If the areas of two similar triangles are equal, prove that they are congruent Let us assume two similar triangle as ΔABC ~ ΔPQR `(ar(triangleABC))/(ar(trianglePQR))=((AB)/(PQ))^2 = ((BC)/(QR))^2=((AC)/(PR))^2 ......... (1)` Given that, ar(ΔABC) = ar(ΔABC) `=>(ar(triangleABC))/(ar(trianglePQR)) =1` Putting this value in equation (1) we obtain `1=((AB)/(PQ))^2= ((BC)/(QR))^2=((AC)/(PR))^2` =>AB = PQ, BC = QR and AC = PR :.ΔABC ≅ ΔPQR (By SSS congruence criterion) Concept: Areas of Similar Triangles Is there an error in this question or solution?
Solution: We know that two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. As we know if three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent. Let's consider two similar triangles ΔABC and ΔDEF Thus, ΔABC ∼ ΔDEF ⇒ AB/DE = BC/EF = CA/FD (SSS criterion) But, Area of ΔABC = Area of ΔDEF (According to the given question) ⇒ Area of ΔABC / Area of ΔDEF = 1 ................. (1) But, Area of ΔABC / Area of ΔDEF = (AB)2 / (DE)2 = (BC)2 / (EF)2 = (CA)2 / (FD)2 (According to theorem 6.6) ............... (2) From equation (1) and (2), (AB)2 / (DE)2 = 1 ⇒ (AB)2 = (DE)2 ⇒ AB = DE ............... (3) Similarly, ⇒ BC = EF ....(4) ⇒ CA = FD .....(5) Now, In ΔABC and ΔDEF ⇒ AB = DE [from equation(3)] ⇒ BC = EF [from equation (4)] ⇒ CA = FD [from equation (5)] Thus, ΔABC ≅ ΔDEF (SSS congruency) ☛ Check: NCERT Solutions Class 10 Maths Chapter 6 Video Solution: If the areas of two similar triangles are equal, prove that they are congruent. |