Inequalities Concise Class-9th Selina ICSE Maths Solution
Inequalities Concise Class-9th Selina ICSE Maths Solution Chapter-11. We provide step by step Solutions of Exercise / lesson-11 Inequalities for ICSE Class-9 Concise Selina Mathematics by R K Bansal.
Our Solutions contain all type Questions with Exe-11 A to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics .
Inequalities Concise Class-9th Selina ICSE Maths Solution Chapter-11
Exercise – 11
From the following figure, prove that: AB > CD.
In a triangle PQR; QR = PR and ∠P = 36o. Which is the largest side of the triangle?
Since ∠R is the greatest, therefore, PQ is the largest side.
If two sides of a triangle are 8 cm and 13 cm, then the length of the third side is between a cm and b cm. Find the values of a and b such that a is less than b.
The sum of any two sides of the triangle is always greater than third side of the triangle.
Third side < 13+8 =21 cm.
The difference between any two sides of the triangle is always less than the third side of the triangle.
Third side > 13-8 =5 cm.
Therefore, the length of the third side is between 5 cm and 9 cm, respectively.
The value of a =5 cm and b= 21cm.
In each of the following figures, write BC, AC and CD in ascending order of their lengths.
Arrange the sides of ∆BOC in descending order of their lengths. BO and CO are bisectors of angles ABC and ACB respectively.
D is a point in side BC of triangle ABC. If AD > AC, show that AB > AC.
n the following figure, ∠BAC = 60o and ∠ABC = 65o.
(i) CF > AF
(ii) DC > DF
In the following figure; AC = CD; ∠BAD = 110o and ∠ACB = 74o.
Prove that: BC > CD
From the following figure; prove that:
(i) AB > BD
(ii) AC > CD
(iii) AB + AC > BC
In a quadrilateral ABCD; prove that:
(i) AB+ BC + CD > DA
(ii) AB + BC + CD + DA > 2AC
(iii) AB + BC + CD + DA > 2BD
Const: Join AC and BD.
(i) In ΔABC,
AB + BC > AC….(i)[Sum of two sides is greater than the
AC + CD > DA….(ii)[ Sum of two sides is greater than the
Adding (i) and (ii)
AB + BC + AC + CD > AC + DA
AB + BC + CD > AC + DA – AC
AB + BC + CD > DA …….(iii)
CD + DA > AC….(iv)[Sum of two sides is greater than the
Adding (i) and (iv)
AB + BC + CD + DA > AC + AC
AB + BC + CD + DA > 2AC
(iii) In ΔABD,
AB + DA > BD….(v)[Sum of two sides is greater than the
BC + CD > BD….(vi)[Sum of two sides is greater than the
Adding (v) and (vi)
AB + DA + BC + CD > BD + BD
AB + DA + BC + CD > 2BD
In the following figure, ABC is an equilateral triangle and P is any point in AC; prove that:
(i) BP > PA
(ii) BP > PC
P is any point inside the triangle ABC. Prove that:
∠BPC > ∠BAC.
Prove that the straight line joining the vertex of an isosceles triangle to any point in the base is smaller than either of the equal sides of the triangle.
We know that exterior angle of a triangle is always greater than each of the interior opposite angles.
∠ADC > ∠B ……..(i)
AB = AC
∠B = ∠C …..(ii)
From (i) and (ii)
∠ADC > ∠ C
(i) In ΔADC,
∠ADC > ∠C
AC > AD ………(iii) [side opposite to greater angle is greater]
(ii) In ΔABC,
AB = AC
AB > AD[ From (iii)]
In the following diagram; AD = AB and AE bisects angle A. Prove that:
(i) BE = DE
(ii) ∠ABD > ∠C
The sides AB and AC of a triangle ABC are produced; and the bisectors of the external angles at B and C meet at P. Prove that if AB > AC, then PC > PB.
In the following figure; AB is the largest side and BC is the smallest side of triangle ABC.
Write the angles xo, yo and zo in ascending order of their values.
Since AB is the largest side and BC is the smallest side of the triangle ABC
Since AB is the largest side and BC is the smallest side of the triangle ABC.
AB > AC > BC
⇒ 180° – z° > 180° – y° > 180° – x°
⇒ – z° > -y° > – x°
⇒ z° > y° > x°.
In quadrilateral ABCD, side AB is the longest and side DC is the shortest.
In triangle ABC, side AC is greater than side AB. If the internal bisector of angle A meets the opposite side at point D, prove that: ∠ADC is greater than ∠ADB.
In isosceles triangle ABC, sides AB and AC are equal. If point D lies in base BC and point E lies on BC produced (BC being produced through vertex C), prove that:
(i) AC > AD
(ii) AE > AC
(iii) AE > AD
We know that the bisector of the angle at the vertex of an isosceles triangle bisects the base at right angle.
Using Pythagoras theorem in AFB,
AB2 = AF2 + BF2…………..(i)
AD2 = AF2 + DF2…………..(ii)
We know ABC is isosceles triangle and AB = AC
AC2 = AF2 + BF2 ……..(iii)[ From (i)]
Subtracting (ii) from (iii)
AC2 – AD2 = AF2 + BF2 – AF2 – DF2
AC2 – AD2 = BF2 – DF2
Let 2DF = BF
AC2 – AD2 = (2DF)2 – DF2
or AC2 – AD2 = 4DF2 – DF2
AC2 = AD2 + 3DF2
hence AC2 > AD2
AC > AD
Similarly, AE > AC and AE > AD.
Given: ED = EC
Prove: AB + AD > BC.
The sum of any two sides of the triangle is always greater than the third side of the triangle.
In triangle ABC, AB > AC and D is a point in side BC. Show that: AB > AD.
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