Answer:
ASA
ΔFGH ≅ ΔIHG ⇒ answer B
Step-by-step explanation:
* Lets revise the cases of congruence
- SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ
- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and
including angle in the 2nd Δ
- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ
≅ 2 angles and the side whose joining them in the 2nd Δ
- AAS ⇒ 2 angles and one side in the first triangle ≅ 2 angles
and one side in the 2ndΔ
- HL ⇒ hypotenuse leg of the first right angle triangle ≅ hypotenuse
leg of the 2nd right angle Δ
* Lets prove the two triangles FGH and IHG are congruent by on of
the cases above
∵ FG // HI and GH is transversal
∴ m∠FGH = m∠IHG ⇒ alternate angles
- In the two triangles FGH and IHG
∵ m∠FHG = m∠IGH ⇒ given
∵ m∠FGH = m∠IHG ⇒ proved
∵ GH = HG ⇒ common side
∴ ΔFGH ≅ ΔIHG ⇒ ASA
* ASA
ΔFGH ≅ ΔIHG
This is the answer 7 x -3
7 x -3 = -21
Length: 2w + 59
width: w
diagonal: (2w + 59) + 2 = 2w + 61
Length² + width² = diagonal²
(2w + 59)² + (w)² = (2w + 61)²
(4w² + 118w + 3481) + w² = 4w² + 122w + 3721
5w² + 118w + 3481 = 4w² + 122w + 3721
w² + 118w + 3481 = 122w + 3721
w² - 4w + 3481 = 3721
w² - 4w - 240 = 0
a = 1, b = -4, c = -240
w = ![[-(b) +/- \sqrt{(b)^{2} - 4(a)(c) }]/2(a)](https://tex.z-dn.net/?f=%5B-%28b%29%20%2B%2F-%20%5Csqrt%7B%28b%29%5E%7B2%7D%20%20-%204%28a%29%28c%29%20%7D%5D%2F2%28a%29)
= ![[-(-4) +/- \sqrt{(-4)^{2} - 4(1)(-240) }]/2(1)](https://tex.z-dn.net/?f=%5B-%28-4%29%20%2B%2F-%20%5Csqrt%7B%28-4%29%5E%7B2%7D%20%20-%204%281%29%28-240%29%20%7D%5D%2F2%281%29)
=
=
=
=
since width cannot be negative, disregard 1 - 2√61
w = 1 + 2√61 ≈ 16.62
Length: 2w + 59 = 2(1 + 2√61) + 59 = 2 + 4√61 + 59 = 61 + 4√61 ≈ 92.24
Answer: width = 16.62 in, length = 92.24 in
Answer:
3/2
Sorry for the bad handwriting.