Inequalities — ICSE Class 9 Maths Important Questions

Answer: (c) $\angle C$C.

The greatest angle lies opposite the greatest side. Comparing the sides, $AB=7>CA=6>BC=5$AB=7>CA=6>BC=5, so the greatest side is $AB$AB. Side $AB$AB is opposite vertex $C$C, hence the greatest angle is $\angle C$C.

Answer: (b) $5$5 cm, $6$6 cm, $10$10 cm.

A triangle exists only if the sum of any two sides exceeds the third. Here $5+6=11>10$5+6=11>10, $5+10>6$5+10>6, $6+10>5$6+10>5, so all conditions hold. In (a) $3+4=7<8$3+4=7<8, in (c) $2+3=5$2+3=5 (not $>5$>5), in (d) $4+4=8<9$4+4=8<9 fail.

Answer: (b) $QR<RP<PQ$QR<RP<PQ.

$\angle R=180^\circ-50^\circ-60^\circ=70^\circ$R=180^-50^-60^=70^. Ordering angles: $\angle P(50^\circ)<\angle Q(60^\circ)<\angle R(70^\circ)$P(50^)< Q(60^)< R(70^). The side opposite the smaller angle is smaller. Side opposite $\angle P$P is $QR$QR, opposite $\angle Q$Q is $RP$RP, opposite $\angle R$R is $PQ$PQ. Hence $QR<RP<PQ$QR<RP<PQ.

Answer: (b) $3<x<13$3<x<13.

By the triangle inequality the third side is less than the sum and greater than the difference of the other two: $8-5<x<8+5$8-5<x<8+5, i.e. $3<x<13$3<x<13.

Answer: (a) $\angle B$B is opposite side $AC$AC and $\angle C$C is opposite side $AB$AB. Since $\angle B>\angle C$B> C, the side opposite the greater angle is longer, so $AC>AB$AC>AB. R correctly explains A.

Since $AB=AC$AB=AC, the triangle is isosceles, so $\angle B=\angle C=70^\circ$B= C=70^.

Then $\angle A=180^\circ-70^\circ-70^\circ=40^\circ$A=180^-70^-70^=40^.

The smallest angle is $\angle A=40^\circ$A=40^, opposite side $BC$BC, so $BC$BC is the smallest. As $\angle B=\angle C$B= C, sides $AC=AB$AC=AB.

Ascending order: $BC<AB=CA$BC<AB=CA.

For a triangle, the sum of any two sides must be greater than the third side.

Check the two smaller sides: $6+7=13$6+7=13 cm.

But the third side is $14$14 cm, and $13<14$13<14.

Since the sum of two sides is not greater than the third side, no such triangle can be drawn.

In $\triangle ABD$ABD, the sum of two sides is greater than the third side:

$AB+BD>AD \quad\text{...(1)}$AB+BD>AD ...(1)

In $\triangle ACD$ACD, similarly:

$AC+DC>AD \quad\text{...(2)}$AC+DC>AD ...(2)

Adding (1) and (2):

$AB+BD+AC+DC>2AD$AB+BD+AC+DC>2AD

But $BD+DC=BC$BD+DC=BC (since $D$D lies on $BC$BC).

Therefore $AB+AC+BC>2AD$AB+AC+BC>2AD, i.e. $AB+BC+CA>2AD$AB+BC+CA>2AD. Hence proved.

Produce $BO$BO to meet $AC$AC at point $D$D.

In $\triangle ABD$ABD, the sum of two sides exceeds the third:

$AB+AD>BD=BO+OD \quad\text{...(1)}$AB+AD>BD=BO+OD ...(1)

In $\triangle ODC$ODC:

$OD+DC>OC \quad\text{...(2)}$OD+DC>OC ...(2)

Adding (1) and (2):

$AB+AD+OD+DC>BO+OD+OC$AB+AD+OD+DC>BO+OD+OC

Cancel $OD$OD from both sides:

$AB+AD+DC>BO+OC$AB+AD+DC>BO+OC

Since $AD+DC=AC$AD+DC=AC, we get $AB+AC>OB+OC$AB+AC>OB+OC, i.e. $OB+OC<AB+AC$OB+OC<AB+AC. Hence proved.

Since $AD\perp BC$AD BC, in right $\triangle ABD$ABD the hypotenuse $AB$AB is the longest side, so

$AB>AD \quad\text{...(1)}$AB>AD ...(1)

Similarly in right $\triangle ACD$ACD, the hypotenuse $AC$AC is the longest side, so

$AC>AD \quad\text{...(2)}$AC>AD ...(2)

(Here we use that the perpendicular is the shortest segment from $A$A to $BC$BC, so any slant segment such as $AB$AB or $AC$AC is greater than $AD$AD.)

Adding (1) and (2):

$AB+AC>2AD$AB+AC>2AD

Hence the sum of the two sides is greater than twice the altitude. Proved.

Given: $\triangle ABC$ABC. To prove: $BA+AC>BC$BA+AC>BC.

Construction: Produce $BA$BA to $D$D such that $AD=AC$AD=AC, and join $DC$DC.

Proof: In $\triangle ADC$ADC, $AD=AC$AD=AC (by construction), so it is isosceles and the base angles are equal:

$\angle ADC=\angle ACD \quad\text{...(1)}$ADC= ACD ...(1)

Now $\angle BCD=\angle BCA+\angle ACD$BCD= BCA+ ACD, so

$\angle BCD>\angle ACD$BCD> ACD

Using (1), $\angle BCD>\angle ADC$BCD> ADC, i.e. $\angle BCD>\angle BDC$BCD> BDC.

In $\triangle BDC$BDC, the side opposite the greater angle is greater. Since $\angle BCD>\angle BDC$BCD> BDC, the side opposite $\angle BCD$BCD (which is $BD$BD) is greater than the side opposite $\angle BDC$BDC (which is $BC$BC):

$BD>BC$BD>BC

But $BD=BA+AD=BA+AC$BD=BA+AD=BA+AC (since $AD=AC$AD=AC).

Therefore $BA+AC>BC$BA+AC>BC. Hence proved.

Let the diagonals $AC$AC and $BD$BD intersect at $O$O.

In any triangle the sum of two sides is greater than the third side. Apply this to the four triangles formed at $O$O.

In $\triangle OAB$OAB: $OA+OB>AB \quad\text{...(1)}$OA+OB>AB ...(1)

In $\triangle OBC$OBC: $OB+OC>BC \quad\text{...(2)}$OB+OC>BC ...(2)

In $\triangle OCD$OCD: $OC+OD>CD \quad\text{...(3)}$OC+OD>CD ...(3)

In $\triangle ODA$ODA: $OD+OA>DA \quad\text{...(4)}$OD+OA>DA ...(4)

Adding (1), (2), (3) and (4):

$2(OA+OB+OC+OD)>AB+BC+CD+DA$2(OA+OB+OC+OD)>AB+BC+CD+DA

Now $OA+OC=AC$OA+OC=AC and $OB+OD=BD$OB+OD=BD, so $OA+OB+OC+OD=AC+BD$OA+OB+OC+OD=AC+BD.

Therefore $2(AC+BD)>AB+BC+CD+DA$2(AC+BD)>AB+BC+CD+DA.

This gives $AB+BC+CD+DA<2(AC+BD)$AB+BC+CD+DA<2(AC+BD).

To obtain the required stricter form, apply the triangle inequality directly in $\triangle ABC$ABC and $\triangle ACD$ACD etc.:

In $\triangle ABC$ABC: $AB+BC>AC$AB+BC>AC; in $\triangle ACD$ACD: $CD+DA>AC$CD+DA>AC; in $\triangle ABD$ABD: $AB+DA>BD$AB+DA>BD; in $\triangle BCD$BCD: $BC+CD>BD$BC+CD>BD.

Adding all four: $2(AB+BC+CD+DA)>2(AC+BD)$2(AB+BC+CD+DA)>2(AC+BD), hence

$AB+BC+CD+DA>AC+BD.$AB+BC+CD+DA>AC+BD.

Hence proved.

(i) $\angle C=180^\circ-80^\circ-45^\circ=55^\circ$C=180^-80^-45^=55^.

(ii) The larger the angle, the longer the opposite side. Angles: $\angle A=80^\circ>\angle C=55^\circ>\angle B=45^\circ$A=80^> C=55^> B=45^. Opposite sides: $a=BC$a=BC (opp $\angle A$A), $b=CA$b=CA (opp $\angle B$B), $c=AB$c=AB (opp $\angle C$C).

So descending order of sides: $BC>AB>CA$BC>AB>CA, i.e. $a>c>b$a>c>b.

(iii) The longest side is $BC$BC ($=a$=a), because it lies opposite the largest angle $\angle A=80^\circ$A=80^.

(iv) The perpendicular from a point to a line is the shortest segment. Since $PM\perp\ell$PM and $PN$PN is a slant segment, $PM<PN$PM<PN.