Question 1: In ΔABC, if ∠A = 40° and ∠B = 60°. Determine the longest and shortest sides of the triangle.
Solution:
Given that in ΔABC, ∠A = 40° and ∠B = 60°
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° – 100° = 80°
∠C = 80°
Now,
⟹ 40° < 60° < 80° = ∠A < ∠B < ∠C
⟹ ∠C is greater angle and ∠A is smaller angle.
Now, ∠A < ∠B < ∠C
⟹ 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 ΔABC, if ∠B = ∠C = 45°, which is the longest side?
Solution:
Given that in ΔABC,
ΔABC, ∠B = ∠C = 45°
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°
Question 3: In ΔABC, side AB is produced to D so that BD = BC. If ∠B = 60° and ∠A = 70°. Prove that: (i) AD > CD (ii) AD > AC
Solution:
Given that, in Δ ABC, side AB is produced to D so that BD = BC.
∠B = 60°, and ∠A = 70°
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)
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 ΔADC.
∠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
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 ΔABC. 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 ΔABC
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 Δ ABC
AB + BC > AC
BC + AC > AB
AC + AB > BC
In ΔOBC
OB + OC > BC … (i)
In ΔOAC
OA + OC > AC … (ii)
In ΔOAB
OA + OB > AB … (iii)
Now, extend BO to meet AC in D.
Now, in ΔABD, we have
AB + AD > BD
AB + AD > BO + OD … (iv) [BD = BO + OD]
Similarly, in ΔODC, 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)
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 ⊥ 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
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 ΔABC, we have
AB + BC > AC …. (i)
In ΔADC, 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 ΔADC, 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.