Triangles — CBSE Class 10 Maths Important Questions

Answer: (b) $2$2 cm.

By the Basic Proportionality Theorem, $\dfrac{AD}{DB} = \dfrac{AE}{EC} \Rightarrow \dfrac{1.5}{3} = \dfrac{1}{EC} \Rightarrow EC = \dfrac{3}{1.5} = 2$AD/DB = AE/EC 1.5/3 = 1/EC EC = 3/1.5 = 2 cm.

Answer: (c) $40$40 cm.

For similar triangles, the ratio of perimeters equals the ratio of corresponding sides. So $\dfrac{24}{P} = \dfrac{3}{5} \Rightarrow P = \dfrac{24 \times 5}{3} = 40$24/P = 3/5 P = 24 × 5/3 = 40 cm.

Answer: (b) $50^\circ$50^.

Since $\triangle ABC \sim \triangle DEF$ABC DEF, corresponding angles are equal, so $\angle B = \angle E = 83^\circ$B = E = 83^. Then $\angle C = 180^\circ - \angle A - \angle B = 180^\circ - 47^\circ - 83^\circ = 50^\circ$C = 180^ - A - B = 180^ - 47^ - 83^ = 50^.

Answer: (a) $2.1$2.1 cm.

By BPT, $\dfrac{AD}{DB} = \dfrac{AE}{EC} = \dfrac{3}{5}$AD/DB = AE/EC = 3/5, so $AE = \dfrac{3}{3 + 5}\,AC = \dfrac{3}{8} \times 5.6 = 2.1$AE = 3/3 + 5\,AC = 3/8 × 5.6 = 2.1 cm.

Answer: (b) Both A and R are true but R is not the correct explanation of A.

A states the AAA (AA) similarity criterion, while R states the SSS similarity criterion. Both are true, but R describes a different criterion and therefore does not explain A.

Since $DE \parallel BC$DE BC, by BPT: $\dfrac{AD}{AB} = \dfrac{AE}{AC}$AD/AB = AE/AC.

$\dfrac{2}{6} = \dfrac{AE}{9} \Rightarrow AE = \dfrac{2 \times 9}{6} = 3$2/6 = AE/9 AE = 2 × 9/6 = 3 cm.

In $\triangle AOB$AOB and $\triangle COD$COD: $\angle AOB = \angle COD$AOB = COD (vertically opposite angles) and $\angle OAB = \angle OCD$OAB = OCD (alternate angles, $AB \parallel DC$AB DC). So $\triangle AOB \sim \triangle COD$AOB COD by AA.

Hence $\dfrac{AO}{OC} = \dfrac{AB}{CD} = \dfrac{2\,CD}{CD} = 2$AO/OC = AB/CD = 2\,CD/CD = 2. Therefore $AO : OC = 2 : 1$AO : OC = 2 : 1.

In $\triangle ABC$ABC, since $LM \parallel CB$LM CB, by BPT: $\dfrac{AM}{MB} = \dfrac{AL}{LC}$AM/MB = AL/LC. Adding $1$1 to both sides gives $\dfrac{AB}{AM} = \dfrac{AC}{AL}$AB/AM = AC/AL, i.e. $\dfrac{AM}{AB} = \dfrac{AL}{AC}$AM/AB = AL/AC ...(i)

In $\triangle ACD$ACD, since $LN \parallel CD$LN CD, by BPT: $\dfrac{AN}{ND} = \dfrac{AL}{LC}$AN/ND = AL/LC. Similarly this gives $\dfrac{AN}{AD} = \dfrac{AL}{AC}$AN/AD = AL/AC ...(ii)

From (i) and (ii): $\dfrac{AM}{AB} = \dfrac{AN}{AD}$AM/AB = AN/AD. Hence proved.

By BPT, $\dfrac{AD}{DB} = \dfrac{AE}{EC} \Rightarrow \dfrac{8x - 7}{5x - 3} = \dfrac{4x - 3}{3x - 1}$AD/DB = AE/EC 8x - 7/5x - 3 = 4x - 3/3x - 1.

Cross-multiplying: $(8x - 7)(3x - 1) = (4x - 3)(5x - 3)$(8x - 7)(3x - 1) = (4x - 3)(5x - 3)

$24x^2 - 29x + 7 = 20x^2 - 27x + 9$24x^2 - 29x + 7 = 20x^2 - 27x + 9

$4x^2 - 2x - 2 = 0 \Rightarrow 2x^2 - x - 1 = 0 \Rightarrow (2x + 1)(x - 1) = 0$4x^2 - 2x - 2 = 0 2x^2 - x - 1 = 0 (2x + 1)(x - 1) = 0.

So $x = 1$x = 1 (rejecting $x = -\tfrac{1}{2}$x = -12, since lengths must be positive).

Since $\triangle ABC \sim \triangle PQR$ABC PQR: $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$AB/PQ = BC/QR and $\angle B = \angle Q$B = Q.

As $AD$AD and $PM$PM are medians, $BD = \dfrac{1}{2}BC$BD = 1/2BC and $QM = \dfrac{1}{2}QR$QM = 1/2QR, so $\dfrac{BD}{QM} = \dfrac{BC}{QR} = \dfrac{AB}{PQ}$BD/QM = BC/QR = AB/PQ.

Now in $\triangle ABD$ABD and $\triangle PQM$PQM: $\dfrac{AB}{PQ} = \dfrac{BD}{QM}$AB/PQ = BD/QM and the included angles $\angle B = \angle Q$B = Q. By SAS similarity, $\triangle ABD \sim \triangle PQM$ABD PQM.

Therefore corresponding sides are proportional, giving $\dfrac{AB}{PQ} = \dfrac{AD}{PM}$AB/PQ = AD/PM. Hence proved.

Given: In $\triangle ABC$ABC, a line $DE \parallel BC$DE BC meets $AB$AB at $D$D and $AC$AC at $E$E.

To Prove: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$AD/DB = AE/EC.

Construction: Join $BE$BE and $CD$CD. Draw $EM \perp AB$EM AB and $DN \perp AC$DN AC.

Proof: $\text{ar}(\triangle ADE) = \dfrac{1}{2} \times AD \times EM$ar( ADE) = 1/2 × AD × EM and $\text{ar}(\triangle DBE) = \dfrac{1}{2} \times DB \times EM$ar( DBE) = 1/2 × DB × EM, so

$\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DBE)} = \dfrac{AD}{DB}$ar( ADE)/ar( DBE) = AD/DB ...(i)

Similarly, using altitude $DN$DN, $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DEC)} = \dfrac{AE}{EC}$ar( ADE)/ar( DEC) = AE/EC ...(ii)

Now $\triangle DBE$DBE and $\triangle DEC$DEC lie on the same base $DE$DE and between the same parallels $DE$DE and $BC$BC, so they are equal in area: $\text{ar}(\triangle DBE) = \text{ar}(\triangle DEC)$ar( DBE) = ar( DEC) ...(iii)

From (i), (ii) and (iii): $\dfrac{AD}{DB} = \dfrac{AE}{EC}$AD/DB = AE/EC. Hence proved.

Construction: Join $AC$AC, meeting $EF$EF at $O$O.

Proof: Since $EF \parallel AB$EF AB and $AB \parallel DC$AB DC, we have $EF \parallel DC$EF DC as well.

In $\triangle ADC$ADC, $EO \parallel DC$EO DC. By BPT: $\dfrac{AE}{ED} = \dfrac{AO}{OC}$AE/ED = AO/OC ...(i)

In $\triangle CAB$CAB, $OF \parallel AB$OF AB. By BPT: $\dfrac{CO}{OA} = \dfrac{CF}{FB}$CO/OA = CF/FB, i.e. $\dfrac{BF}{FC} = \dfrac{AO}{OC}$BF/FC = AO/OC ...(ii)

From (i) and (ii): $\dfrac{AE}{ED} = \dfrac{BF}{FC}$AE/ED = BF/FC. Hence proved.

Each object with its shadow forms a right triangle. The Sun's rays are parallel, so the angles of elevation are equal, and both triangles have a right angle at the ground.

(i) By AA similarity the two triangles are similar.

(ii) $\dfrac{\text{height of stick}}{\text{shadow of stick}} = \dfrac{\text{height of tower}}{\text{shadow of tower}} \Rightarrow \dfrac{6}{4} = \dfrac{h}{28}$height of stick/shadow of stick = height of tower/shadow of tower 6/4 = h/28.

(iii) $h = \dfrac{6 \times 28}{4} = 42$h = 6 × 28/4 = 42 m, so the tower is $\mathbf{42}$42 m tall.

OR When the tower's shadow is $21$21 m: $\dfrac{6}{s} = \dfrac{42}{21} \Rightarrow s = \dfrac{6 \times 21}{42} = 3$6/s = 42/21 s = 6 × 21/42 = 3 m. The stick's shadow is $\mathbf{3}$3 m.