Question

the length of a buildings shadow is 20 meters at the same time that a 2-meter stick casts a shadow of 4 meters. using similar triangles, what is the height of the building?
a. 15 meters

b. 10 meters

c. 20 meters

d. 5 meters

in triangle δabc, ab = 8 cm, bc = 10 cm, and ac = 12 cm. in triangle δdef, de = 16 cm and ef = 20 cm. what is the length of df if δabc ~ δdef?
a. 24 cm

b. 25 cm

c. 23 cm

d. 22 cm

two fences are intended to be parallel. one fence divides the triangles side into 9 m and 15 m segments, and the other divides the opposite side into 12 m and x segments. what is x to confirm the fences are parallel?
a. 20 m

b. 24 m

c. 17 m

d. 16 m

Explanation:

First Question (Building Shadow)

Step1: Set up proportion

Let \( h \) be the height of the building. Since the triangles are similar, \(\frac{\text{height of building}}{\text{length of building's shadow}}=\frac{\text{height of stick}}{\text{length of stick's shadow}}\), so \(\frac{h}{20}=\frac{2}{4}\).

Step2: Solve for \( h \)

Cross - multiply: \( 4h = 20\times2 \). Then \( 4h=40 \), divide both sides by 4: \( h = 10 \).

Step1: Find the scale factor

For similar triangles \(\triangle ABC\sim\triangle DEF\), the ratio of corresponding sides is equal. The ratio of \( AB\) to \( DE\) is \(\frac{AB}{DE}=\frac{8}{16}=\frac{1}{2}\), and the ratio of \( BC\) to \( EF\) is \(\frac{BC}{EF}=\frac{10}{20}=\frac{1}{2}\). So the scale factor is \(\frac{1}{2}\) (from \(\triangle ABC\) to \(\triangle DEF\) it is 2).

Step2: Find \( DF \)

Since \(\frac{AC}{DF}=\frac{1}{2}\) (because \( AC\) corresponds to \( DF\)), and \( AC = 12\) cm, then \( DF=12\times2 = 24\) cm.

Step1: Apply Basic Proportionality Theorem

By the Basic Proportionality Theorem (Thales' theorem), if two lines are parallel, then \(\frac{9}{15}=\frac{12}{x}\).

Step2: Solve for \( x \)

Cross - multiply: \( 9x=15\times12 \). \( 9x = 180 \), divide both sides by 9: \( x = 20 \).

Answer:

b. 10 meters

Second Question (Similar Triangles)