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:
Step-by-step explanation:
If you plot J and K in a coordinate plane, you see that the line formed is a perfectly horizontal line through y = 2. In order for this triangle to be an isosceles, the third x-coordinate would have to be located midway through the x-coordinates of the base. The midpoint between the x-coordinates is found by adding the 2 x-coordinates and dividing the sum by 2. -6+\3 = -3 and -3/2 = -3/2. So the x-coordinate is -3.2 or -1.5
Answer:
no it can't because
Step-by-step explanation:
coz the sum of any two angle should be greater than 3rd side
- 13 + 5 = 18 which is greater than 7
- 13 + 7 = 20 which is greater than 5
- 5 + 7 = 12 which is not greater than 13
so triangle cant be made
Becouse k and n are 'neighbours' so their sum are 180
so k=180-42=138