Question

points a, b, and c are midpoints of the sides of right triangle def. which statements are true? select three options. (the formula for the area of a triangle is ( a=\frac{1}{2} mathrm{bh} )) ( square b c=6 mathrm{~cm} ) ( square a c=5 mathrm{~cm} ) ( square b a=4 mathrm{~cm} ) ( square ) the perimeter of triangle ( a b c=12 mathrm{~cm} ). ( square ) the area of triangle ( a b c ) is ( \frac{1}{3} ) the area of triangle def. ( d e=10 mathrm{~cm} ) ( f d=6 mathrm{~cm} ) ( f e=8 mathrm{~cm} )

Explanation:

Step1: Use the mid - segment theorem

The mid - segment theorem states that the segment connecting the mid - points of two sides of a triangle is parallel to the third side and half its length.
Since \(A\), \(B\), and \(C\) are mid - points:

• \(BC=\frac{1}{2}FD\). Given \(FD = 6\mathrm{cm}\), then \(BC=\frac{1}{2}\times6 = 3\mathrm{cm}\) (so the statement \(BC = 6\mathrm{cm}\) is false).
• \(AC=\frac{1}{2}DE\). Given \(DE=10\mathrm{cm}\), then \(AC = 5\mathrm{cm}\).
• \(BA=\frac{1}{2}FE\). Given \(FE = 8\mathrm{cm}\), then \(BA=4\mathrm{cm}\).

Step2: Calculate the perimeter of \(\triangle ABC\)

\(P_{\triangle ABC}=AB + BC+AC\). Substituting the values \(AB = 4\mathrm{cm}\), \(BC = 3\mathrm{cm}\), \(AC = 5\mathrm{cm}\), we get \(P_{\triangle ABC}=4 + 3+5=12\mathrm{cm}\).

Step3: Calculate the areas

The area of \(\triangle DEF\) using \(A=\frac{1}{2}bh\) (let \(FD\) be the base \(b = 6\mathrm{cm}\) and \(FE\) be the height \(h = 8\mathrm{cm}\)), \(A_{\triangle DEF}=\frac{1}{2}\times6\times8=24\mathrm{cm}^2\).
The area of \(\triangle ABC\) using \(A=\frac{1}{2}bh\) (let \(BC\) be the base \(b = 3\mathrm{cm}\) and \(AB\) be the height \(h = 4\mathrm{cm}\)), \(A_{\triangle ABC}=\frac{1}{2}\times3\times4 = 6\mathrm{cm}^2\). And \(\frac{A_{\triangle ABC}}{A_{\triangle DEF}}=\frac{6}{24}=\frac{1}{4}\) (so the statement that the area of \(\triangle ABC\) is \(\frac{1}{3}\) the area of \(\triangle DEF\) is false).

Answer:

AC = 5 cm, BA = 4 cm, The perimeter of triangle ABC = 12 cm.