NTSE Stage 2 SAT 2019 Exam Question Paper

Question : 56 of 100

Marks: +1, -0

Let ABC be a triangle with sides a, b, c. Then lengths of medians of the triangle formed by the medians of the triangle ABC are

Solution:

According to parallelogram law, Lengths of medians of ΔABC is given as:

M_{a}=\;\;{1}/{2}√{2b^{2}+2c^{2}-a^{2}}

M_{b}=\;\;{1}/{2}√{2c^{2}+2a^{2}-b^{2}}

M_{c}=\;\;{1}/{2}√{2a^{2}+2b^{2}-c^{2}}

Let length of median of triangle formed by these medians p, q, r

p=\;\;{1}/{2}√{2M_{b}^{2}+2M_{c}^{2}-M_{a}^{2}}

=\;\;{1}/{2}√{\;\;{1}/{2}(2c^{2}+2a^{2}-b^{2})+\;\;{1}/{2}(2a^{2}+2b^{2}-c^{2})-\;\;{1}/{4}(2b^{2}+2c^{2}-a^{2})}

=\;\;{1}/{2}√{c^{2}+a^{2}-\;\;{b^{2}}/{2}+a^{2}+b^{2}-\;\;{c^{2}}/{2}-\;\;{b^{2}}/{2}-\;\;{c^{2}}/{2}+\;\;{a^{2}}/{4}}

=\;\;{1}/{2}√{\;\;{9a^{2}}/{4}}

⇒p=3/4a

Similarly

q=3/4b

and

r=3/4c

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