Frank Solutions for Class 9 Maths Chapter 13 Inequalities in Triangles are provided here, which will improve the problem-solving skills of students and also will help them learn the techniques to solve difficult questions with speed and accuracy. Practising these solutions on a daily basis is the most convenient way to score desired marks in the final exams. Students who find Maths a complex subject are advised to practise Frank Solutions, as the answers are available in an interactive manner.
Chapter 13 deals with the study of Inequality in Triangles. This chapter provides formulas and properties of Inequalities in Triangles which are important from an exam perspective. By practising these solutions, students can identify their weak areas and work on them. Besides, it will boost the problem-solving and time-management skills of students, irrespective of their intelligence levels. For in-depth knowledge of the concepts, access Frank Solutions for Class 9 Maths Chapter 13 Inequalities in Triangles, from the link provided below.
Frank Solutions for Class 9 Maths Chapter 13 Inequalities in Triangles Download PDF
Access Frank Solutions for Class 9 Maths Chapter 13 Inequalities in Triangles
1. Name the greatest and the smallest sides in the following triangles:
(a) △ABC, ∠A = 560, ∠B = 640 and ∠C = 600
(b) △DEF, ∠D = 320, ∠E = 560 and ∠F = 920
(c) △XYZ, ∠X = 760, ∠Y = 840
Solution:
(a) In the given â–³ABC,
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 â–³DEF,
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 â–³XYZ,
∠X + ∠Y + ∠Z = 1800
760 + 840 + ∠Z = 1800
1600 + ∠Z = 1800
∠Z = 1800 – 1600
We get,
∠Z = 200
Therefore,
∠X = 760, ∠Y = 840, ∠Z = 200
In the given â–³XYZ,
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 ascending order:
(a) △ABC, ∠A = 450, ∠B = 650
(b) △DEF, ∠D = 380, ∠E = 580
Solution:
(a) In â–³ABC,
∠A + ∠B + ∠C = 1800
450 + 650 + ∠C = 1800
1100 + ∠C = 1800
∠C = 1800 – 1100
We get,
∠C = 700
Hence,
∠A = 450, ∠B = 650, ∠C = 700
450 < 650 < 700
Hence,
Ascending order of the angles in the given triangle is,
∠A < ∠B < ∠C
Therefore,
Ascending order of sides in the triangle is,
BC, AC, AB
(b) In â–³DEF,
∠D + ∠E + ∠F = 1800
380 + 580 + ∠F = 1800
960 + ∠F = 1800
∠F = 1800 – 960
We get,
∠F = 840
Hence,
∠D = 380, ∠E = 580, ∠F = 840
380 < 580 < 840
Hence,
Ascending order of the angles in the given triangle is,
∠D < ∠E < ∠F
Therefore,
Ascending order of sides in the triangle is,
EF, DF, DE
3. Name the smallest angle in each of these triangles:
(i) In â–³ABC, AB = 6.2 cm, BC = 5.6 cm and AC = 4.2 cm
(ii) In â–³PQR, PQ = 8.3 cm, QR = 5.4 cm and PR = 7.2 cm
(iii) In â–³XYZ, XY = 6.2 cm, XY = 6.8 cm and YZ = 5 cm
Solution:
(i) We know that,
In a triangle, the angle opposite to the smallest side is the smallest
In â–³ABC,
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 â–³PQR,
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 â–³XYZ,
YZ = 5 cm is the smallest side
Therefore,
∠X is the smallest angle
4. In a triangle ABC, BC = AC and ∠A = 350. Which is the smallest side of the triangle?
Solution:
In â–³ABC,
BC = AC (given)
∠A = ∠B = 350 [angles opposite to the two equal sides are equal]
Let ∠C = x0
In â–³ABC,
∠A + ∠B + ∠C = 1800
350 + 350 + x0 = 1800
700 + x0 = 1800
x0 = 1800 – 700
We get,
x0 = 1100
Hence,
∠C = x0 = 1100
Therefore,
Angles in a triangle are ∠A = ∠B = 350 and ∠C = 1100
In â–³ABC,
The greatest angle is ∠C
Here, the smallest angles are ∠A and ∠B
Hence,
The smallest sides are BC and AC
5. In △ABC, the exterior ∠PBC > exterior ∠QCB. Prove that AB> AC.
Solution:
In the given triangle,
Given that, ∠PBC > ∠QCB ……. (1)
∠PBC + ∠ABC = 1800 (linear pair angles)
∠PBC = 1800 – ∠ABC …. (2)
Similarly,
∠QCB = 1800 – ∠ACB ……(3)
Now,
From (2) and (3), we get,
1800 – ∠ABC > 1800 – ∠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. â–³ABC is isosceles with AB = AC. If BC is extended at D, then prove that AD > AB.
Solution:
In â–³ACD,
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 â–³ABD,
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.
Solution:
Given
In â–³ABC,
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.
Solution:
Let â–³ABC be a right-angled triangle in which the right angle is at B
By the angle sum property of a triangle,
∠A + ∠B + ∠C = 1800
∠A + 900 + ∠C = 1800
We get,
∠A + ∠C = 900
Hence,
The other two angles must be acute i.e less than 900
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 of â–³ABC
But AC is the hypotenuse of â–³ABC
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 â–³ABC. Prove that the perimeter of â–³ABC
is greater than twice of AD.
Solution:
Construction: Join AD
In â–³ACD,
AC + CD > AD ….. (1)
∵ The sum of two sides of a triangle is greater than the third side
Similarly,
In â–³ADB,
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.
Solution:
Given
PQRS is a quadrilateral
PR and QS are the diagonals of the quadrilateral
To prove: PQ + QR + SR + PS > PR + QS
Proof: In â–³PQR,
PQ + QR > PR (Sum of two sides of a triangle is greater than the third side)
Similarly,
In â–³PSR,
PS + SR > PR
In â–³PQS,
PS + PQ > QS and
In â–³QRS,
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)
Solution:
We know that,
The sum of two sides of a triangle is greater than the third side
Hence,
In â–³AOB,
OA + OB > AB ……… (1)
In â–³BOC,
OB + OC > BC ……… (2)
In â–³COD,
OC + OD > CD ……… (3)
In â–³AOD,
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 â–³ABC, P, Q and R are points on AB, BC and AC, respectively. Prove that AB + BC + AC > PQ + QR + PR.
Solution:
We know that,
The sum of two sides of a triangle is greater than the third side
Hence,
In â–³APR,
AP + AR > PR …….. (1)
In â–³BPQ,
BQ + PB > PQ …….. (2)
In â–³QCR,
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 â–³PQR, PR > PQ and T is a point on PR such that PT = PQ. Prove that QR > TR.
Solution:
In â–³PQT,
We have
PT = PQ ……. (1) (given)
In â–³PQR,
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:
(i) CD + DA + AB + BC > 2AC
(ii) CD + DA + AB > BC
Solution:
(i) In â–³ABC,
We have,
AB + BC > AC …… (1)
In â–³ACD,
We have,
AD + CD > AC ………. (2)
Adding equations (1) and (2), we get,
AB + BC + AD + CD > 2AC
Hence, proved
(ii) In â–³ACD,
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 = 1250 and ∠ACD = 1050. Show that BC > CD.
Solution:
In â–³ACD,
AC = CD … (given)
∠CDA = ∠DAC …….. (△ACD is an isosceles triangle)
Let ∠CDA = ∠DAC = x0
∠CDA + ∠DAC + ∠ACD = 1800
x0 + x0 + 1050 = 1800
2x0 + 1050 = 1800
2x0 = 1800 – 1050
We get,
2x0 = 750
x = (750 / 2)
We get,
x = 37.50
∠CDA = ∠DAC = x0 = 37.50 ………… (1)
∠DAB = ∠DAC + ∠BAC
1250 = 37.50 + ∠BAC {from equation (1)}
1250 – 37.50 = ∠BAC
87.50 = ∠BAC
Also,
∠BCA + ∠ACD = 1800
∠BCA + 1050 = 1800
We get,
∠BCA = 750
Now,
In â–³BAC,
∠ACB + ∠BAC + ∠ABC = 1800
750 + 87.50 + ∠ABC = 1800
∠ABC = 1800 – 750 – 87.50
We get,
∠ABC = 17.50
Since 87.50 > 17.50
Hence,
∠BAC > ∠ABC
BC > AC
Therefore,
BC > CD …. (Since AC = CD)
Hence, proved
Comments