# Class 9 RD Sharma Solutions – Chapter 10 Congruent Triangles- Exercise 10.6

**Question 1: In** Î”**ABC, if âˆ A = 40Â° and âˆ B = 60Â°. Determine the longest and shortest sides of the triangle.**

**Solution:**

Given that in Î”ABC, âˆ A = 40Â° and âˆ B = 60Â°

To find: longest and shortest side

We know that,

Sum of angles in a triangle 180Â°

âˆ A + âˆ B + âˆ C = 180Â°

40Â° + 60Â° + âˆ C = 180Â°

âˆ C = 180Â° – 100Â° = 80Â°

âˆ C = 80Â°

Now,

âŸ¹ 40Â° < 60Â° < 80Â° = âˆ A < âˆ B < âˆ C

âŸ¹ âˆ C is greater angle and âˆ A is smaller angle.

Now, âˆ A < âˆ B < âˆ C

âŸ¹ BC < AC < AB [Side opposite to greater angle is larger and side opposite to smaller angle is smaller]

AB is longest and BC is the shortest side.

**Question 2: In a **Î”**ABC, if âˆ B = âˆ C = 45Â°, which is the longest side?**

**Solution:**

Given that in Î”ABC,

Î”ABC, âˆ B = âˆ C = 45Â°

To find: longest side

We know that.

Sum of angles in a triangle =180Â°

âˆ A + âˆ B + âˆ C = 180Â°

âˆ A + 45Â° + 45Â° = 180Â°

âˆ A = 180Â° – (45Â° + 45Â°) = 180Â° – 90Â° = 90Â°

âˆ A = 90Â°

**Question 3: In **Î”**ABC, side AB is produced to D so that BD = BC. If âˆ B = 60Â° and âˆ A = 70Â°. Prove that: (i) AD > CD (ii) AD > AC**

**Solution:**

Given that, in Î” ABC, side AB is produced to D so that BD = BC.

âˆ B = 60Â°, and âˆ A = 70Â°

To prove:

(i) AD > CD (ii) AD > AC

First join C and D

We know that,

Sum of angles in a triangle =180Â°

âˆ A + âˆ B + âˆ C = 180Â°

70Â° + 60Â° + âˆ C = 180Â°

âˆ C = 180Â° – (130Â°) = 50Â°

âˆ C = 50Â°

âˆ ACB = 50Â° … (i)

Also in Î” BDC

âˆ DBC =180 – âˆ ABC [ABD is a straight angle]

180 – 60Â° = 120Â°

and also BD = BC[given]

âˆ BCD = âˆ BDC [Angles opposite to equal sides are equal]

Now,

âˆ DBC + âˆ BCD + âˆ BDC = 180Â° [Sum of angles in a triangle =180Â°]

âŸ¹ 120Â° + âˆ BCD + âˆ BCD = 180Â°

âŸ¹ 120Â° + 2âˆ BCD = 180Â°

âŸ¹ 2âˆ BCD = 180Â° – 120Â° = 60Â°

âŸ¹ âˆ BCD = 30Â°

âŸ¹ âˆ BCD = âˆ BDC = 30Â° …. (ii)

Now, consider Î”ADC.

âˆ BAC âŸ¹ âˆ DAC = 70Â° [given]

âˆ BDC âŸ¹ âˆ ADC = 30Â° [From (ii)]

âˆ ACD = âˆ ACB + âˆ BCD

= 50Â° + 30Â°[From (i) and (ii)] = 80Â°

Now, âˆ ADC < âˆ DAC < âˆ ACD

AC < DC < AD [Side opposite to greater angle is longer and smaller angle is smaller]

AD > CD and AD > AC

Hence proved

**Question: 4 Is it possible to draw a triangle with sides of length 2 cm, 3 cm**,** and 7 cm?**

**Solution:**

Given lengths of sides are 2cm, 3cm and 7cm.

To check whether it is possible to draw a triangle with the given lengths of sides

We know that,

A triangle can be drawn only when the sum of any two sides is greater than the third side.

So, let’s check the rule.

2 + 3 < 7 [the sum of any two sides is not greater than the third side, so not satisfying the triangle condition]

2 + 7 > 3

and 3 + 7 > 2

So, the triangle does not exit.

**Question: 5 O is any point in the interior of **Î”**ABC. Prove that.**

**(i) AB + AC > OB + OC**

**(ii) AB + BC + CA > OA + QB + OC**

**(iii) OA + OB + OC > (1/2)(AB + BC +CA)**

**Solution:**

Given that O is any point in the interior of Î”ABC

To prove

(i) AB + AC > OB + OC

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

(iii) OA + OB + OC > (1/2)(AB + BC +CA)

We know that in a triangle the sum of any two sides is greater than the third side.

So, we have

In Î” ABC

AB + BC > AC

BC + AC > AB

AC + AB > BC

In Î”OBC

OB + OC > BC … (i)

In Î”OAC

OA + OC > AC … (ii)

In Î”OAB

OA + OB > AB … (iii)

Now, extend BO to meet AC in D.

Now, in Î”ABD, we have

AB + AD > BD

AB + AD > BO + OD … (iv) [BD = BO + OD]

Similarly, in Î”ODC, we have

OD + DC > OC … (v)

(i) Adding (iv) and (v), we get

AB + AD + OD + DC > BO + OD + OC

AB + (AD + DC) > OB + OC

AB + AC > OB + OC … (vi)

Similarly, we have

BC + BA > OA + OC … (vii)

and CA+ CB > OA + OB … (viii)

(ii) Adding equation (vi), (vii) and (viii), we get

AB + AC + BC + BA + CA + CB > OB + OC + OA + OC + OA + OB

âŸ¹ 2AB + 2BC + 2CA > 2OA + 2OB + 2OC

âŸ¹ 2(AB + BC + CA) > 2(OA + OB + OC)

âŸ¹ AB + BC + CA > OA + OB + OC

(iii) Adding equations (i), (ii) and (iii)

OB + OC + OA + OC + OA + OB > BC + AC + AB

2OA + 2OB + 2OC > AB + BC + CA

We get = 2(OA + OB + OC) > AB + BC +CA

(OA + OB + OC) > (1/2)(AB + BC +CA)

**Question: 6 Prove that the perimeter of a triangle is greater than the sum of its altitudes.**

Proof:We know that of all the segments that can be drawn to a given line, from a point not lying on it, the perpendicular distance i.e, the perpendicular line segment is the shortest.

Therefore

AD âŠ¥ BC

AB > AD and AC > AD

AB + AC > 2AD …. (i)

BE âŠ¥ AC

BA > BE and BC > BE

BA + BC > 2BE … (ii)

CF âŠ¥ AB

CA > CF and CB > CF

CA + CB > 2CF … (iii)

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

AB + AC + BA + BC + CA + CB > 2AD + 2BE + 2CF

2AB + 2BC + 2CA > 2(AD + BE + CF)

AB + BC + CA > AD + BE + CF

The perimeter of the triangle is greater than that the sum of its altitudes

Hence proved

**Question 7: In Fig., prove that:**

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

**(ii) CD + DA + AB > BC**

**Solution:**

To prove

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

(ii) CD + DA+ AB > BC

We know that, in a triangle sum of any two sides is greater than the third side

(i) So,

In Î”ABC, we have

AB + BC > AC …. (i)

In Î”ADC, we have

CD + DA > AC …. (ii)

Adding (i) and (ii), we get

AB + BC + CD + DA > AC + AC

AB + BC + CD + DA > 2AC

(ii) Now, In Î” ABC, we have,

AB + AC > BC … (iii)

And in Î”ADC, we have

CD + DA > AC

Add AB on both sides

CD + DA + AB > AC + AB

From equation (iii) and (iv), we get,

CD + DA + AB > AC + AB > BC

CD + DA + AB > BC

Hence 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 drown 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.**

**Solution:**

(i)False because sum of three sides of a triangle is greater than sum of its three altitudes

(ii)True

(iii)True

(iv)False because the difference of any two sides of a triangle is less than third side.

(v)True because the side opposite to the greatest angle is longest in a triangle.

(vi)True because the perpendicular distance is the shortest distance from a point to a line not containing it.

**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.**

**Solution:**

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

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

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

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

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

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

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