Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions -Ex-10.6, Congruent Triangles, Class 9, Maths

Ex-10.6, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q. (1) In Δ ABC, if ∠ A = 40° and ∠ B = 60°. Determine the longest and shortest sides of the triangle.

Solution:

Given that in Δ ABC, ∠ A = 40° and ∠ B = 60°

Ex-10.6, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

We have to find longest and shortest side

We know that,

Sum of angles in a triangle 180°

  • ∠ A + ∠ B + ∠ C = 180°

  • 40° + 60° + ∠ C = 180°

  • ∠ C = 180° – ((10)0°) = 80°

  • ∠ C = 80°

Now,

=>  40° < 60° < 80° = ∠ A < ∠ B < ∠ C

=> ∠ C is greater angle and ∠ A is smaller angle.

Now, ∠ A < ∠ B < ∠ C

=> BC < AAC < AB [Side opposite to greater angle is larger and side opposite to smaller angle is smaller]

AB is longest and BC is smallest or shortest side.

 

Q.  (2) In a Δ ABC, if ∠ B = ∠ C = 45°, which is the longest side?

Solution: Given that in Δ ABC,

∠ B = ∠ C = 45°

Ex-10.6, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

We have to find longest side

We know that.

Sum of angles in a triangle =180°

  • ∠ A + ∠ B + ∠ C = 180°

  • ∠ A + 45° + 45° = 180°

  • ∠ A = 180° – (45° + 45°) = 180° – 90° = 90°

  • ∠ A = 90°

 

Q. (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

 Sol: Given that, in Δ ABC, side AB is produced to D so that BD = BC.

∠ B = 60°, and ∠ A = 70°

Ex-10.6, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

To prove,

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

First join C and D

We know that,

Sum of angles in a triangle =180°

  • ∠ A + ∠ B + ∠ C = 180°

  • 70° + 60° + ∠ C = 180°

  • ∠ C = 180° – (130°) = 50°

  • ∠ C = 50°

  • ∠ ACB = 50°……(i)

And also in Δ BDC

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

  • 180 – 60° = 120°

and also BD = BC[given]

  • ∠ BCD = ∠ BDC [Angles opposite to equal sides are equal]

Now,

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

=> 120° + ∠ BCD + ∠ BCD = 180°

=> 120° + 2∠ BCD = 180°

=> 2∠ BCD = 180° – 120° = 60°

=> ∠ BCD = 30°

=> ∠ BCD = ∠ BDC = 30°  ….(ii)

Now, consider Δ ADC.

∠ BAC => ∠ DAC = 70° [given]

∠ BDC => ∠ ADC = 30° [From (ii)]

∠ ACD = ∠ ACB+ ∠ BCD

= 50° + 30°[From (i) and (ii)] =80°

Now, ∠ ADC < ∠ DAC < ∠ ACD

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

  • AD > CD and AD > AC

Hence proved

Or,

We have,

∠ ACD > ∠ DAC and ∠ ACD > ∠ ADC

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

 

Q. (4) Is it possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?

Sol:

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 or 2 + 3 < 7

2 + 7 > 3

and 3 + 7 > 2

Here 2 + 3 ≯ 7

So, the triangle does not exit.

 

Q. (5) O is any point in the interior of  Δ 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 Δ ABC

To prove

(i) AB + AC > OB + OC

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

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

Ex-10.6, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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

So, we have

In Δ ABC

AB + BC > AC

BC + AC > AB

AC + AB > BC

In Δ OBC

OB + OC > BC…….(i)

In Δ OAC

OA + OC > AC……(ii)

In Δ OAB

OA + OB > AB……(iii)

Now, extend (or) produce BO to meet AC in D.

Now, in Δ ABD, we have

  • AB + AD > BD

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

Similarly in Δ 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 => 2AB + 2BC + 2CA > 2OA + 2OB + 2OC

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

=> 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)

 

Q. (6) Prove that the perimeter of a triangle is greater than the sum of its altitudes.

Solution:

Given, Δ ABC in which AD ⊥ BC, BE ⊥ AC and CF ⊥ AB.

To prove,

AD + BE + CF < AB + BC + AC

Figure:

 Ex-10.6, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

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 ⊥ BC

  • AB > AD and AC > AD

  • AB + AC > 2AD ……(i)

BE ⊥ AC

  • BA > BE and BC > BE

  • BA + BC > 2BE ……(ii)

CF ⊥ 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

 

Q. (7) In Fig., prove that:

Ex-10.6, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

(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

From the given figure,

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

(i) So,

In Δ ABC, we have

AB + BC > AC ……(i)

In Δ 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 Δ ABC, we have,

AB +AC > BC……(iii)

And in Δ 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

 

Q. (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 (F)

Reason: Sum of these sides of a triangle is greater than sum of its three altitudes

(ii) True (T)

(iii) True (T)

(iv) False (F)

Reason: The difference of any two sides of a triangle is less than third side

(v) True (T)

Reason: The side opposite to greater angle is longer and smaller angle is shorter in a triangle

(vi) True (T).

Reason: The perpendicular distance is the shortest distance from a point to a line not containing it.

 

Q. (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

Reason: Since a triangle can have only one right angle, other two angles must be less than 90

  • The right angle is the largest angle.

  • 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.

The document Ex-10.6, Congruent Triangles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on Ex-10.6, Congruent Triangles, Class 9, Maths RD Sharma Solutions - RD Sharma Solutions for Class 9 Mathematics

1. What are congruent triangles?
Ans. Congruent triangles are two triangles that have exactly the same shape and size. In other words, all corresponding sides and angles of congruent triangles are equal.
2. How can we prove two triangles congruent?
Ans. Two triangles can be proven congruent by using different criteria such as SSS (side-side-side), SAS (side-angle-side), ASA (angle-side-angle), AAS (angle-angle-side), RHS (right angle-hypotenuse-side), etc. These criteria involve comparing the corresponding sides and angles of the triangles.
3. What is the importance of congruent triangles in mathematics?
Ans. Congruent triangles play a crucial role in various mathematical concepts and applications. They are used to prove various geometric theorems, solve problems related to angles and sides of triangles, and establish properties of other geometric shapes. They are also used in real-life applications such as architecture, engineering, and navigation.
4. How do congruent triangles help in solving problems involving triangles?
Ans. Congruent triangles help in solving problems involving triangles by providing a systematic approach to analyze and compare the corresponding sides and angles of triangles. By establishing congruence between two triangles, we can deduce that all their corresponding parts are equal. This allows us to find unknown angles or sides, solve for missing measurements, or prove geometric theorems.
5. What are some real-life examples where congruent triangles are used?
Ans. Congruent triangles have various real-life applications. For example, in architecture and construction, congruent triangles are used to ensure that structures are symmetrical and stable. In navigation, congruent triangles are used to determine distances and angles between locations. In art and design, congruent triangles are used to create visually pleasing patterns and shapes.
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