True, ABC is congruent to PQR by the transitive property.
false
false
PQ ST
Here is the answer to your query. Consider two ∆ABC and ∆PQR. In these two triangles ∠B = ∠Q = 90�, AB = PQ and AC = PR. We can prove the R.H.S congruence rule i.e. to prove ∆ABC ≅ ∆PQR We need the help of SSS congruence rule. We have AB = PQ, and AC = PR So, to prove ∆ABC ≅ ∆PQR in SSS congruence rule we just need to show BC = QR Now, using Pythagoras theorems in ∆ABC and ∆PQR Now, in ∆ABC and ∆PQR AB = PQ, BC = QR, AC = PR ∴ ∆ABC ≅ ∆PQR [Using SSS congruence rule] So, we have AB = PQ, AC = PR, ∠B = ∠Q = 90� and we have proved ∆ABC ≅ ∆PQR. This is proof of R.H.S. congruence rule. Hope! This will help you. Cheers!!!
(-2,4)
1/5
True, ABC is congruent to PQR by the transitive property.
An equilateral and right triangle are contradictory.
false
False. If ABC definitely equals DEF equals MNO and MNO equals PQR then ABC does not equal PQR by the transitive property.
false
m = pqr/s Multiply both sides by s: ms = pqr Divide both sides by pq: ms/pq = r
True
Determine the relation between the lengths or arcs pqr pab and bcr?
PQ ST
yes because when you take pqr and divide it by two you will get an answer of 15874