RD Sharma Solutions Class 9 Chapter 10 Ex 10.6
(1) In \(\Delta\)
Solution:
Given that in \(\Delta\)
We have to find longest and shortest side
We know that,
Sum of angles in a triangle 180Â°

\(\angle\)
A + \(\angle\) B + \(\angle\) C = 180Â° 
40Â° + 60Â° + \(\angle\)
C = 180Â° 
\(\angle\)
C = 180Â° – ((10)0Â°) = 80Â° 
\(\angle\)
C = 80Â°
Now,
=> Â 40Â° < 60Â° < 80Â° = \(\angle\)
=> \(\angle\)
Now, \(\angle\)
=> 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.
(2) In a \(\Delta\)
Solution: Given that in \(\Delta\)
\(\angle\)
We have to find longest side
We know that.
Sum of angles in a triangle =180Â°

\(\angle\)
A + \(\angle\) B + \(\angle\) C = 180Â° 
\(\angle\)
A + 45Â° + 45Â° = 180Â° 
\(\angle\)
A = 180Â° – (45Â° + 45Â°) = 180Â° – 90Â° = 90Â° 
\(\angle\)
A = 90Â°
(3) In \(\Delta\)
Â Sol: Given that, in \(\Delta\)
\(\angle\)
To prove,
(i) AD > CD (ii) AD > AC
First join C and D
We know that,
Sum of angles in a triangle =180Â°

\(\angle\)
A + \(\angle\) B + \(\angle\) C = 180Â° 
70Â° + 60Â° + \(\angle\)
C = 180Â° 
\(\angle\)
C = 180Â° – (130Â°) = 50Â° 
\(\angle\)
C = 50Â° 
\(\angle\)
ACB = 50Â°â€¦â€¦(i)
And also in \(\Delta\)
\(\angle\)

180 – 60Â° = 120Â°
and also BD = BC[given]

\(\angle\)
BCD = \(\angle\) BDC [Angles opposite to equal sides are equal]
Now,
\(\angle\)
=> 120Â° + \(\angle\)
=> 120Â° + 2\(\angle\)
=> 2\(\angle\)
=> \(\angle\)
=> \(\angle\)
Now, consider \(\Delta\)
\(\angle\)
\(\angle\)
\(\angle\)
= 50Â° + 30Â°[From (i) and (ii)] =80Â°
Now, \(\angle\)

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,
\(\angle\)

AD > DC and AD> AC Â Â Â [Side opposite to greater angle is longer and smaller angle is smaller]
(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 \(\not>\)
2 + 7 > 3
and 3 + 7 > 2
Here 2 + 3 \(\not>\)
So, the triangle does not exit.
(5) O is any point in the interior ofÂ \(\Delta\)
(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 \(\Delta\)
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 \(\Delta\)
AB + BC > AC
BC + AC > AB
AC + AB > BC
In \(\Delta\)
OB + OC > BCâ€¦â€¦.(i)
In \(\Delta\)
OA + OC > ACâ€¦â€¦(ii)
In \(\Delta\)
OA + OB > ABâ€¦â€¦(iii)
Now, extend (or) produce BO to meet AC in D.
Now, in \(\Delta\)

AB + AD > BD

AB + AD > BO + OD â€¦â€¦.(iv) [BD = BO + OD]
Similarly in \(\Delta\)
OD + DC > OCâ€¦â€¦(v)
(i) Adding (iv) and (v), we get
AB + AD + OD + DC > BO + OD + OC

AB + (AD + DC) > OB + OC

AB + AC > OB + OC â€¦â€¦(vi)
Similarly, we have
BC + BA > OA + OCâ€¦â€¦(vii)
and CA+ CB > OA + OBâ€¦â€¦(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)
(6) Prove that the perimeter of a triangle is greater than the sum of its altitudes.
Solution:
Given, \(\Delta\)
To prove,
AD + BE + CF < AB + BC + AC
Figure:
Â
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 \(\perp\)

AB > AD and AC > AD

AB + AC > 2AD â€¦â€¦(i)
BE \(\perp\)

BA > BE and BC > BE

BA + BC > 2BE â€¦â€¦(ii)
CF \(\perp\)

CA > CF and CB > CF

CA + CB > 2CF â€¦â€¦(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
(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
From the given figure,
We know that, in a triangle sum of any two sides is greater than the third side
(i) So,
In \(\Delta\)
AB + BC > ACÂ â€¦â€¦(i)
In \(\Delta\)
CD + DA > ACâ€¦â€¦(ii)
Adding (i) and (ii), we get
AB + BC + CD + DA > AC + AC
AB + BC + CD + DA > 2AC
(ii) Now, In \(\Delta\)
AB +AC > BCâ€¦â€¦(iii)
And in \(\Delta\)
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
(8) Which of the following statements are true (T) and which are false (F)?
Â (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.
(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.