## Class 8: Triangles – Exercise 30C

Q2. State giving reasons, whether the following pairs of triangles are congruent or Not.

$i) \ \ \Delta ABC \ in \ which \angle A = 50^{\circ}, \ \angle B=60^{\circ} \ and \ BC = 4.5 \ cm \ and \ \Delta DEF \ in \ which \ \angle E = 60^{\circ}, \ EF = 4.5 cm, \ \angle F = 70^{\circ}$

$\angle C = 70^{\circ}, \angle D^{\circ} = 50$

$\angle B = \angle E,\ \angle C = \angle F$

$BC = EF = 4.5$

Congruent by ASA

$ii)\ \ \Delta DEF \ in\ which\ \angle E=48^{\circ},\ DE=6cm\ and\ EF=8cm\ and \ \Delta MNR \ in \ which\ \angle R=48^{\circ},\ MN = 6 cm\ and\ MR = 8cm$

$DE = MN,\ EF = MR \ but \angle E \ne \angle M \ Hence \ Not \ Congruent.$

$iii)\ \ \Delta KLM \ in \ which \ KM=4 cm, \ \angle K=75^{\circ}, \ \angle M=40^{\circ} \ and\ \Delta PQR \ in \ which \ PR=4 cm,\ \angle Q=65^{\circ},\ \angle R=40^{\circ}$

$\angle L = 65^{\circ} and \angle P = 75^{\circ} \angle K = \angle P,$

$\angle M = \angle R \ and \ KM = PR$

ASA Theorem applied, Triangles are Congruent.

$iv) \\ \Delta ABC\ in\ which\ AB=3 cm,\ \angle A=90^{\circ},\ BC=5 cm\ and\ \Delta KLM \ in\ which\ KM = 3 cm,\ \angle K=90^{\circ},\ \angle M=5cm$

Applying Pythagoras theorem,

$AC = 4 cm \ KL = 4 cm$

$AB=KM,\ AC=KL\ and\ \angle A=\angle K$

Hence Triangles are congruent by SAS

Q3. In the adjoining figure, P in the mid point of AB and $\angle PAC = \angle PBD$. Prove that: $\Delta PAC \cong \Delta PBD$

$\angle PAC = \angle PBD$, (given)

$AP = PB$  (given)

$\angle APC = \angle DPB$ (opposite angles)

Hence $\Delta PAC \cong \Delta ADC$

Q4. In the adjoining figure, $DC \parallel AB \ and \ \angle B = \angle D$. Prove that $\Delta ABC \cong \Delta ADC$

Given $\angle B = \angle D$

$\angle DCA = \angle CAB (alternate \ angles)$

$\angle DAC = \angle ACB (alternate \ angles)$

$AC = Common$

Therefore, by ASA $\Delta ABC \cong \Delta ADC$

Q5. In the adjoining figure, $BC = AD, \angle CAB = \angle ABD \ and\ \angle ACB = \angle BDA.$

$Prove \Delta ABC \cong \Delta BAD$

In $\Delta ABD \ and\ \Delta ABC$

$\angle CAB = \angle DBA$

$\angle ADB = \angle ACB$

and DA = CB

Therefore, by ASA, $\Delta ABC \cong \Delta BAD$

Q6. In the adjoining figure, ABC is a triangle in which $AB=AC,\ BL \perp AC \ and \ CM \perp AB. \ Prove\ that\ BL=CM$.

Take $\Delta ABL \ and\ \Delta ACM$

$AB = AC$

$\angle A$ is common and

$\angle ABL = \angle ACM$

Therefore, by ASA, $\Delta ABL \cong \Delta ACM \ Hence \ BL = CM$

Q7. $In \ \Delta ABC AB=AC. \ If \ D \ in \ the \ mid-point \ of \ BC,$

$Prove \ that: \ AD is \ bisector \ of \angle A \ and \ AD \perp BC$

$In\ \Delta ABD \ and\ \Delta ADC$

$AB=AC$

$BD = DC$

$and AD \ is \ Common$

By SSS, $\Delta ABD \cong \Delta ADC$

Therefore $\angle BAD = \angle DAC$

$i) AD \ in \ bisector \ of \angle A$

$ii) \angle ADB + \angle ADC = 180^{\circ}\ \rightarrow \angle ADB = \angle ADC = 90^{\circ}$

Q8. In $\Delta ABC$, it in given that $AB = AC \ and\ AD$ in bisector of $\angle A$, meeting BC at D. Prove: i) $\Delta ABD \cong \Delta ACD \ ii) AD \perp BC$

$Given \ AB = AC$

$\angle ABD = \angle ACD$ (angles opposite equal sides of a triangle)

$\angle BAD = \angle DAC$ (angle bisector)

Therefore by ASA:  $\Delta ABD \cong \Delta ACD$

Hence $\angle ADB = \angle ADC = 90^{\circ}\ Therefore,\ AD \perp BC$

Q9. In the adjoining figure, $\Delta ABC$ in such that $AB = AC and \angle OBC = \angle OCB$. Prove that: $i) \Delta ABO \cong \Delta ACO \ ii) AO \ in \ the \ bisector \ of \ \angle A$

$\angle ABC = \angle ACB \ given \ AB = AC$

$\angle ABO + \angle OBC = \angle ACO+\angle OCB$

Given $\angle OBC = \angle OCB$

Therefore,

$\angle ABO = \angle ACO$

AB = AC (given)

AO is Common

Because  $\angle OBC=\angle OCB \rightarrow OB=OC$

Therefore, SAS applies, Hence $\Delta ABO \cong \Delta ACO$

Since $\Delta ABO \cong \Delta ACO, \ i) \angle BAO = \angle CAO \ ii) AO \ in \ Bisector \ of \ \angle A$

Q10. In the adjoining figure, $AB \parallel GF \ and \ AC \parallel DE \ and \ BF = CE.$

$Prove \ that \ \Delta BDE \cong \Delta FGC$

In $\Delta BDE \ and\ \Delta FGC$

$\angle DBE = \angle GFC$  (alternate angles)

$\angle DEB = \angle GCF$  (alternate angles)

$BF + FE = FE + EC$

Hence, $BE = FC$

Given $BF=EC \ and \ FE$  is common.

Therefore, $BE = FC$

Applying ASA, $\Delta BDE \cong \Delta FGC$

Q11. In the adjoining figure, ABCD as a square and CEB is an isosceles triangle in which EC = EB show: $\ i) \Delta DCE \cong \Delta ABE \ ii) AE = DE$

Given $DC = AB$ (Sides of a square)

$CE = EB$ (Sides of an isosceles triangle)

$\angle ECB = \angle EBC$

$Add \ 90^{\circ} on \ both \ side, \angle ECB + 90^{\circ} = \angle EBC + 90^{\circ} or \angle DCE = \angle ABE$

Hence SAS applies,

Therefore, $i)\ \Delta DCE \cong \Delta ABE \ ii) \ Because \ of i)\ AE = DE$

Q12. Find the vales of x and y in each of the following cases:

i)

Given $PN \parallel LM,$

$PQ = KM$

$\angle PQN = \angle MKL$

$\angle PNL = \angle NLM$(alternate angles)

Therefore, AAS applies Hence, $\Delta PQN \cong \Delta LMK$

Therefore $19=3x+4 \Rightarrow x=5 2y-3=55 \Rightarrow y=29$

ii)

In $\Delta ABC \ \& \ \Delta ADC$

$AC = Common$

$AB = AD \ and \ BC = DC$

Therefore by SSS, $\Delta ABC \cong \Delta ADC$

Hence, $7y + 1 = 180-103-34$

$7y+1=43 \Rightarrow y=6 \ and\ 2x-5=103 \Rightarrow x=54$

iii)

In $\Delta PQS \ \&\ \Delta PQR$

$PR = SQ$

$PS = QR$

And PQ in common

Therefore, by SSS, $\Delta PQS \cong \Delta PQR$

Hence $2y-5=71 \Rightarrow y = 38$

$\Delta SPO \cong \Delta ORQ$

Therefore $\ \angle SPO = \angle RQO = 34 \ \& \ \angle SOP = 75^{\circ} \ \&\ \angle POQ = 105^{\circ}$

Since, $\angle SPQ = \angle RQP \angle OPQ = (3x+3)$

Now Calculate, $105 + 2(3x+3) = 180 \Rightarrow x = 11.5$

iv)

In $\Delta RMP \ and\ \Delta RQN$

$\angle RPQ = \angle RQP$

$\angle RPM = \angle RQN$

$RP = RQ \ \&\ \angle MRP = \angle QRN$

Hence, ASA applies, Therefore, $\Delta RPM \cong \Delta RQN$

Hence $x -3=14 \Rightarrow x = 15 \$

And $5y-3=3x+1 \ or \ 5y = 55 \Rightarrow y=11$

$\\$

Q13. In adjoining figure, the sides BA and CA of  have been produced to D and E such that BA = AD and CA = AE Prove ED ∥ BC

In $\Delta AED \ and\ \Delta ABC$

$\angle EAD = \angle BAC$(Opposite Angles)

$BA = AD \ and\ EA = AC$

Hence $\Delta AED \cong \Delta ABC$

$\Rightarrow \angle ABC = \angle ADE \ and\ \angle ACB = \angle AED$

Hence $ED \parallel BC$

Q14. Equilateral triangle ABC and ACE have been drawn on the sides AB and AC respectively of $\Delta ABC$, as shown in the adjoining figure, prove: i) $\angle DAC = \angle EAB \ ii) \ DC = BE$

In $\Delta DAC \ \&\ \Delta EAB$

$DA = AB$

$AC = AE$

$\angle DAC = 60^{\circ} +\angle BAC =\angle BAE = 60^{\circ} +\angle BAC$

Hence $\Delta DAC \cong \Delta EAB$  Therefore,

$i)\ \angle DAC =\angle BAE$

$ii) \ DC = BE$

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Q15. In a regular pentagon ABCDE, prove that $\Delta ABD \ \$ is isosceles.

Since ABCD in a regular pentagon, all sides are equal and all internal angles are $108^{\circ}$.

In $\Delta ADE \ and\ \Delta BDC$

$ED = DC \ \&\ EA = CB \ \&\ \angle EA = \angle DCB =108^{\circ}$

Therefore $\Delta AED \cong \Delta DCB$

Hence $DA = DB$  Therefore,$\Delta ABD =$ isosceles triangle

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Q16. In the adjoining figure, $ABCD$ in a square, and $P,\ Q,\ R,\$ are points on the side $AB,\ BC,\ and\ CD$  respectively such that,$AP = BQ = CR \ and\ \angle PQR = 90^{\circ}.$

$Prove: \ i) \ PB = QC \ ii)\ PQ = QR \ iii) \ \angle QPR = 45^{\circ}$

Given ABCD in a square

$\therefore DC = AB$

$DR + RC = AP + PB$

$RC = AP$(given)

$\therefore DR = PB$

Similarly

$DC =BC$

$DR + RC = CQ + QB$

$RC = QB (Given)$

$\therefore DR = CQ$

Now Consider $\Delta PQB \ \&\ \Delta RCQ$

$\angle PBQ = \angle QCR = 90^{\circ}$ (Square)

$\therefore RQ = PQ$

$RC=QB \ \& \ PB=CQ$

$\therefore applying \ SSS, \Delta PQB \cong \Delta RCQ,$

$PQ = QR \Rightarrow \angle QPR = \angle QRP = 45^{\circ}$

Q17. In the adjoining figure, $QK \parallel ML, \ QM \parallel KL \ and \ RL=LP.$

$Prove:\ i)\ \angle MPL=\angle KLR \ ii)\ ML=QK$

Since $ML \parallel QK \angle PLM= \angle LRK$

And Similarly, $KL \parallel QM \angle LKR=\angle PML$

$PL = LR$ Given

Therefore, using AAS, $\Delta PML \cong \Delta KLR$

$i) \therefore \angle KLR= \angle MPL$

$ii) \ Since \ QM \parallel KL, \ ML = QK$

Q18. In the adjoining figure ABCD is a square, $EF \parallel AC$, and R is mid point of EF. Prove:  $i) \ AE=CF \ ii) \ DE=DF \ iii)\ DR \ bisect\ \angle EDF$

In $\Delta DER \ \&\ \Delta DFR$,

$\angle EDR = \angle FDR$ (Median would bisect the angle)

$ER = FR \ \& \ DR =$ Common

$\therefore \ \Delta DER \cong \Delta DFR$

$\Rightarrow DE = DF$

In $\Delta DAE \ \& \ \Delta DCF$

$DA = DC$ (Side of a square)

$angle DAE = \angle DCF = 90^{\circ} DE = DF$ (Proved above)

Because $\angle EDR=\angle FDR \Rightarrow \angle ADE = \angle CDF$

Applying SAS, Proves that$\Delta DAE \cong \Delta DCF$

Therefore, $AE = CF$. Hence Proven.

Q19. In the adjoining figure, $\angle TPQ=\angle SQP,\ \angle SRP=\angle TRQ \ and \ R$ in the Mid point of PQ Prove: $\ i) \ \Delta PRT \cong \Delta QRS \ ii)\ PT = SQ \ iii)\ \angle PTR = \angle QSR$

In $\Delta PRT \ \&\ \Delta QRS$

$\angle TPR = \angle SQR$  (given)

$PR = QR$   (given)

$\angle TRP = \angle SRQ (\angle SRT \ is\ Common)$

$\therefore \ By \ applying \ ASA \ \ \Delta PRT \cong \Delta QRS$

Since $\Delta PRT \cong \Delta QRS, PT = SQ \ \&\ \angle PTR = \angle QSR$

Q20. In adjoining figure, $\angle PON = 90^{\circ} and MO = ON.$

$Prove:\ i) \ PM = PN \ ii)\ \angle OMQ = \angle ONQ$

Consider $\Delta MPO \ \&\ \Delta NPO$

$PO$ is common

$MO = ON$ (given)

$\angle POM = \angle PON$

Hence, $\Delta MPO \cong \Delta NPO$

Therefore,  $i) \ PM = PN$

Now Consider $\Delta MOQ \ \&\ \Delta NOQ$

$MO = ON$

$\angle MOQ = \angle NOQ$

$OQ$ is common

Therefore, $\Delta MOQ \cong \Delta NOQ \ Hence, \ \angle OMQ = \angle ONQ$

## Class 8: Triangles – Exercise 30D

Q.1 $\Delta ABC$, right angled at A, find BC when:

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i)  $AB=24 \ cm\ \ and\ AC= 7 \ cm\ ,$

$BC= \sqrt{(24^2+7^2 } =25 \ cm\$

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ii)  $AB=28 \ cm\ \ and\ AC=45 \ cm\$

$BC=\sqrt{(28^2+45^2 } =53 \ cm\$

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iii)  $AB=1.6 \ cm\ \ and\ AC=3 \ cm\$

$BC=\sqrt{(1.6^2+3^2 } =3.4 \ cm\$

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iv)  $AB=5.6 \ cm\ \ and\ AC=4.2 \ cm\$

$BC=\sqrt{(5.6^2+4.2^2 } =7 \ cm\$

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Q.2. In $\Delta PQR$, right angled at Q, find PQ

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i)  $PR= 29 \ cm\ \ and\ QR = 21\ cm\$

$PQ=\sqrt{(29^2-21^2 } =20\ cm\$

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ii)  $PR=5\ cm\ \ and\ QR=4.8\ cm\$

$PQ=\sqrt{(5^2-4.8^2 } =1.4\ cm\$

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iii)  $PR=3.4\ cm\ \ and\ QR= 3\ cm\$

$PQ=\sqrt{(3.4^2-3^2 } =1.6\ cm\$

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iv)  $PR=5.3\ cm\ \ and\ QR= 4.5 \ cm\$

$PQ=\sqrt{(5.3^2-4.5^2 } =2.8\ cm\$

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Q.3. The length of the side of some triangles is given below which one of them is right angled? In case of a right angled triangle, find which angle measures $90^{\circ}$

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i)  $AB=7 \ cm\ , BC=8 \ cm\ \ and\ AC=15 \ cm\ .$

$BC^2+AC^2=8^2+15^2=289=17^2=AB^2$

$\angle C=90^{\circ}$

$\\$

ii)  $PQ=24 \ cm\ , QR=20 \ cm\ \ and\ PR=32 \ cm\ .$

Not a right angled triangle

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iii)  $XY=24 \ cm\ , YZ=26 \ cm\ \ and\ ZX=10 \ cm\$

$XY^2+ZX^2=24^2+10^2=676=26^2=YZ$

$\angle X=90^{\circ}$

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iv) $LM=1.6\ cm\ , MN=1.4 \ cm\ \ and\ LN=1.8 \ cm\ .$

Not a right angled triangle

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Q.4. Find the unknown side of each of the following figures;

1. $ZY=\sqrt{(32^2+24^2 } =40 \ cm\$
2. $LN=\sqrt{(2^2+1.5^2 } =2.5 \ cm\$
3. $RQ=\sqrt{(34^2-16^2 } =30 \ cm\$
4. $AC=\sqrt{(20^2-12^2 } =16 \ cm\$
5. $DF=\sqrt{(0.7^2+2.4^2 } =2.5 \ cm\$
6. $KL=\sqrt{(2.5^2+6^2 } =6.5 \ cm\$

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Q.5. If the adjoining figure, it is given that:

$AD \perp BC, AB=25, AD=15\ cm\ \ and\ AC=17\ cm\$.Find the length of  $i)\ BD, ii)\ DC, iii)\ BC$

$BD=\sqrt{(25^2-15^2 } =20 \ cm\$

$DC=\sqrt{(17^2-15^2 } =8 \ cm\$

$BC=BD+DC=28 \ cm\$

$\\$

Q.6. If the length of the diagonal of a rectangle is 37 cm. if the length of the shorter side is 12 cm, Find: i) The length of its longer side ii) Perimeter of rectangle iii) Area of rectangle

$AB=\sqrt{(37^2-12^2 } =35 \ cm\$

Perimeter$= 2\times 35+2\times 12=70+24=94 \ cm\$

Area $=35\times 12=420 \ cm\ ^2$

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Q.7. The Base of an 150 sales triangle in 28 cm long and $AB=AC=50\ cm\$. Let $AD\perp BC$ find i) Length of AD ii)  Area of ABC

$BD = 14 \ cm\$

$AD=\sqrt{(20^2-14^2 )} =48$

Area of ABC $=1/2\times 28\times 48=672 \ cm\ ^2$

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Q.8. A diagonal of a rhombus in 16 cm long  and each of its aides measures 10 cm. Find the length of the other diagonal.

Diagonal of a rhombus bisect each other and also interest each other at right angle.

$AO=\sqrt{(10^2-8^2 )} =6$

Shorter diagonal = 12 cm.

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Q.9. The supporting wire to the top of a vertical pole in 13 cm long and it in fastened to the ground at a state 5 m among from the foot of the pole. How high is the pole?

Let AB be the pole.

$h=\sqrt{(13^2-5^2 )} =13m$

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Q.10. A ladder 26 m long rust against a vertical wall with its foot 10m against from the wall how high up the wall will the ladder reach.

$h=\sqrt{(26^2-10^2)} =24 m$

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Q.11. A 15 meter against a vertical wall 21 reach a window at a height of 12m from the ground. How for in the foot of the ladder?

Let the distance of the foot of the ladder from the wall=d

$d=\sqrt{(15^2-12^2) } =9 m$

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Q.12. The height of thus to were are 34m and 10m respectively. If the distance between there fact in 32m, find the distance between their tops.

Let l be the distance between the tops

$l=\sqrt{(24^2 +32^2) } =40 m$

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Q.13. In the adjoining figure, ABC in a triangle in which $\angle B=90^{\circ}$.  If D  in the mid point of BC. Prove that $AC^2=AD^2+3CD^2$

$AC^2=AB^2+BC^2$

$=AB^2+(2CD) ^2$

$=(AD^2-BD^2) +4CD^2$

$=AD^2-CD^2+4CD^2$

$AC^2=AD^2+CD^2$

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Q.14. In the adjoining figure, it in given of that  $AB=27 \ cm\ , CD=12 \ cm\ , AC=36 \ cm\ \angle BAC=\angle DCA=90^{\circ} \ and\ AM=\ CM\ Find \ i)\ BM^2 \ ii)\ MD^2, \ iii)\ BD^2$

$BM^2=(15+12) ^2+18^2=1053$

$MD^2=18^2+ 12^2=468$

$BD^2=15^2+36^2=1521$

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Q.15. In the following figure, it in being given that:$\angle PQR=90^{\circ}, \angle MNR=90^{\circ} PM=RM, PQ=6 \ cm\ , QR=8 \ cm\ , MN=12 \ cm\$.

Find the Perimeter of: $\ i)\ \Delta PMR \ ii)\ \Delta MNR \ iii)\ Quadrilateral PQRM$

$PR= \sqrt{(6^2+8^2 )} =10 \ cm\$

$PN=NR= 5 \ cm\$

$MP=MR=\sqrt{(5^2+12^2 )} =13 \ cm\$

Perimeter of $\Delta PMR=13+13+10=36 \ cm\$

Perimeter of  $\Delta MNR=12+5+13=30 \ cm\$

Perimeter of $PQRM=13+6+8+13=40 \ cm\$

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Q.16. In the adjoining figure,  $\angle PQR=\angle QRS=90^{\circ}. Prove PR^2-PQ^2=QR^2- SR^2$

$PR^2=PQ^2+QR^2 ...i)$

$QS^2=SR^2+QR^2 ...ii)$

Subtract ii) from i) ,

$PR^2-PQ^2=QR^2- SR^2$

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Q.17. In the Adjoining figure,$\angle ABC=90^{\circ}. Prove: AC^2+PQ^2=AQ^2=AQ^2+PC^2$

$AC^2=AB^2+BC^2$

$= (AQ^2-BQ^2) +(PC^2-PB^2)$

$= AQ^2+PC^2-(BQ^2+PB^2)$

$= AQ^2+PC^2-PQ^2$

$\Rightarrow AC^2+PQ^2=AQ^2+PC^2$

$\\$

Q.18. In a  $\Delta ABC$, if $D \ and\ E$ are mid point of $AB \ and\ AC$, respectively and $\ \angle BAC=90^{\circ}, Prove BE^2 +CD^2=5DE^2$

$BE^2=AB^2+AE^2 ...i)$

$CD^2=AC^2+AD^2 ...ii)$

$BE^2+CD^2=(AB^2+AC^2) +(AE^2+AD^(2)$

$= (2AD) ^2+(2AE) ^2+AE^2+AD^2$

$= 5AD^2+5AE^2=5DE^2$

$\\$

Q.19. In quadrilateral $ABCD, \angle B=\angle D=90^{\circ}. \ Prove \ that \ AB^2-AD^2=CD^2-CB^2$

$AB^2=AC^2-CB^2 ...i)$

$AD^2=AC^2-CD^2 ...ii)$

Subtracting ii)  from i)

$AB^2-AD^2=CD^2-CB^2$

## Class 8: Perimeter and Area of Planes (Lecture Notes)

TRIANGLES

Perimeter of a Triangle

If $a, \ b, \ c\$  are the lengths of the sides of any triangle. Then:

1. Perimeter of a triangle $= (a+b+c)$ units
2. Area of the triangle $= \sqrt{(s(s-a)(s-b)(s-c)} \ units.$

where $s=\frac{1}{2}(a+b+c)$.This is also known as  $Heron's \ Formula$

Area of a Triangle

Refer to the adjoining figure. If b is the base and h is the height, then

$Area\ of\ triangle = \frac{1}{2}(base \times height) = \frac{1}{2}(b \times h)$

Note: If you could take any side as the base, then the corresponding height is the

would be the length of the perpendicular to this side from the opposite vertex.

Area of Right Angled Triangle

Let the triangle be ABC, with B = 90. Please refer to the adjoining figure.

$Area \ of \ triangle = \frac{1}{2}(BC \times AB)$

Area of an Equilateral triangle

In an equilateral triangle, all three sides are equal.

Let us say the side $= a$  unit.

1. Height of an equilateral triangle = $\frac{\sqrt{3}}{2}a$  units
2. Area of an equilateral triangle with side $a \ is\ \frac{\sqrt{3}}{2}a^2$ units

RECTANGLE & SQUARE

Perimeter and Area of Rectangle

If the sides of a rectangle are$l$ units and $b$units (refer to adjoining figure), then

1. Perimeter of rectangle $= 2(l+b)$ units
2. Area of rectangle $= (l \times b)$  sq. units
3. Diagonal of a rectangle $(d) = \sqrt{(l^2+b^2)}$ units

Perimeter and Area of a Square

If the sides of a square is $a$ units, then

1. Perimeter  $= 4a$ units
2. Area $= a^2$ units
3. Diagonal of a square $= a \sqrt{2}$ units

PARALLELOGRAM, RHOMBUS AND TRAPEZIUM

Area of a Parallelogram

Let ABCD be a Parallelogram with base b  and height h units. Let AC be the diagonal. Refer to the adjoining figure.

$Area = (b \times h)$ sq. units.

Area of a Rhombus

Please refer to the adjoin diagram. Let $d1 \ and\ d2$ are the diagonals of the Rhombus. We know that the diagonals intersect at right angles and bisect each other.

$Area\ of\ Rhombus= \frac{1}{2} \times (product\ of\ diagonals) = \frac{1}{2} \times (AC \times DB)$ sq. units

Area of a Trapezium

Please refer to the adjoining figure.  $AB \parallel DC$

$Area\ of\ Trapezium \ ABCD \\= \frac{1}{2} \times (sum \ of \ the \ parallel \ sides) \times (distance\ between\ them) \\= \frac{1}{2} \times (AB + DC) \times (h)$ sq. units

CIRCLE

Circumference and Area of a Circle

Let the circle be of radius r

1. Circumference of the circle $= 2\pi r = \pi d$
2. Area of the circle $= \pi r^2$

Area of a Ring (shaded area)

Refer to the adjoining figure. The radius of the larger circle is R and that of the smaller circle is r.  Area of the ring is the shaded area.

1. $Area\ of\ the\ ring=\pi R^2 - \pi r^2 = \pi(R^2-r^2) = \pi(R+r)(R-r)$

## Class 8: Volume and Surface Area of Solids – Exercise 37

Q.1. Find the volume, the total surface area and the lateral surface area of the cuboid having:

• $Length (l)= 24 cm, \ breadth(b)= 16 cm\ and \ height(h) = 7.5 cm$
• $Length (l)= 10 m, \ breadth(b)= 35 cm \ and \ height(h) = 1.2 m$

a)

Volume of a cuboid $=(l\times b\times h)=24\times 16\times 7.5= 2880 cm^3$

Total surface Area of a cuboid $= 2(lb+bh+lh) = 2(24\times 16+16\times 7.5+24\times 7.5) cm^2=1368 cm^2$

Lateral surface Area of a cuboid $= 2(l+b)\times h= 2(24+16)\times 7.5 cm^2= 600 cm^2$

b)

Volume of a cuboid $=(l\times b\times h)=10\times 0.35\times 1.2= 4.2 m^3$

Total surface Area of a cuboid $=2(lb+bh+lh) = 2(10\times 0.35+0.35\times 1.2+10\times 1.2) cm^2=31.84 m^2$

Lateral surface Area of a cuboid  $=2(l+b)\times h= 2(10+0.35)\times 1.2 cm^2=24.84 m^2$

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Q.2. Find the capacity of a rectangular tub whose length $= 6 m$, breadth $=2.5 m$ and depth $= 1.4 m$. Also find the area of the iron sheet required to make the tub.

Volume of the tub $=(l\times b\times h)=6\times 2.5\times 1.4= 21 m^3$

Total surface Area of a cuboid $=2(lb+bh+lh) = 2(6\times 2.5+2.5\times 1.4+1.4\times 6) m^2=53.8 m^2$

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Q.3. A wall of length $13.5 m$, width $60 cm$ and height $1.6 m$ is to be constructed by using bricks of dimensions $22.5 cm \ by \ 12 cm \ by\ 8 cm$. How many bricks would be needed.

Volume of the wall $=(l\times b\times h)=13.5\times 0.60\times 1.6= 12.96 m^3$

Volume of the brick $=(l\times b\times h)=0.225\times 0.12\times 0.08= 0.00216 m^3$

Number of bricks needed $= (Volume \ of \ the \ wall)/(Volume \ of \ the \ brick)=12.96/0.00216=6000$

$\\$

Q.4. How many planks each measuring $5 m \ by\ 24 cm \ by\ 10 cm$ can be stored in a place $15 m \ long, \ 4 m \ wide \ and\ 60 cm$ deep?

Volume of the place $=(l\times b\times h)=15\times 4\times 0.60= 36 m^3$

Volume of the plank $=(l\times b\times h)=5\times 0.24\times 0.10= 0.12 m^3$

Number of planks stored $= (Volume \ of \ the \ place)/(Volume \ of \ the \ plank)=36/0.12=300$

$\\$

Q.5. A classroom is $10 m \ long, \ 6.4 m \ broad \ and\ 5 m$ height. If each student is given $1.6 m^2$  of the floor area, how many students can be accommodated in the room? How many cubic meters of air would each student get?

Area of the floor of the classroom $= (l\times b)=10\times 6.4= 64 m^2$

Area given to each student $= 1.6 m^2$

Number of students that can be accommodated in the room $= 64/1.6=40$

Cubic meters of air would each student get $= 1.6 m^2\times 5m=8m^3$

$\\$

Q.6. Find the length of the longest pole that can be placed in a room $12 m \ long, 8 m$ broad and $9 m$ high.

Diagonal of a cuboid $=\sqrt{12^2+8^2+9^2}=17m$ is the longest pole that can be placed in the room

$\\$

Q.7. The volume of the cuboid is $972 m^3$. If its length and breadth be $16 m \ and\ 13.5 m$ respectively, find its height.

Volume of a cuboid $=(l\times b\times h)$

$\Rightarrow 972=16\times 13.5\times h$

$\Rightarrow h= 4.5m$

$\\$

Q.8. The volume of the cuboid is $1296 m^3$. Its length is $24 m$ and its breadth and height Are in the ratio of $3:2$. Find the breadth and height of the cube.

Volume of a cuboid $=(l\times b\times h)$

$\Rightarrow 1296=24\times 3x\times 2x$

$\Rightarrow x= 3 m$

$\Rightarrow Breadth=9 m \ and \ Height \ is \ 6 m$

$\\$

Q.9. The surface area of the cuboid is $468 cm^2$. Its length and breadth are $12cm \ and\ 9 cm$ respectively. Find its height.

Surface Area of a cuboid $= 2(lb+bh+lh)$

$\Rightarrow 468= 2(12\times 9+9\times h+h\times 12)$

$\Rightarrow h=6 m$

$\\$

Q.10. The length, breadth and height of the room are $8 m, \ 6.5 m \ and\ 3.5 m$ respectively.  Find: i) the area of the four walls of the room ii)the area of the floor of the room.

$l=8m, \ b=6.5 m, \ h=3.5m$

1. i) Area of four walls would be $=(l\times h+b\times h)\times 2$

$=(8\times 3.5+6.5\times 3.5)\times 2=101.5 m^2$

1. ii) The area of the floor of the room $=l\times b=8\times 6.5=52 m^2$

$\\$

Q.11. A room $9 m \ long, \ 6 m \ wide \ and\ 3.6 m$ high has one door $1.4 m \ by\ 2 m$ and two windows each $1.6 m \ by\ 75 cm$. Find the: i) area of four walls, excluding the doors and the windows. ii) cost of painting the wall from inside at a rate of $22.50 Rs/m^2$. iii) the cost of painting the ceiling at $25 Rs/m^2$.

1. i) Area of walls excluding the doors are windows

$= (l\times h+b\times h)\times 2-(Area \ of \ Doors)\times 1-(Area \ of \ Window)\times 2$

$=(9\times 3.6+6\times 3.6)\times 2-1.4\times 2-(1.6\times 0.75)\times 2=102.8 m^2$

ii) Cost of painting the wall $=102.8\times 22.50=2313 Rs.$

iii) Costof painting the ceiling $=(9\times 6)\times 25=1350 Rs.$

$\\$

Q.12. An assembly hall is $45 m \ long, \ 30 m \ broad \ and\ 16 m$ height. It has five doors, each measuring $4 m \ by\ 3.5 m$ and four windows $2.5 m \ by \ 1.6 m$ each. Find the

1. i) cost of wall paper at a rate of $35Rs/m^2$
2. ii) cost of carpeting the floor at the rate of $154 Rs/m^2$.

Wall dimensions: $l=45 m, \ b=30 m, \ h=16 m$

Door dimensions $=4m \ by \ 3.5 m$

Window dimensions $=2.5 m \ by\ 1.6 m$

Area of walls excluding the doors are windows

$= (l\times h+b\times h)\times 2-(Area \ of \ Doors)\times 5-(Area \ of \ Window)\times 4$

$=(45\times 16+30\times 16)\times 2-4\times 3.5\times 5-(2.5\times 1.6)\times 4=2314 m^2$

1. i) Cost of painting the wall $=2314\times 35=80990 Rs.$
2. ii) cost of carpeting the floor $=45\times 30\times 154=207900 Rs.$

$\\$

The length, breadth and height of the cuboid are in the ratio of  $7:6:5$. If the surface area of the cuboid is $1926 cm^2$, find its dimensions. Also find the volume of the cuboid.

Wall dimensions: $l=7x , \ b=6x , \ h=5x$

Surface Area of a cuboid $= 2(lb+bh+lh)$

$\Rightarrow 2(42x^2+30x^2+35x^2 ) cm^2=1926 cm^2$

$\Rightarrow x=3$

$\Rightarrow l=21 cm, \ b=18 cm \ and \ h=15 cm$

Volume $=21\times 18\times 15=5670 cm^3$

$\\$

Q.13. If the area of the three adjacent faces of a cuboidal box are $120cm^2, 72 cm^2 \ and\ 60 cm^2$  respectively, then find the volume of the box.

Let the  dimensions: $l ,\ b , \ h$

$l\times b=120$

$b\times h=72$

$h\times l=60$

Multiplying the above three expressions we get

$l^2\times b^2\times h^2=120\times 72\times 60 \Rightarrow Volume= \sqrt{(120\times 72\times 60)}=720cm^3$

$\\$

Q.14. A river $2 m$ deep and $40 m$ wide is flowing at a rate of $4.5 km/hr$. How many cubic meters of water runs into the sea per minute?

Rate of flow $=(4.5\times 1000)/3600 m/s=1.25 m/s$

Volume of water flowing $=2\times 40\times 1.25\times 60 =6000 m^3$

$\\$

Q.15. A closed wooded box $80 cm \ long, \ 65 cm \ wide, \ and\ 45 cm$ high, is made up of wood 2.5 cm thick. Find i) the capacity of the box, ii) weight of the box if $100cm^3$  of wood weighs $8$ grams.

External Volume of the Box $=(l\times b\times h)=80\times 65\times 45= 234000 cm^3$

Internal Length $=[80-(2.5+2.5)]=75 cm$

Internal Breadth $=[65-(2.5+2.5)]=60 cm$

Internal Height $=[45-(2.5+2.5)]=40 cm$

Internal Volume $=75\times 60\times 40= 180000 cm^3$

Volume of Wood $=234000-180000=4320 gm=4.32 kg$

$\\$

The external dimensions of a wooden box, open at the top are $54cm \ by \ 30 cm \ by \ 16 cm$. It is made up of wood $2 cm$ thick. Calculate i) the capacity of the box ii) the volume of the wood.

External Volume of the Box $=(l\times b\times h)$

$=54\times 30\times 16= 25920 cm^3$

Internal Length $=[54-(2+2)]=50 cm$

Internal Breadth $=[30-(2+2)]=26 cm$

Internal Height $=[16-(2)]=14 cm$

Internal Volume $=50\times 26\times 14= 18200 cm^3$

Volume of Wood $=25920-18200=7720 gm=7.72 kg$

$\\$

Q.16. The internal dimension of the closed box, made up of iron $1 cm$ thick, are $24 cm$ by $18 cm by 12 cm$. Find the volume of the iron in the box.

Internal Volume of the Box $=(l\times b\times h)$

$=24\times 18\times 12= 5184 cm^3$

External Length $=[24+(1+1)]=26 cm$

External Breadth $=[18+(1+1)]=20 cm$

External Height $=[12+(1+1)]=14 cm$

External Volume $=26\times 20\times 14=7280 cm^3$

Volume of Iron $=7280-5184=2096 cm^3$

$\\$

Q.17. Find the volume, the total surface area and the lateral surface area and the diagonal of each cube whose edges measures: i) $8 m$ ii) $6.5 cm$ iii) $2 cm 6 mm$

i)

Volume of a cube $= 8^3=512 m^3$

Total surface Area of a cube $= 6\times 8^2=384 m^2$

Lateral surface Area of a cube $= 4\times 8^2=256 m^2$

Diagonal of a cube $= a\sqrt{3}=8\times 1.732=13.856 m$

$\\$

ii)

Volume of a cube $= (6.5)^3=274.625 cm^3$

Total surface Area of a cube $= 6\times (6.5)^2=253.5 cm^2$

Lateral surface Area of a cube $= 4\times (6.5)^2=169 cm^2$

Diagonal of a cube $= a\sqrt{3}=6.5\times 1.732=11.258 cm$

$\\$

iii)

Volume of a cube $= (2.6)^3=17.576 cm^3$

Total surface Area of a cube $= 6\times (2.6)^2=40.56 cm^2$

Lateral surface Area of a cube $= 4\times (2.6)^2=27.04 cm^2$

Diagonal of a cube $= a\sqrt{3}=2.6\times 1.732=4.5032 cm$

$\\$

Q.18. The surface area of the cube is $1176 cm^2$. Find its volume.

Surface Area of a cube $= 6a^2=1176$

$\Rightarrow a=14$

Therefore Volume of cube $= 14^3=2744 cm^3$

$\\$

Q.19. The volume of the cube is $216 cm^3$. Find its surface area.

Volume of a cube $= a^3=216$

$\Rightarrow a=6$

Surface area $=6\times 6^2=216 cm^2$

$\\$

Q.20. The volume of a cube is $343 cm^3$. Find its surface area.

Volume of a cube $= a^3=343$

$\Rightarrow a=7$

Surface area $=6\times 7^2=294 cm^2$

$\\$

Q.21. A solid piece of metal in the form of cuboid of dimensions $24 cm \ by\ 18 cm \ by\ 4 cm$ is melted down and re-casted into a cube. Find the length of each edge of the cube.

Volume of a cuboid $=(l\times b\times h)=24\times 18\times 4=1728$

Let the dimension of cube $=a$

Volume of Cube $= a^3=1728 \Rightarrow 12 cm$

Q.22. Three cubes of metal with edges $5 cm \ by\ 4 cm \ by\ 3 cm$ are melted to form a single cube. Find the lateral surface area of the new cube formed.

Let the dimension of the large cube $=a$

Volume of Large Cube $=5^3+4^3+3^3=216= a^3$

Therefore the dimension of the large cube $=6 cm$

Lateral surface Area of a cube $= 4\times (6)^2=144 cm^2$

## Class 8: Circle – Exercise 33

Q.1. Fill in the blanks

1. A line segment joining any point on the circle to its center is called a radius  of the circle.
2. All the radii of a circle are equal.
3. A line segment having its end points on a circle is called chord  of a circle.
4. A chord that passes through the center of the circle is called a diameter  of the circle.
5. Diameter of a circle is twice  its radius.
6. A diameter is the largest  chord of the circle.
7. The interior of a circle together with the circle is called the area of the circle.
8. A chord of a circle divides the whole circular region into two parts, each called a  segment.
9. Half of a circle is called a semicircle.
10. A segment of a circle containing the center is called the major segment of the circle.
11. The mid point of the diameter of a circle is the center  of the circle.
12. The perimeter of the circle is called its circumference.

$\\$

Q.2. State which of the following statements are true or false:

1. Diameter of a circle is a part of a semi-circle of a circle : True
2. Two semi-circles of a circle together make the whole circle: True
3. Two semi-circular regions of a circle together make the whole circular region:  True
4. An infinite number of chords may be drawn in a circle: True
5. A line can meet a circle at the most at two points: True
6. An infinite number of diameters can be drawn in a circle: True
7. A circle has an infinite number of radii: True
8. A circle consists of an infinite number of points: True
9. Center of a circle lies on a circle: True

$\\$

Q.3. From an external point P, 29 cm away from the center of a circle, a tangent PT of length 21 cm is drawn. Find the radius of the circle.

Radius $= \sqrt{(29^2-21^2 )}= 20 cm$

$\\$

Q.4. Two tangents  $PM \ and\ PN$ are drawn from an exterior point p to a circle with center O. Prove that: $\angle OPM \cong \angle OPN$

In  $\Delta PMO \ and\ \Delta PNO,$

PO is common, OM=ON (radius of the circle)  and

$PN=PM$ (Tangents to a circle from one point  circle are equal.)

Hence   $\Delta PMO \cong \Delta PNO$

$\\$

Q.5. In the given figure,  $\Delta ABC$is inscribed in a circle with center O. If  $\angle ACB=40^{\circ}$, find angle

$\angle ABC+ \angle BCA+ \angle BAC=180^{\circ}$

$\angle BAC=90$  (angle in a semicircle is a right angle).

$\Rightarrow \angle ABC=180-90-40=50^{\circ}$

$\\$

Q.6. In the given figure, O is the centre of a circle.  $\Delta ABC$ is inscribed in this circle. If  $AB = AC, \ find\ \angle ABC \ and\ \angle ACB$.

$\angle BAC=90^{\circ}$  (angle in a semicircle is a right angle).

$AB=AC$

$\Rightarrow \angle CBA= \angle CBA=x^{\circ}$

$\Rightarrow 2x+90=180$

$\Rightarrow x=45^{\circ}$

$\therefore \angle CBA= \angle CBA=45^{\circ}$

$\\$

Q.7. In the given figure, O is the centre of a circle. If  $\angle ABC = 54^{\circ}$ , find $\angle ACB$. Also, if $\angle BCD = 43^{\circ}$  , find $\angle CBD$.

$\angle CBA+ \angle BAC+ \angle ACB=180$

$\angle BAC=90^{\circ}$  (angle in a semicircle is a right angle).

$\Rightarrow \angle ACB=180-90-54=36^{\circ}$

Similarly

$\angle CBD+ \angle CDB+ \angle BCD=180^{\circ}$

$\angle BDC=90^{\circ}$   (angle in a semicircle is a right angle).

$\Rightarrow \angle CBD=180-90-43=47^{\circ}$

$\\$

Q.8. In the given figure, $\Delta ABC$ is inscribed in a circle with center O. If $\angle ABC=(3x - 7)^{\circ} \ and\ \angle ACB=(x -7)^{\circ}$ , find the value of  $x$.

$\angle CBA+ \angle BAC+ \angle ACB=180$

$\angle BAC=90^{\circ}$ (angle in a semicircle is a right angle).

$\Rightarrow (3x-7)+90+(x+13)=180 \Rightarrow x=21^{\circ}$

$\\$

Q.9. In the adjoining figure, PRT is a tangent to the circle with center O. QR is a diameter of the circle. If  $\angle QPR = 53^{\circ} \ and\ \angle PQR = x^{\circ}$ , then find the value of x.

$\angle RPQ+ \angle PQR+ \angle QRP=180$

$\angle QRP=90^{\circ}$   (tangent is perpendicular to the line drawn from the center to the point of contact)

$\Rightarrow \angle PQR=x=37^{\circ}$

$\\$

Q.10. In the adjoining figure, PRT is a tangent to the circle with centre O. OR is the radius of the circle at the point of contact. P, O are joined and produced to the point Q on the circle. If  $\angle RPO = 28^{\circ} , \angle POR =x^{\circ} \ and\ \angle ORQ =y^{\circ}$ , then find the values of $x and y$.

$\angle PRO=90$  (tangent is perpendicular to the line drawn from the center to the point of contact)

$\Rightarrow 90+x+28=180\Rightarrow x=62^{\circ}$

Also  $2 \angle RQO= \angle ROP$

$\Rightarrow \angle ORQ+\angle RQO+ \angle QOR=180$

$\Rightarrow 31+118+y=180$

$\Rightarrow y=31^{\circ}$

$\\$

Q.11. In the adjoining figure, PT is a tangent to the circle with center O, QT is a diameter of the circle. If  $PT = QT \ and\ \angle QPT = x^{\circ}$ , then find the value of $x$.

Given  $PT = QT$

$\Rightarrow \angle QPT= \angle PQT = x$

$\Rightarrow 2x+90=180 \Rightarrow x=45^{\circ}$

$\\$

Q.12. In the adjoining figure, PX and PY are tangents drawn from an exterior point P to a circle with centre O and radius 8 cm. If PX = 15 cm, OP = a cm, PY = b cm,  $\angle POX = 56^{\circ} \ and\ \angle OPY = x^{\circ}$  then find the value of $a,\ b\ and\ x$.

$a= OP= \sqrt{(8^2+15^2 )}= 17 cm$

$b= PY= \sqrt{(17^2-8^2 )}= 15 cm$

$\angle POX = \angle POY=56$

$\Rightarrow 56+90+x=180$

$\Rightarrow x=34^{\circ}$

$\\$

Q.13. In the adjoining figure, AB is the diameter of the circle with centre O. If  $\angle ABM = 124^{\circ} \ and\ CAB = x^{\circ}$, then find the value of $x$.

$180- 124+90+x=180 \Rightarrow$

$x=34^{\circ}$

$\\$

Q.14. In the adjoining figure, PQ is a diameter of a circle with center O.  $\Delta PQR$ is an isosceles triangle with RP=RQ.  PQ is produced to a point S such that RQ = QS. If  $\angle QPR = x \ and\ \angle QSR =y$, then find the values of $x \ and\ y$.

In  $\Delta PQR, RP=RQ \Rightarrow \angle RPQ= \angle RQP=x$

And we know  $\angle PRQ=90 \Rightarrow x=45^{\circ}$

$\angle RQS=180-45=135^{\circ}$

Given  $RQ=QS \Rightarrow \angle QRS=y$

$\Rightarrow In \Delta PRS, 45+y+y+90=180 \Rightarrow y=22 \frac{1}{2}^{\circ}$

## Class 8: Area Propositions-Exercise 32

Q.1. In the given figure, $\Delta ABC$ is right angled at B in which BC = 15 cm and CA = 17 cm. Find the area of acute angled $\Delta DBC$, it being given that $AD \parallel BC$

$AB= \sqrt{(17^2- 15^2 )}=8\ cm$ (since $\Delta ABC$ is a right angled triangle)

Area of $\Delta DBC = \frac{1}{2} \times 15\times 8=60 \ cm^2$

$\\$

Q.2. In the adjoining figure, area

$(\parallel gm ABCD) = 48 cm^2 \ and\ FC \parallel AB \ Find: \ i) \ Area \ of \parallel gm ABEF \ ii)\ Area \ of \Delta EAB$

Area of $\parallel gm ABEF= Area \ of \parallel gm \ ABCD= 48 cm^2$

This is because $\parallel gm ABCD \ and\ \parallel gm \ ABEF$ are between the same parallels and have the same base.

Also, since $\Delta EAB \ and \ the \parallel gm \ ABCD$ are on the same base and between the same parallels, so the

Area of $\Delta EAB= \frac{1}{2} \times Area \ of \parallel gm \ ABCD=24 \ cm^2$

$\\$

Q.3. In the adjoining figure, area $\Delta ABC = 27 cm^2 \ and\ EF \parallel BC. \ Find : \ i) Area \ of \parallel gm \ ABCF \ ii) \ Area \ of rect. \ BCDE.$

Also, since $\Delta ABC \ and \ the \parallel gm \ ABCF$ are on the same base and between the same parallels, so the

Area of $\Delta ABC= 1/2\times \ Area \ of \parallel gm \ ABCF$

$\\Rightarrow Area \ of \parallel gm \ ABCF=2\times ( Area \ of \Delta ABC)=54cm^2$

Area of $\parallel gm \ BCDE= Area \ of \parallel gm ABCF= 54 cm^2$

This is because $\parallel gm \ ABCF \ and\ \parallel gm \ BCDE$ are between the same parallels and have the same base.

$\\$

Q.4. In trapezium $ABCD$, it is being given that $AB \parallel DC$ and diagonals $AC \ and\ BD$ intersect at O. Prove that: $\ i) \ Area (\Delta DAB) =Area (\Delta CAB) \ ii) \ Area (\Delta AOD) =Area (\Delta BOC)$

Since $\Delta EAB \ and\ \Delta CAB$ are on the same base and between the same parallels, the area of the two $\Delta$‘s will be equal.

$\Rightarrow \ Area (\Delta DAB) =Area (\Delta CAB)$

Also Since $\Delta ACD \ and\ \Delta CBD$ are on the same base and between the same parallels,the area of the $\ two \Delta$‘s will be equal.

$\Rightarrow \ Area (\Delta ACD)=Area (\Delta CBD)$

Now, subtract the area of $\Delta ODC$ from both sides we get

$\Rightarrow \ Area (\Delta AOD) =Area (\Delta BOC)$

$\\$

Q.5. In the adjoining figure, $ABCD$ is a parallelogram, P is a point on $DC \ and \ Q \ is \ a \ point \ on \ BC$. Prove that : $i) \Delta APB \ and\ \Delta AQD$ are equal in area $ii) Area (\Delta AQD) = Area (\Delta ADP) + Area (\Delta BCP)$

Since $\Delta APB \ and \ the \parallel gm ABCD$ are on the same base and between the same parallels, so the

Area of $\Delta APB= \frac{1}{2} \times \ Area \ of\ \parallel gm \ ABCD$

Similarly, $\Delta AQD \ and \ the \ \parallel gm \ ABCD$ are on the same base and between the same parallels, so the

Area of $\Delta AQD= \frac{1}{2} \times \ Area \ of \parallel gm \ ABCD$

$\Rightarrow \Delta APB \ and\ \Delta AQD \ are \ equal \ in \ area$

Area $(\Delta ADP)+ Area (\Delta BCP)= \frac{1}{2} \times DP\times (height)+\frac{1}{2} \times PC\times (height)$

$\Rightarrow Area (\Delta ADP)+ Area (\Delta BCP)= \frac{1}{2} \times (DP+PC)\times (height)$

$= \frac{1}{2} \times (DC)\times (height)$

$= \frac{1}{2} \times \ Area \ of\ \parallel gm \ ABCD = Area \ of \Delta AQD$

Hence Proved.

$\\$

Q.6. In the adjoining figure, $ABCD$ is a quadrilateral. A line through D, parallel to AC, meets BC produced in P. Prove that $Area (\Delta ABP) = Area (quad. ABCD)$.

$Area (quad. ABCD)= Area (\Delta ABC)+ Area (\Delta ACD)$

$Area (\Delta ACD)= \frac{1}{2} \times \ Area \ of \parallel gm \ ADPC$

Similary, $Area (\Delta PCD)= \frac{1}{2} \times \ Area \ of \parallel gm \ ADPC$

$\Rightarrow Area (\Delta ACD)= Area (\Delta PCD)$

Hence $Area (quad. ABCD)= Area (\Delta ABC)+ Area (\Delta PCD)$

But $Area (\Delta PCD)= Area (\Delta PCA)$

Hence Proved that $Area (\Delta ABP) = Area (quad. ABCD)$

$\\$

Q.7. $PQRS$ is any quadrilateral. Line segments passing through the vertices are drawn parallel to the diagonals of this quadrilateral so as to obtain a parallelogram $ABCD$ as shown in the adjoining figure. Prove that: $Area (quad. PQRS) = \frac{1}{2} \times \ Area (\parallel gm ABCD)$

$Area (quad. PQRS)=Area (\Delta PQR)+ Area (\Delta PRS)$

$= \frac{1}{2} \times \ Area \ of \parallel gm \ PRAB+ \frac{1}{2} \times \ Area \ of \parallel gm \ PRCD$

Since $\parallel gm \ ABCD= \parallel gm \ ABRP+ \parallel gm \ PRCD$

$=\frac{1}{2} \times \ Area (\parallel gm \ ABCD)$

Hence Proven.

## Class 8: Triangles – Exercise 30 B

Q.1. Find the value of $x$ in each of the following figures:

i)

$\angle ABC=\angle BAC=35^{\circ} \ (since \ BD=DA)$

$\angle BDA = 180 - 35 - 35 = 110^{\circ}$

Similarly, $\angle ECA=\angle AEC=28^{\circ} (since EC=EA)$

$\angle AEC=180-28-28=124^{\circ}$

$x+ \angle ADE+ \angle AED=180$

$x+(180-110)+(180-124)=180$

$\Rightarrow x=54^{\circ}$

ii)

$\angle OBC= \angle OCB=40^{\circ} (Since \ OB=OC)$

$\angle BOC=180-40-40=100$

$\angle OCA= \angle OAC=30^{\circ} (Since \ OA=OC)$

$\angle AOC=180-30-30=120^{\circ}$

We know $\angle BOC+ \angle COA+ \angle AOB=360$

$\Rightarrow \angle AOB=(360-100-120)=140^{\circ})$

$\Rightarrow 2x+140=180$

$\Rightarrow x=20^{\circ}$

$\\$

Q.2. State giving reasons, whether it is possible to construct a triangle or not with sides of lengths:

• $3 \ cm, \ 4 \ cm,\ 7 \ cm$
• $9 \ cm, \ 8 \ cm,\ 16 \ cm$
• $7 \ cm, \ 9.2 \ cm, \ 6.7 \ cm$
• $3 \ cm, \ 6.2 \ cm, \ 10.8 \ cm$

Answer (Note: The sum of any two sides of a triangle is always greater than the third side)

i) Not Possible to construct the triangle because (3+4 not greater than 7

ii) Possible to construct the triangle as sum of any two sides is greater then the third side.

iii) Possible to construct the triangle as sum of any two sides is greater then the third side.

iv) Not possible to construct a triangle as sum of $4.3+6.2<10.8$

$\\$

Q.3. In $\Delta ABC, \angle B=60^{\circ} \ and\ \angle C=45^{\circ}$. Find $\angle A$. Name i) the largest side of $\Delta ABC$, ii) the smallest side of $\Delta ABC$ iii) write the sides of $\Delta ABC$ in ascending order of their lengths.

$\angle A=180-60-45=75^{\circ}$   .   Therefore

i) the largest side of $\Delta ABC=BC$

ii) the smallest side of $\Delta ABC=AB$

iii) $AB < AC < AB$

$\\$

Q.4. In $\Delta XYZ, \angle X=53^{\circ} \ and\ \angle Y=67^{\circ}$   .  Name i) the smallest side of $\Delta XYZ$, ii) the largest side of $\Delta XYZ$ iii) write the sides of $\Delta XYZ$ in ascending order of their lengths.

$\angle Z=180-53-67=60^{\circ}$   .   Therefore

i) the largest side of $\Delta XYZ=XZ$

ii) the smallest side of $\Delta XYZ=YZ$

iii) $YZ < XY

$\\$

Q.5. In $\Delta PQR, \angle P= \angle R=64{\circ}$   .  Name the smallest side and the equal sides.

$\angle Q=180-64-64=52^{\circ}$

Smallest side $=PR$

Equal Sides $=QR \ \&\ PQ$

$\\$

Q.6 In $\Delta ABC, \angle A= \angle B \ and\ \angle C=100^{\circ}$. Name the largest side and the equal sides.

$x+x+100=180 (let \ \angle A= \angle B=x)$

$\Rightarrow x=40^{\circ}$

Largest side $=AB$, Equal Sides $=AC \ \&\ BC$

$\\$

Q.7. In $\Delta DEF, \angle D \colon \angle E \colon \angle F=7 \colon 3\colon 5$. Name the smallest side and the largest sides.

$7x+3x+5x=180 \Rightarrow x=12$

$\angle D=84^{\circ} , \angle E=36^{\circ} \ and\ \angle F=60^{\circ}$

Largest side $=EF$, Smallest Side $=DF$

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Q.8. In the adjoining figure, $AC=DC, \angle CAD=50^{\circ} \ and\ \angle BAD=23^{\circ}$. Find $\angle ADC, \angle ACD, \angle ADB \ and\ \angle ABD$. Also show that $AD>AC, \ AD>BD \ and\ AB>BC$

$AD=AD \Rightarrow \angle ADC= \angle DAC=50^{\circ}$

$\angle ACD=180-50-50=80^{\circ}$

Therefore $\angle ABD=180-50=130^{\circ}$

$\Rightarrow \angle ABD=180-130-23=27^{\circ}$

Since $\angle ACD> \angle ADC \Rightarrow AD>AC$

Since $\angle ABD> \angle BAD \Rightarrow AD>BD$

Since $\angle ACB> \angle BAC\Rightarrow AB>BC$

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Q.9. In the adjoining figure, $BA \parallel CD \ and\ AB=BC$. If $\angle BAC=56^{\circ} \ and \ \angle COD=85^{\circ}$, find the values of $x, \ y \ and \ z$.

Since $AB=BC, \angle BAC= \angle ACB=56^{\circ}$

$x+y=180-112=68 ...i)$

$\angle AOB= \angle DOC=85^{\circ}$  (opposite angles)

Therefore $x+56+85=180 \Rightarrow x=39$

Which implies that $y=68-39=29$

Becasue $AB \parallel CD, \angle ABC=\angle DCX=x+y=39+29=68^{\circ}$

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Q.10. In the adjoining figure, $BA \parallel CD, BA=BC and CE=DE$. If $\angle ABC=66^{\circ} , \angle DCE=48^{\circ} \ and\ \angle DAC=63^{\circ}$  , find the value of $x, \ y \ and\ z$.

$\angle BAC= \angle BCA=y$

$2y+66=180 \Rightarrow y=57^{\circ}$

$\angle ECD= \angle EDC=48$

$x+48+48=180 \Rightarrow x=84^{\circ}$

Since $BA \parallel CD, \angle BAC= \angle ACD=y=57$

Therefore $63+57+z=180 \Rightarrow z=60^{\circ}$

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Q.11. In the adjoining figure, $AD=BD=CD$ and $\angle CAD=40^{\circ}$. Show that $AC>AD, AB>AD, AC>AB$.

Since $AD=CD, \angle DAC= \angle DCA=40^{\circ}$

$\angle ADB=180-(180-40-40)=80^{\circ}$

Since $AD=BD, \angle DAB= \angle DBA$

$\angle ADC=100 \Rightarrow \angle ADB=80^{\circ}$

$\angle DAB= \angle DBA=50^{\circ}$

Since$\angle ADC> \angle ACD \Rightarrow AC>AD$

Since $\angle ADB> \angle ABD \Rightarrow AB>AD$

Since $\angle ABD> \angle ACB\Rightarrow AC>AB$

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Q.12. If $D$ is a point on side $BC of \Delta ABC$, prove that: $i) \ AB+BD>AD \ ii)\ AC+CD>AD \ iii)\ AB+BC+AC>2AD$

In $\Delta ABD, because it is a valid triangle, AB+BD>AD$

In $\Delta ADC, because it is a valid triangle, AC+CD>AD$

In $\Delta ABC$, because it is a valid triangle,

$AB+(BD+CD)+AC>2 AD$

$\Rightarrow AB+BC+AC>2AD$

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Q.13. In the adjoing figure, $ABCD$ is a quadrilateral. Prove that: $i)AD+DC>AC ii)AB+BC>AC iii) AB+BC+AD+DC>2AC iv) AD+AB+BC>CD$

In $\Delta ADC, AD+DC>AC$

In $\Delta ABC AB+BC>AC$

Adding the above two expresions we get

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

$AB+BC>AC and AD+AC>CD$. Add these two expressions we get

$AD+AB+BC>CD$