1(ii)
Assertion (A): In ΔABC and ΔPQR, if ∠BAC = ∠QPR and ∠ABC = ∠PQR, then ΔABC ~ ΔPQR
Statement 2: ΔABC ~ ΔPQR by SSS axiom
- (a)(A) is true, (R) is false.
- (b)(A) is false, (R) is true.
- (c)Both (A) and (R) are true, and (R) is the correct reason for (A).
- (d)Both (A) and (R) are true, and (R) is the incorrect reason for (A).
Show SolutionHide Solution↓
In ΔABC and ΔPQR,
⇒ ∠BAC = ∠QPR [Given]
⇒ ∠ABC = ∠PQR [Given]
∴ ΔABC ~ ΔPQR (By A.A. axiom)
So, assertion (A) is true and reason (R) is false.
Hence, option 1 is the correct option.