller angle has the ..... side opposite to it.

(v) Difference of any two sides of a triangle is ...... than the third side.

(vi) If two sides of a triangle are unequal, then the larger side has ...... angle opposite to it.

Answer 9:

(1)

(2)

(3)

(4)

(5)

(6)

Question 10:

O is any point in the interior of Δ ABC. Prove that

(i) AB + AC > OB + OC

(ii) AB + BC + CA > OA + OB + OC

(iii) OA + OB + OC > $\frac{1}{2}$(AB + BC + CA)

Answer 10:

It is given that, is any point in the interior of

We have to prove that

(1) Produced to meet at.

In we have

    .........(1)

And in we have

            .........(2)

Adding (1) & (2) we get

HenceProved.

(2) We have to prove that

From the first result we have 

       ..........(3)

And

     .........(4)

Adding above (4) equation

HenceProved.

(3) We have to prove that

In triangles, and we have

Adding these three results

HenceProved.

Question 11:

Prove that in a quadrilateral the sum of all the sides is greater than the sum of its diagonals.

Answer 11:

We have to prove that the sum of four sides of quadrilateral is greater than sum of diagonal.

Since the sum of two sides of triangle is greater than third side.

In we have

..........(1)

In we have 

..........(2)

In we have 

.........(3)

In we have 

.........(4)

Adding (1) & (2) & (3) and (4) we get

Hence Proved.

Question 12:

Prove that in a triangle, other than an equilateral triangle, angle opposite to the longest side is greater than $\frac{2}{3}$of a right angle.

Answer 12:

Let AC be the longest side in the ∆ABC.

Now,

AC > AB

⇒ ∠B > ∠C         .....(1)             (In a triangle, greater side has greater angle opposite to it)

Also,

AC > BC

⇒ ∠B > ∠A         .....(2)             (In a triangle, greater side has greater angle opposite to it)

From (1) and (2), we have

∠B + ∠B > ∠A + ∠C

⇒ ∠B + ∠B + ∠B > ∠A + ∠B + ∠C

⇒ 3∠B > 180º                      (Using angle sum property of a triangle)

⇒ ∠B > $\frac{1}{3}$ × 180º

Or ∠B > $\frac{2}{3}$ × 90º

Thus, the angle opposite to the longest side is greater than $\frac{2}{3}$of a right angle.

Hence proved.

Question 13:

D is any point on side AC of a ∆ABC with AB = AC. Show that CD < BD.

Answer 13:

It is given that, D is any point on side AC of a ∆ABC with AB = AC.



In ∆ABC,

AB = AC      (Given)

∴ ∠ACB = ∠ABC         (In a triangle, equal sides have equal angles opposite to them)

Now, ∠ABC > ∠DBC

⇒ ∠ACB > ∠DBC            (∠ACB = ∠ABC)

In ∆BCD,

∠DCB > ∠DBC

⇒ BD > CD                       (In a triangle, greater angle has greater side opposite to it)

Or CD < BD