Triangles — Important Questions

Answer: (b) $2$ cm.

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

Answer: (c) $40$ cm.

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

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

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

Answer: (a) $2.1$ cm.

By BPT, $\dfrac{AD}{DB} = \dfrac{AE}{EC} = \dfrac{3}{5}$, so $AE = \dfrac{3}{3 + 5}\,AC = \dfrac{3}{8} \times 5.6 = 2.1$ 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 $DE \parallel BC$, by BPT: $\dfrac{AD}{AB} = \dfrac{AE}{AC}$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

From (i) and (ii): $\dfrac{AE}{ED} = \dfrac{BF}{FC}$. 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) $\dfrac{\text{height of stick}}{\text{shadow of stick}} = \dfrac{\text{height of tower}}{\text{shadow of tower}} \Rightarrow \dfrac{6}{4} = \dfrac{h}{28}$.

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

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