Question

The vertices of triangle ABC are A(2,3), B(–1,2), and C(3,1). If ΔDEF ∼ ΔABC and DE = 4AB, what is the length of EF? √10

Answer 1

Answer:

The length of EF is $4\sqrt{17}$ units.

Step-by-step explanation:

The vertices of triangle ABC are A(2,3), B(–1,2), and C(3,1).

Using distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$BC=\sqrt{(3+1)^2+(1-2)^2}=\sqrt{17}$

It is given that ΔDEF ∼ ΔABC.

If two triangles are similar then the corresponding sides are proportional.

$\frac{AB}{DE}=\frac{BC}{EF}$

$\frac{AB}{4AB}=\frac{BC}{EF}$        (DE = 4AB)

$\frac{1}{4}=\frac{BC}{EF}$

$EF=4\times BC$

$EF=4\times \sqrt{17}$

$EF=4\sqrt{17}$

Therefore the length of EF is $4\sqrt{17}$ units.

Answer 2

If the length is 4 times from DE=4AB then the corresponding length of EF which is BC is also 4 times larger. Length of BC=sqrt((2-1)^2+(-1-3)^2)=sqrt of 17 Hence, length of EF is 1/4 of sqrt 17. Hope it helps