The triangles that are similar would be ΔGCB and ΔPEB due to Angle, Angle, Angle similarity theorem.
<h3>How to identify similar triangles?</h3>
From the image attached, we see that we are given the Parallelogram GRPC. Thus;
A. The triangles that are similar would be ΔGCB and ΔPEB due to Angle, Angle, Angle similarity theorem.
B. The proof of the fact that ΔGCB and ΔPEB are similar pairs of triangles is as follow;
∠CGB ≅ ∠PEB (Alternate Interior Angles)
∠BPE ≅ ∠BCG (Alternate Interior Angles)
∠GBC ≅ ∠EBP (Vertical Angles)
C. To find the distance from B to E and from P to E, we will first find PE and then BE by proportion;
225/325 = PE/375
PE = 260 ft
BE/425 = 225/325
BE = 294 ft
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8/7 each side of the expression both go up by 1
Answer:
I think is scalene and acute
Answer: I am not sure if this is correct but I got 2.431619300243096x10^22, 2.6746457133 x 10^13, and the the last one I also get 2.4316193 x 10^10. I AM NOT TO SURE IF I DID THIS CORRECT OR NOT, PLEASE DO NOT REPORT ME IF THIA IS WRONG, I TRIED TO HELP.
Step-by-step explanation: I AM NOT SO GOOD WITH MATH!!
The answer is B
Since you know that BOD is a right angle add 12 And 15 and subtract your answer from 90