Frank Solutions for Chapter 13 Inequalities in Triangles Class 9 Mathematics ICSE
Exercise 13.1
1. Name the greatest and the smallest sides in the following triangles:
(a) △ABC, ∠A = 56°, ∠B = 64° and ∠C = 60°
(b) △DEF, ∠D = 32°, ∠E = 56° and ∠F = 92°
(c) △XYZ, ∠X = 76°, ∠Y = 84°
Answer
(a) In the given △ABC,
The greatest angle is ∠B and the opposite side to the ∠B is AC
Therefore,
The greatest side is AC
The smallest angle in the △ABC is ∠A and the opposite side to the ∠A is BC
Therefore,
The smallest side is BC
(b) In the given △DEF,
The greatest angle is ∠F and the opposite side to the ∠F is DE
Therefore,
The greatest side is DE
The smallest angle in the △DEF is ∠D and the opposite side to the ∠D is EF
Therefore,
The smallest side is EF
(c) In △XYZ,
∠X + ∠Y + ∠Z = 180°
⇒ 76° + 84° + ∠Z = 180°
⇒ 160° + ∠Z = 180°
⇒ ∠Z = 180° – 160°
We get,
∠Z = 20°
Therefore,
∠X = 76°, ∠Y = 84°, ∠Z = 20°
In the given △XYZ,
The greatest angle is ∠Y and the opposite side to the ∠Y is XZ
Therefore,
The greatest side is XZ
The smallest angle in the △XYZ is ∠Z and the opposite side to the ∠Z is XY
Therefore,
The smallest side is XY
2. Arrange the sides of the following triangles in an ascending order:
(a) △ABC, ∠A = 45°, ∠B = 65°
(b) △DEF, ∠D = 38°, ∠E = 58°
Answer
(a) In △ABC,
∠A + ∠B + ∠C = 180°
⇒ 45° + 65° + ∠C = 180°
⇒ 110° + ∠C = 180°
⇒ ∠C = 180° – 110°
We get,
∠C = 70°
Hence,
∠A = 45°, ∠B = 65°, ∠C = 70°
⇒ 45° < 65° < 70°
Hence,
Ascending order of the angles in the given triangle is,
∠A < ∠B < ∠C
Therefore,
Ascending order of sides in triangle is,
BC, AC, AB
(b) In △DEF,
∠D + ∠E + ∠F = 180°
⇒ 38° + 58° + ∠F = 180°
⇒ 96° + ∠F = 180°
⇒ ∠F = 180° – 96°
We get,
∠F = 84°
Hence,
∠D = 38°, ∠E = 58°, ∠F = 84°
⇒ 38° < 58° < 84°
Hence,
Ascending order of the angles in the given triangle is,
∠D < ∠E < ∠F
Therefore,
Ascending order of sides in triangle is,
EF, DF, DE
3. Name the smallest angle in each of these triangles:
(i) In △ABC, AB = 6.2 cm, BC = 5.6 cm and AC = 4.2 cm
(ii) In △PQR, PQ = 8.3 cm, QR = 5.4 cm and PR = 7.2 cm
(iii) In △XYZ, XY = 6.2 cm, XY = 6.8 cm and YZ = 5 cm
Answer
(i) We know that,
In a triangle, the angle opposite to the smallest side is the smallest
In △ABC,
AC = 4.2 cm is the smallest side
Hence,
∠B is the smallest angle
(ii) We know that,
In a triangle, the angle opposite to the smallest side is the smallest
In △PQR,
QR = 5.4 cm is the smallest side
Hence,
∠P is the smallest angle
(iii) We know that,
In a triangle, the angle opposite to the smallest side is the smallest
In △XYZ,
YZ = 5 cm is the smallest side
Therefore,
∠X is the smallest angle
4. In a triangle ABC, BC = AC and ∠A = 35°. Which is the smallest side of the triangle?
In △ABC,
BC = AC (given)
∠A = ∠B = 35° [angles opposite to the two equal sides are equal]
Let ∠C = x°
In △ABC,
∠A + ∠B + ∠C = 180°
⇒ 35° + 35° + x° = 180°
⇒ 70° + x° = 180°
⇒ x° = 180° – 70°
We get,
x° = 110°
Hence,
∠C = x° = 110°
Therefore,
Angles in a triangle are ∠A = ∠B = 35° and ∠C = 110°
In △ABC,
The greatest angle is ∠C
Here, the smallest angles are ∠A and ∠B
Hence,
Smallest sides are BC and AC
5. In △ABC, the exterior ∠PBC > exterior ∠QCB. Prove that AB > AC.
In the given triangle,
Given that, ∠PBC > ∠QCB …(1)
∠PBC + ∠ABC = 180° (linear pair angles)
∠PBC = 180° – ∠ABC …(2)
Similarly,
∠QCB = 180° – ∠ACB …(3)
Now,
From (2) and (3), we get,
180° – ∠ABC > 180° – ∠ACB
⇒ – ∠ABC > – ∠ACB
Therefore,
∠ABC < ∠ACB or ∠ACB > ∠ABC
We know that,
In a triangle, the greater angle has the longer side opposite to it
Therefore,
AB > AC
Hence, proved
6. △ABC is isosceles with AB = AC. If BC is extended at D, then prove that AD > AB.
In △ACD,
We have,
∠ACB = ∠CDA + ∠CAD [Using exterior angle property]
∠ACB > ∠CDA …(1)
Now,
AB = AC (given)
∠ACB = ∠ABC …(2)
From equations (1) and (2), we get,
∠ABC > ∠CDA
We know that,
In a triangle, the greater angle has the longer side opposite to it.
Now,
In △ABD,
We have,
∠ABC > ∠CDA
Therefore,
AD > AB
Hence, proved
7. Prove that the perimeter of a triangle is greater than the sum of its three medians.
Answer
Given
In △ABC,
AD, BE and CF are its medians
We know that,
The sum of any two sides of a triangle is greater than twice the median bisecting the third side.
Hence,
AD is the median bisecting BC
So,
AB + AC > 2 AD …(1)
BE is the median bisecting AC
AB + BC > 2BE …(2)
CF is the median bisecting AB
BC + AC > 2CF …(3)
Now,
Adding equations (1), (2) and (3), we get,
(AB + AC) + (AB + BC) + (BC + AC) > 2AD + 2BE + 2CF
⇒ 2 (AB + BC + AC) > 2 (AD + BE + CF)
We get,
AB + BC + AC > AD + BE + CF
Hence, proved
8. Prove that the hypotenuse is the longest side in a right angled triangle.
Answer
By angle sum property of a triangle,
∠A + ∠B + ∠C = 180°
⇒ ∠A + 90° + ∠C = 180°
We get,
∠A + ∠C = 90°
Hence,
The other two angles must be acute i.e. less than 90°
Therefore,
∠B is the largest angle in △ABC
∠B > ∠A and
∠B > ∠C
AC > BC and
AC > AB
∵ In any triangle, the side opposite to the greater angle is longer
Hence,
AC is the largest side in △ABC
But, AC is the hypotenuse of △ABC
Therefore,
Hypotenuse is the longest side in a right angled triangle
Hence, proved
9. D is a point on the side of the BC of △ABC. Prove that the perimeter of △ABC is greater than twice of AD.
Construction: Join AD
In △ACD,
AC + CD > AD …(1)
∵ Sum of two sides of a triangle is greater than the third side
Similarly,
In △ADB,
AB + BD > AD …(2)
Adding equations (1) and (2), we get,
AC + CD + AB + BD > 2 AD
⇒ AB + BC + AC > 2AD (since, CD + BD = BC)
Hence, proved
10. For any quadrilateral, prove that its perimeter is greater than the sum of its diagonals.
Answer
PQRS is a quadrilateral
PR and QS are the diagonals of quadrilateral
To prove: PQ + QR + SR + PS > PR + QS
Proof: In △PQR,
PQ + QR > PR (Sum of two sides of a triangle is greater than the third side)
Similarly,
In △PSR,
PS + SR > PR
In △PQS,
PS + PQ > QS and
In △QRS,
QR + SR > QS
Now,
We have,
PQ + QR > PR
PS + SR > PR
PS + PQ > QS
QR + SR > QS
After adding all the above inequalities, we get,
PQ + QR + PS + SR + PS + PQ + QR + SR > PR + PR + QS + QS
⇒ 2PQ + 2QR + 2PS + 2SR > 2PR + 2QS
⇒ 2 (PQ + QR + PS + SR) > 2 (PR + QS)
We get,
PQ + QR + PS + SR > PR + QS
Hence, proved
11. ABCD is a quadrilateral in which the diagonals AC and BD intersect at O. Prove that AB + BC + CD + AD < 2 (AC + BD)
We know that,
Sum of two sides of a triangle is greater than the third side
Hence,
In △AOB,
OA + OB > AB …(1)
In △BOC,
OB + OC > BC …(2)
In △COD,
OC + OD > CD …(3)
In △AOD,
OA + OD > AD …(4)
Now,
Adding equations (1) (2) and (3) and (4), we get,
2 (OA + OB + OC + OD) > AB + BC + CD + AD
⇒ 2[(OA + OC) + (OB + OD)] > AB + BC + CD + AD
We get,
2 (AC + BD) > AB + BC + CD + AD
∵ OA + OC = AC and OB + OD = BD
Therefore,
AB + BC + CD + AD < 2 (AC + BD)
Hence, proved
12. In △ABC, P, Q and R are points on AB, BC and AC respectively. Prove that AB + BC + AC > PQ + QR + PR.
We know that,
Sum of two sides of a triangle is greater than the third side
Hence,
In △APR,
AP + AR > PR …(1)
In △BPQ,
BQ + PB > PQ …(2)
In △QCR,
QC + CR > QR …(3)
Now,
Adding equations (1), (2) and (3), we get,
AP + AR + BQ + PB + QC + CR > PR + PQ + QR
⇒ (AP + PB) + (BQ + QC) + (CR + AR) > PR + PQ + QR
We get,
AB + BC + AC > PQ + QR + PR
Hence, proved
13. In △PQR, PR > PQ and T is a point on PR such that PT = PQ. Prove that QR > TR.
In △PQT,
We have
PT = PQ …(1) (given)
In △PQR,
PQ + QR > PR
⇒ PQ + QR > PT + TR (∵ PR = PT + TR)
⇒ PQ + QR > PQ + TR [Using equation (1)]
We get,
QR > TR
Hence, proved
14. ABCD is a trapezium. Prove that:
(ii) CD + DA + AB > BC
Answer
(i) In △ABC,
We have,
AB + BC > AC ...(1)
In △ACD,
We have,
AD + CD > AC …(2)
Adding equations (1) and (2), we get,
AB + BC + AD + CD > 2AC
Hence, proved
(ii) In △ACD,
We have,
CD + DA > CA
⇒ CD + DA + AB > CA + AB
⇒ CD + DA + AB > BC (since, AB + AC > BC)
Hence, proved
15. In △ABC, BC produced to D, such that, AC = CD; ∠BAD = 125° and ∠ACD = 105°. Show that BC > CD.
In △ACD,
AC = CD …(given)
∠CDA = ∠DAC …(△ACD is an isosceles triangle)
Let ∠CDA = ∠DAC = x°
∠CDA + ∠DAC + ∠ACD = 180°
⇒ x° + x° + 105° = 180°
⇒ 2x° + 105° = 180°
⇒ 2x° = 180° – 105°
We get,
2x° = 75°
⇒ x = (75°/2)
We get,
x = 37.5°
∠CDA = ∠DAC = x° = 37.5° …(1)
∠DAB = ∠DAC + ∠BAC
⇒ 125° = 37.5° + ∠BAC [from equation (1)]
⇒ 125° – 37.5° = ∠BAC
⇒ 87.5° = ∠BAC
Also,
∠BCA + ∠ACD = 180°
⇒ ∠BCA + 105° = 180°
We get,
∠BCA = 75°
Now,
In △BAC,
∠ACB + ∠BAC + ∠ABC = 180°
⇒ 75° + 87.5° + ∠ABC = 180°
⇒ ∠ABC = 180° – 75° – 87.5°
We get,
∠ABC = 17.5°
Since, 87.5° > 17.5°
Hence,
∠BAC > ∠ABC
BC > AC
Therefore,
BC > CD …(Since AC = CD)
Hence, proved
16. In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that:
(a) PN < RN
(b) SN < SR
Answer
(a) In the given △PQR,
PS < PR …(Of all the straight lines that can be drawn to a given straight line from a point outside it, the perpendicular is the shortest.)
PN < PR …(i) (∵ PN < PS)
Also,
RT < PR …(Of all the straight lines that can be drawn to a given straight line from a point outside it, the perpendicular is the shortest.)
RN < PR …(ii) (∵ RN < RT)
Dividing (i) by (ii),
PN/RN < PR/PR
⇒ PN/RN < 1
⇒ PN < RN
(a) In △RTO,
∠RTQ + ∠TQR + ∠TRQ = 180°
⇒ 90° + 60° + ∠TRQ = 180°
⇒ 150° + ∠TRQ = 180°
⇒ ∠TRQ = 180° - 150°
⇒ ∠TRQ = 30°
∠TRQ = ∠SRN = 30° …(iii)
In △NSR,
∠RNS + ∠SRN = 90° (∵ ∠NSR = 90°)
∠RNS + 30° = 90° [from (ii)]
∠RNS = 90° - 30°
∠RNS = 60° …(iv)
∠SRN < ∠RNS …(from (iii) and (iv))
SN < SR
17. (a) In △PQR, PS ⊥ QR; prove that; PQ > QS and PQ > PS
(a) In △PQS,
PS < PQ …(Of all the straight lines that can be drawn to a given straight line from a point outside it, the perpendicular is the shortest.)
i.e., PQ > PS
Also, QS < QP …(Of all the straight lines that can be drawn to a given straight line from a point outside it, the perpendicular is the shortest.)
(b) In △PQS,
PS ⊥ QR …(Given)
PS < PR …(Of all the straight lines that can be drawn to a given straight line from a point outside it, the perpendicular is the shortest.)
i.e., PR > PS
(c) In △PQR,
PQ + PR > QR (∵ Sum of two sides of a triangle is always greater than third side.)
In △PQS,
PQ + QS > PS (∵ Sum of two sides of a triangle is always greater than third side.)…(i)
In △PRS,
PR + SR > PS (∵ Sum of two sides of a triangle is always greater than third side.) …(ii)
Adding (i) and (ii),
PQ + QS + PR + SR > 2PS
⇒ PQ + (QS + SR) + PR > 2PS
⇒ PQ + QR + PR > 2PS
Since PQ + PR > QR
⇒ PQ + QR > 2PS
18. In the given figure, T is a point on the side PR of triangle PQR. Show that
(a) PT < QT
(b) RT > QT
Note: The question is incomplete.
Question should be:
In the given figure, T is a point on the side PR of an equilateral triangle PQR,
Show that:
(a) PT < QT
(b) RT < QT
Answer
(a) In △PQR,
PQ = QR = PR
⇒ ∠P = ∠Q = ∠R = 60°
In △PQT,
∠PQT < 60°
∴ ∠PQT < ∠P
∴ PT < QT
(b) In △TQR,
∠TQR < 60°
∴ ∠TQR < ∠R
∴ RT < QT
19. In △PQR is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR.
In △PQR,
PQ + PR > QR ...(∵ Sum of the two sides of a triangles is always greater than the third side.)…(i)
Also, In △SQR,
SQ + SR > QR …(∵ Sum of the two sides of a triangle is always greater than the third side.)…(ii)
Dividing (i) and (ii),
(PQ + PR)/(SQ + SR) > QR/QR
⇒ (PQ + PR)/(SQ + SR) > 1
⇒ PQ + PR > SQ + SR
i.e., SQ + SR < PQ + PR
20. Prove that in an isosceles triangle any of its equal sides is greater than the straight line joining the vertex on any point on base of the triangle.
Answer
PS ⊥ QR, the straight line joining vertex P to the line QR.
To prove: PQ > PT and PR > PT
In △PSQ,
PS2 + SQ2 = PQ2 ...(Pythagoras theorem)
PS2 = PQ2 – SQ2 ...(i)
In △PST,
PS2 + ST2 = PT2 …(Pythagoras theorem)
⇒ PS2 = PT2 - ST2 …(ii)
⇒ PQ2 – SQ2 = PT2 – ST2 …[from (i) and (ii)]
⇒ PQ2 – (ST + TQ)2 = PT2 – ST2
⇒ PQ2 - (ST2 + 2ST × TQ + TQ2) = PT2 – ST2
⇒ PQ2 – ST2 – 2ST × TQ – TQ2 = PT2 – ST2
⇒ PQ2 – PT2 = TQ2 + 2ST × TQ
⇒ PQ2 – PT2 = TQ × (2ST + TQ)
As, TQ × (2ST + TQ) > 0 always
⇒ PQ2 – PT2 > 0
⇒ PQ2 > PT2
⇒ PQ > PT
Also, PQ = PR
PR > PT
21. △ABC, in a isosceles triangle with AB = AC. D is a point on BC produced. ED intersects AB at E and AC at F. Prove that AF > AE.
∠AEF > ∠ABC …(Exterior angle property)
∠AFE = ∠DFC
∠ACB > DFC …(Exterior angle property)
⇒ ∠ACB > ∠AFE
Since AB = AC
⇒ ∠ACB = ∠ABC
So, ∠ABC > ∠AFE
⇒ ∠AFE > ∠ABC > ∠AFE that is, ∠AEF > ∠AFE
⇒ AF > AE
22. In △ABC, AE is the bisector of ∠BAC. D is a point on AC such that AB = AD. Prove that BE = DE and ∠ABD > ∠C.
