Question

please answer 11,12 questions.please Answer it in detailed manner

Answer 1

11.)Given right DABC, right angled at B.

D and E are points of trisection of the side BC

Let BD = DE = EC = k

Hence we get BE = 2k and BC = 3k

In ΔABD, by Pythagoras theorem, we get

AD^2= AB^2+ BD^2

AD^2= AB^2+ k^2

Similarly, in ΔABE we get

AE^2= AB^2+ BE^2

Hence AE^2= AB^2+ (2k)^2

= AB^2+ 4k^2 and

AC^2= AB^2+ BC^2

= AB + (3k)^2

AC^2= AB^2+ 9k^2

Consider, 3AC^2+ 5AD^2= 3(AB^2+ 9k^2) + 5(AB^2 + 4k^2)

= 8AB^2+ 32k^2

= 8(AB^2+ 4k^2)

∴ 3AC^2+ 5AD^2= 8AE^2

12.)Given ΔABC is an isosceles triangle in which ∠B = 90°

⇒ AB = BC

By Pythagoras theorem, we have AC^2= AB^2+ BC^2

⇒AC^2= AB^2+ AB^2[Since AB = BC]

∴ AC^2= 2AB^2→ (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

Hence ar(tri.ABE)/ar(tri.ACD) = AB^2/AC^2

ar(tri.ABE)/ar(tri.ACD) = AB^2/2AB^2

ar(tri.ABE)/ar(tri.ACD) = 1/2

∴ ar(ΔABE) : ar(ΔACD) = 1 : 2

Answer 2

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