Answer:
The perimeter of triangle PQR is 17 ft
Step-by-step explanation:
Consider the triangles PQR and STU
1. PQ ≅ ST = 4 ft (Given)
2. ∠PQR ≅ ∠STU (Given)
3. QR ≅ TU = 6 ft (Given)
Therefore, the two triangles are congruent by SAS postulate.
Now, from CPCTE, PR = SU. Therefore,
Now, side PR is given by plugging in 3 for 'y'.
PR = 3(3) - 2 = 9 - 2 = 7 ft
Now, perimeter of a triangle PQR is the sum of all of its sides.
Therefore, Perimeter = PQ + QR + PR
= (4 + 6 + 7) ft
= 17 ft
Hence, the perimeter of triangle PQR is 17 ft.
Answer:
∠AEC = 139°
Step-by-step explanation:
Since EC bisects ∠BED then ∠BEC = ∠CED = 4x + 1
∠AED = ∠AEB + ∠BEC + ∠CED = 180 ← straight angle
Substitute values into the equation
11x - 12 + 4x + 1 + 4x + 1 = 180, that is
19x - 10 = 180 ( add 10 to both sides )
19x = 190 ( divide both sides by 19 )
x = 10
Hence
∠AEC = ∠AEB + ∠BEC = 11x - 12 + 4x + 1 = 15x - 11, hence
∠AEC = (15 × 10) - 11 = 150 - 11 = 139°
Maintaining the same width of the compass as AB, draw an arc from point X such that it intersects the afternoon drawn from N in a point