Question

prove $\triangle aeb \cong \triangle dcb$, and then solve for $x$.

practice:

1. on a sunny day, sara’s shadow is 5m long while the shadow cast by a tree was 8m long. if sara is 1.6 m tall, determine the height of the tree.
2. john’s eyes are at a height of 1.7 m. he stands 10 m from a shed and holds a 12 cm pencil 80 cm away by extending his arm. the bottom of the pencil is directly in front of his eyes and the top of the pencil is in the line of sight from his eyes to the top the building. determine the height of the shed.

Explanation:

Problem 1: Determine the height of the tree

Step 1: Set up proportion

Since the triangles formed by Sara and her shadow, and the tree and its shadow are similar (by AA similarity, as both have right angles and share the same angle from the sun), we can set up a proportion. Let \( h \) be the height of the tree. Then \(\frac{\text{Height of Sara}}{\text{Length of Sara's shadow}}=\frac{\text{Height of tree}}{\text{Length of tree's shadow}}\), so \(\frac{1.6}{5}=\frac{h}{8}\).

Step 2: Solve for \( h \)

Cross - multiply: \( 5h = 1.6\times8 \). Calculate \( 1.6\times8 = 12.8 \). Then \( h=\frac{12.8}{5}=2.56 \).

Step 1: Convert units

First, convert all units to meters. John's height \( h_{John}=1.7 \) m. The distance from John to the shed \( d = 10 \) m. The length of the pencil \( l_{pencil}=12 \) cm \( = 0.12 \) m, and the distance from the pencil to John's eyes \( d_{pencil}=80 \) cm \( = 0.8 \) m.
Since the triangles formed by the pencil (from John's eyes to the pencil) and the shed (from John's eyes to the shed) are similar (by AA similarity, as both have right angles and share the same angle of sight), we can set up a proportion. Let \( H \) be the height of the shed from John's eyes level. Then \(\frac{l_{pencil}}{d_{pencil}}=\frac{H}{d + d_{pencil}}\) (wait, actually, the distance from the pencil to the shed is \( 10\) m, but the correct proportion is based on the similar triangles: the ratio of the height of the pencil to its distance from John's eyes is equal to the ratio of the height of the shed (from John's eyes level) to the distance from John's eyes to the shed. The distance from John's eyes to the shed is \( 10\) m \( = 1000\) cm, and the distance from John's eyes to the pencil is \( 80\) cm. The height of the pencil is \( 12\) cm. Let \( H\) be the height of the shed from John's eyes level. So \(\frac{12}{80}=\frac{H}{1000}\).

Step 2: Solve for \( H \)

Cross - multiply: \( 80H=12\times1000 \). Calculate \( 12\times1000 = 12000 \). Then \( H=\frac{12000}{80}=150\) cm \( = 1.5\) m.
Then the total height of the shed is the height from John's eyes level plus John's height: \( 1.5 + 1.7=3.2\) m.

Step 1: Prove congruence

• \( \angle AEB=\angle DCB = 90^{\circ}\) (given right angles).
• \( \angle ABE=\angle DBC \) (vertical angles are equal).
• By the AA (Angle - Angle) similarity criterion, \( \triangle AEB\sim\triangle DCB \). Wait, if we assume it's congruence (maybe there is a typo and it's similarity or some other information). Wait, from the diagram, \( AE = 4\), \( EB\) (let's say), \( DC\) (height of \( \triangle DCB\)) and \( CB = 12\), \( CD = 6\)? Wait, maybe the triangles are similar. Let's assume it's similarity. The ratio of sides: In \( \triangle AEB\) and \( \triangle DCB\), \( \angle AEB=\angle DCB = 90^{\circ}\), \( \angle ABE=\angle DBC\), so \( \triangle AEB\sim\triangle DCB\) by AA similarity.

Step 2: Set up proportion for similarity

If \( \triangle AEB\sim\triangle DCB\), then \(\frac{AE}{DC}=\frac{EB}{CB}=\frac{AB}{DB}\). From the diagram, \( AE = 4\), \( DC = 6\), \( CB = 12\), and \( EB\) is related to \( x\) (since \( AE=x\)? Wait, maybe the labels are: \( AE=x\), \( EB = 4\), \( DC = 6\), \( CB = 12\). Then using similarity \(\frac{AE}{DC}=\frac{EB}{CB}\), so \(\frac{x}{6}=\frac{4}{12}\).

Step 3: Solve for \( x \)

Cross - multiply: \( 12x=4\times6 \). Calculate \( 4\times6 = 24 \). Then \( x=\frac{24}{12}=2 \). Wait, maybe the correct proportion is \(\frac{AE}{DC}=\frac{AB}{DB}\) or other. But based on the given \(\frac{x}{4}=\frac{6}{15}\) (from the hand - written \(\frac{x}{4}=\frac{6}{15}\))? Wait, if we take the proportion from the hand - written \(\frac{x}{4}=\frac{6}{15}\) (maybe the sides are \( AE = x\), \( EB = 4\), \( DC = 6\), \( CB = 15\)? Wait, let's recalculate. If \(\frac{x}{4}=\frac{6}{15}\), cross - multiply: \( 15x=4\times6 = 24\), then \( x=\frac{24}{15}=1.6\). Wait, maybe the correct approach is:
Since \( \triangle AEB\sim\triangle DCB\), \(\frac{AE}{DC}=\frac{EB}{CB}\). Let \( AE=x\), \( EB = 4\), \( DC = 6\), \( CB = 12\). Then \(\frac{x}{6}=\frac{4}{12}\), \( 12x = 24\), \( x = 2\).

Answer:

The height of the tree is \( 2.56 \) m.

Problem 2: Determine the height of the shed