plz help me in this its hard
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Given ΔABC is an isosceles triangle in which ∠B = 90°
⇒ AB = BC
By Pythagoras theorem, we have AC2= AB2+ BC2
⇒AC2= AB2+ AB2[Since AB = BC]
∴ AC2= 2AB2→ (1)
It is also given that ΔABE ~ ΔACD
Recall that ratio of areas of similar triangles is equal to ratio of squares of their corresponding sides.

∴ ar(ΔABE) : ar(ΔACD) = 1 : 2
⇒ AB = BC
By Pythagoras theorem, we have AC2= AB2+ BC2
⇒AC2= AB2+ AB2[Since AB = BC]
∴ AC2= 2AB2→ (1)
It is also given that ΔABE ~ ΔACD
Recall that ratio of areas of similar triangles is equal to ratio of squares of their corresponding sides.

∴ ar(ΔABE) : ar(ΔACD) = 1 : 2
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both have equal ares. 1:1
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