QUESTION IMAGE
Question
- in right triangle abc, ac = 4 and bc = 5. a new triangle dec is formed by connecting the midpoints of ac and bc.
a. what is the scale factor of triangle abc to triangle dec?
b. what is the area of triangle abc?
c. what is the area of triangle dec?
d. does the scale factor for the side lengths apply to the area of two similar triangles?
Part a
Step1: Recall Midsegment Theorem
The midsegment of a triangle is parallel to the third side and half its length. Since \( D \) and \( E \) are midpoints, \( DE \parallel AB \) and \( DE=\frac{1}{2}AB \), \( DC = \frac{1}{2}AC \), \( EC=\frac{1}{2}BC \).
Step2: Determine Scale Factor
Scale factor is the ratio of corresponding side lengths. For \( \triangle ABC \) to \( \triangle DEC \), the ratio of \( AC \) to \( DC \) (or \( BC \) to \( EC \)) is \( \frac{AC}{DC}=\frac{4}{2} = 2\) (since \( DC=\frac{1}{2}AC = 2\)). So the scale factor is \( 2:1 \) or \( 2 \).
Step1: Identify Triangle Type
\( \triangle ABC \) is a right triangle with legs \( AC = 4 \) and \( BC = 5 \)? Wait, no, wait. Wait, in a right triangle, the legs are the perpendicular sides. Wait, the right angle is at \( C \), so legs are \( AC \) and \( BC \)? Wait, no, wait the diagram: \( AC \) is vertical, \( BC \) is horizontal, right angle at \( C \). So area of a right triangle is \( \frac{1}{2}\times\text{leg}_1\times\text{leg}_2 \).
Step2: Calculate Area
Here, \( \text{leg}_1 = AC = 4 \), \( \text{leg}_2=BC = 5 \). So area \( A=\frac{1}{2}\times AC\times BC=\frac{1}{2}\times4\times5 = 10 \).
Step1: Recall Similar Triangles Area Ratio
For similar triangles, the ratio of areas is the square of the scale factor. We know from part (a) the scale factor of \( \triangle ABC \) to \( \triangle DEC \) is \( 2 \), so the ratio of areas \( \frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle DEC}=2^2 = 4 \).
Step2: Solve for Area of \( \triangle DEC \)
Let \( A_{DEC} \) be the area of \( \triangle DEC \). We know \( A_{ABC}=10 \), so \( \frac{10}{A_{DEC}} = 4\), then \( A_{DEC}=\frac{10}{4}=2.5 \) or \( \frac{5}{2} \). Alternatively, since \( DC = 2 \) (midpoint, \( DC=\frac{1}{2}AC = 2 \)) and \( EC=\frac{1}{2}BC = 2.5 \), then area of \( \triangle DEC=\frac{1}{2}\times DC\times EC=\frac{1}{2}\times2\times2.5 = 2.5 \).
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The scale factor of triangle \( ABC \) to triangle \( DEC \) is \( 2 \).