Answer/Step-by-sep explanation:
1. Given:
∆NMK ≅ ∆TRP
[tex] \overline{NM} = 20 [/tex]
[tex] \overline{MK} = 15 [/tex]
[tex] \overline{KN} = 25 [/tex]
[tex] \overline{TR} = 3x - 1 [/tex]
a. To complete the congruent statement, thus: ∆MNK ≅ ∆RTP
b. The side that is congruent to [tex] \overline{TR} [/tex] is [tex] \overline{NM} [/tex]. Thus:
[tex] \overline{TR} [/tex] ≅ [tex] \overline{NM} [/tex]
c. Since [tex] \overline{TR} [/tex] ≅ [tex] \overline{NM} [/tex], therefore:
[tex] \overline{TR} = \overline{NM} [/tex]
[tex] 3x - 1 = 20 [/tex] (substitution)
Add 1 to both sides
[tex] 3x = 20 + 1 [/tex]
[tex] 3x = 21 [/tex]
Divide both sides by 3
[tex] x = \frac{21}{3} [/tex]
[tex] x = 7 [/tex]
2. a. Slope of LK = [tex] \frac{rise}{run} = \frac{4}{3} [/tex]
Slope of LM = [tex] \frac{rise}{run} = -\frac{3}{5} [/tex]
b. ✍️Length of LK is the distance between L(-7, 4) and (-4, 8):
[tex] Lk = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex]
[tex] Lk = \sqrt{(-4 -(-7))^2 + (8 - 4)^2} [/tex]
[tex] Lk = \sqrt{(3)^2 + (4)^2} [/tex]
[tex] Lk = \sqrt{9 + 16} [/tex]
[tex] Lk = \sqrt{25} [/tex]
[tex] Lk = 5 [/tex]
✍️Length of LM is the distance between L(-7, 4) and (-2, 1):
[tex] LM = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} [/tex]
[tex] LM = \sqrt{(-2 -(-7))^2 + (1 - 4)^2} [/tex]
[tex] LM = \sqrt{(5)^2 + (-3)^2} [/tex]
[tex] LM = \sqrt{25 + 9} [/tex]
[tex] LM = \sqrt{34} [/tex]
[tex] LM = 5.8 [/tex] (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.
