Question

Unit 4 Test, Part 2: Congruence and Constructions Help with my math please

Answer 1

Answer/Step-by-sep explanation:

1. Given:

∆NMK ≅ ∆TRP

$\overline{NM} = 20$

$\overline{MK} = 15$

$\overline{KN} = 25$

$\overline{TR} = 3x - 1$

a. To complete the congruent statement, thus: ∆MNK ≅ ∆RTP

b. The side that is congruent to $\overline{TR}$ is $\overline{NM}$. Thus:

$\overline{TR}$ ≅ $\overline{NM}$

c. Since $\overline{TR}$ ≅ $\overline{NM}$, therefore:

$\overline{TR} = \overline{NM}$

$3x - 1 = 20$ (substitution)

Add 1 to both sides

$3x = 20 + 1$

$3x = 21$

Divide both sides by 3

$x = \frac{21}{3}$

$x = 7$

2. a. Slope of LK = $\frac{rise}{run} = \frac{4}{3}$

Slope of LM = $\frac{rise}{run} = -\frac{3}{5}$

b. ✍️Length of LK is the distance between L(-7, 4) and (-4, 8):

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

$Lk = \sqrt{(-4 -(-7))^2 + (8 - 4)^2}$

$Lk = \sqrt{(3)^2 + (4)^2}$

$Lk = \sqrt{9 + 16}$

$Lk = \sqrt{25}$

$Lk = 5$

✍️Length of LM is the distance between L(-7, 4) and (-2, 1):

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

$LM = \sqrt{(-2 -(-7))^2 + (1 - 4)^2}$

$LM = \sqrt{(5)^2 + (-3)^2}$

$LM = \sqrt{25 + 9}$

$LM = \sqrt{34}$

$LM = 5.8$ (nearest tenth)

∆KLM is not an isosceles ∆ because it does not has two equal side lengths. This we can see because LK and LM are not equal.

Therefore, Anthony is incorrect. Am isosceles ∆ has two equal sides.