Prove that the ratio of the areas of two similar triangles is equal to the ratio of the
squares of their corresponding sides.
Answers
Answered by
1
Consider two triangles ABC and DEF.
AX and DY are the bisectors of the angles A and D respectively.
Ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. so,
Area (ΔABC) / Area (ΔDEF) = AB2/ DE2 -----(1)
ΔABC ~ ΔDEF ⇒ ∠A = ∠D
1/ 2 ∠A = 1 / 2 ∠D ⇒ ∠BAX = ∠EDY
Consider ΔABX and ΔEDY
∠BAX = ∠EDY
∠B = ∠E
So, ΔABX ~ ΔEDY [By A-A Similarity]
AB/DE = AX/DY
⇒ AB2/DE2 = AX2/DY2 --------- (2)
From equations (1) and (2), we get
Area (ΔABC) / Area (ΔDEF) = AX2/ DY2
Hence proved.
AX and DY are the bisectors of the angles A and D respectively.
Ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. so,
Area (ΔABC) / Area (ΔDEF) = AB2/ DE2 -----(1)
ΔABC ~ ΔDEF ⇒ ∠A = ∠D
1/ 2 ∠A = 1 / 2 ∠D ⇒ ∠BAX = ∠EDY
Consider ΔABX and ΔEDY
∠BAX = ∠EDY
∠B = ∠E
So, ΔABX ~ ΔEDY [By A-A Similarity]
AB/DE = AX/DY
⇒ AB2/DE2 = AX2/DY2 --------- (2)
From equations (1) and (2), we get
Area (ΔABC) / Area (ΔDEF) = AX2/ DY2
Hence proved.
Attachments:
Triyan:
please mark it as brainliest answer if u found this helpful
Similar questions