Triangles - Similarity of Triangles

In $\triangle ABC$, $DE \parallel BC$. If $AD = x$, $DB = x - 2$, $AE = x + 2$ and $EC = x - 1$, find the value of $x$.

By Basic Proportionality Theorem (BPT), since $DE \parallel BC$, we have: $\frac{AD}{DB} = \frac{AE}{EC}$ Substituting the given values: $\frac{x}{x - 2} = \frac{x + 2}{x - 1}$ $x(x - 1) = (x + 2)(x - 2)$ $x^2 - x = x^2 - 4$ $-x = -4$ $x = 4$

We apply Thales Theorem because the line $DE$ is parallel to the base $BC$. The resulting linear equation is solved by cross-multiplication.

A girl of height $90\text{ cm}$ is walking away from the base of a lamp-post at a speed of $1.2\text{ m/s}$. If the lamp is $3.6\text{ m}$ above the ground, find the length of her shadow after $4$ seconds.

Let $AB$ be the lamp-post and $CD$ be the girl. Let $DE$ be the shadow of length $s$. Height of lamp-post $AB = 3.6\text{ m}$. Height of girl $CD = 90\text{ cm} = 0.9\text{ m}$. Distance walked in $4\text{ s}$, $BD = 1.2 \times 4 = 4.8\text{ m}$. In $\triangle ABE$ and $\triangle CDE$: $\angle B = \angle D = 90^\circ$ $\angle E = \angle E$ (Common) Therefore, $\triangle ABE \sim \triangle CDE$ (AA similarity). $\frac{BE}{DE} = \frac{AB}{CD}$ $\frac{4.8 + s}{s} = \frac{3.6}{0.9}$ $\frac{4.8 + s}{s} = 4$ $4.8 + s = 4s$ $3s = 4.8$ $s = 1.6\text{ m}$

By using the AA similarity criterion between the large triangle formed by the lamp and the small triangle formed by the girl, we set up a proportion between heights and base lengths to find the shadow length.

1. In $\triangle POQ$ and $\triangle SOR$: $\angle OPQ = \angle ORS$ (Alternate interior angles as $PQ \parallel RS$) $\angle OQP = \angle OSR$ (Alternate interior angles) $\angle POQ = \angle SOR$ (Vertically opposite angles)
2. Therefore, $\triangle POQ \sim \triangle SOR$ by AA similarity criterion.
3. Since triangles are similar, their sides are proportional: $\frac{PQ}{RS} = \frac{OP}{OS}$ $\frac{3}{9} = \frac{2}{OS}$ $\frac{1}{3} = \frac{2}{OS}$ $OS = 2 \times 3 = 6 \text{ cm}$

We use the property that parallel lines intersected by transversals create equal alternate interior angles, establishing AA similarity. Corresponding sides of similar triangles are then set in a ratio to find the unknown length.

1. Let $AB$ be the pole and $BC$ be its shadow. Let $PQ$ be the tower and $QR$ be its shadow.
2. In $\triangle ABC$ and $\triangle PQR$: $\angle B = \angle Q = 90^{\circ}$ (Vertical objects) $\angle C = \angle R$ (Angle of elevation of the sun is the same at the same time)
3. By AA similarity, $\triangle ABC \sim \triangle PQR$.
4. Thus, $\frac{AB}{PQ} = \frac{BC}{QR}$ $\frac{6}{h} = \frac{4}{28}$ $\frac{6}{h} = \frac{1}{7}$ $h = 6 \times 7 = 42 \text{ m}$

Since the sun's rays are parallel, the angles of elevation are equal, making the triangles formed by the objects and their shadows similar. We use the ratio of height to shadow length to find the tower's height.