In ∆ABC, D and E are the mid-points of AB and AC respectively. Find the ratio of the areas of ∆ADE and ∆ABC.
Answers
SOLUTION :
Given:
In ΔABC, D and E are the midpoints of AB and AC respectively.
Therefore, DE II BC (By Converse of mid-point theorem)
Also, DE = 1/2BC
In ΔADE and ΔABC
∠ADE = ∠B (Corresponding angles)
∠DAE = ∠BAC (common)
ΔADE–ΔABC (By AA Similarity)
We know that the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.
ar(ΔADE)/ar(ΔABC) = AD²/AB²
ar(ΔADE)/ar(ΔABC) = AD² / 2AD²
[AD = 2AB as D is the mid point]
ar(ΔADE)/ar(ΔABC) = 1² /2²
ar(ΔADE)/ar(ΔABC) = 1/4
Hence, the ratio of the areas ΔADE and ΔABC is ar(ΔADE) : ar(ΔABC) = 1 : 4.
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Answer:
Step-by-step explanation:
Given:
In ΔABC, D and E are the midpoints of AB and AC respectively.
Therefore, DE II BC (By Converse of mid-point theorem)
Also, DE = 1/2BC
In ΔADE and ΔABC
∠ADE = ∠B (Corresponding angles)
∠DAE = ∠BAC (common)
ΔADE–ΔABC (By AA Similarity)
We know that the ratio of areas of two similar triangles is equal to the ratio of square of their corresponding sides.
ar(ΔADE)/ar(ΔABC) = AD²/AB²
ar(ΔADE)/ar(ΔABC) = AD² / 2AD²
[AD = 2AB as D is the mid point]
ar(ΔADE)/ar(ΔABC) = 1² /2²
ar(ΔADE)/ar(ΔABC) = 1/4
Hence, the ratio of the areas ΔADE and ΔABC is ar(ΔADE) : ar(ΔABC) = 1 : 4.