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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Answer:
x=-1
Step-by-step explanation:
solution: option B and C both are correct i.e., option C is correct i.e., ∠E ≅∠H and ∠I ≅ ∠F .
option C is correct i.e., ∠E ≅∠H.
explanation:
it is given that ratio of corresponding sides of ΔFGE and ΔIJH are equal
i.e.,
and if ∠E ≅ ∠H
Then ΔFGE and ΔIJH are similar by SAS (side angle side) similarity theorem.
so option C is correct i.e., ∠E ≅ ∠H.
and option B is also correct
explanation:
since it is given that
And if ∠I ≅ ∠F
then ΔFGE and ΔIJH are similar by SAS (side angle side) similarity theorem.
Answer:
Bc,AB,Ac
Step-by-step explanation:
Bc because
Bc is 150
Ab is 150
Ac is 100
We knownthe equation of a line is y = mx + b. where m is slope and b is y intercept.
the point slope equation of a line is y - y1 = m (x - x1)
plug in x1 and y1 and m
y- (5) = (3)(x - (-1))
y - 5 = 3x + 3
y = 3x + 8
