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apoorv

· started a discussion

· 1 Months ago

Wrong calulation (Question: 188648.0 )

Hello team,

AD^2= AE^2+DE^2,but instead you have taken AE^2-DE^2.

so correct option is none of the above.

please correct if i am wrong

Question:
In an equilateral triangle ABC of side 10 cm, the side BC is trisected at D. Then the length (in cm) of AD is –
Options:
A) 3 \(\sqrt{7}\)
B) 7 \(\sqrt{3}\)
C) \(\cfrac{10 \sqrt{7}}{3}\)
D) \(\cfrac{7 \sqrt{10}}{3}\)
Solution:
Ans: (c)


Knowledge Expert

· commented

· 1 Months ago

Dear Student,

Its, AE^2 = AB^2-BE^2 because in right angle triangle ABE, angle E = 90 degree, AB = 10 = Hypotenouse, DE = Base, AE = Perpendicular.

Since, Hypotenouse^2 = Base^2 + Perpendicular^2

AB^2 = DE^2 + AE^2

Hence, AE^2 = AB^2 -DE^2 which has been given.


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