Congruent - SSS
yes
cannot be determined
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!!!
PQ ST
Congruent - SSS
yes
cannot be determined
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!!!
Similar -AA (got it right on apex)
PQ ST
The answer will be AA which is short for (Angle Angle). Hope this helped.
"If two legs of one right triangle are congruent to the corresponding legs of another right triangle, then the two triangles are congruent."Example:Given:
Similar SAS-apex
Conventionally you use the symbol that looks like an "equals" sign but consists of three lines. It is the same symbol as is used for identities. ABC ≡ PQR
1/5
True, ABC is congruent to PQR by the transitive property.