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Pranay SB

· started a discussion

· 1 Months ago

Other way: (Question: 111672.0 )

* HAG is a right angle triangle; HG=root(2) (mentioned in the question)
* As it is symmetric, HA=AG=x (let's say)
* By Pythagoras theorem, x^2+x^2=(sqrt(2))^2, which yields x=1
* Now, GB=8-1=7; Let FCM be an another triangle(M is not mentioned in the given figure) formed at the top right corner, following symmetry, GB=BM=7;
*Now, following Pythagoras theorem for triangleMGB, the lenght of GM can be found (sqrt(98)).
* Multiplying this length (sqrt(98)) with the width (sqrt(2)) will give us 14

Question:
A rectangular plank \(\sqrt[]{2}\)m wide is placed symmetrically on the diagonal of a square of side 8 m as shown in the figure. The area of the plank is:


Options:
A)

\(\big( 16\sqrt[]{2}-3 \big)\)sq.m

B) 7\(\sqrt[]{2}\) sq. m
C) 98 sq.m
D) 14 sq.m
Solution:
Ans: (d)


Here EG = EH = \( \cfrac{\sqrt{2}}{2}=\cfrac{1}{\sqrt{2}}\)

From \(\Delta\)HAG,

tan \(45^o=\cfrac{AE}{EG}\)

or \(1=\cfrac{AE}{\cfrac{1}{\sqrt{2}}}\)

 \( \therefore\) AE = \(\cfrac{1}{\sqrt{2}}\)

Also,  AC = \(\sqrt{8^2+8^2}=\sqrt{128}=8\sqrt{2}\)

So that EF = AC - 2AE  = \(8\sqrt{2}-2\times\cfrac{1}{\sqrt{2}}\)

= \(8\sqrt{2}-\sqrt{2}=7\sqrt{2}\)

Hence area of plank = HG x EF = \(\sqrt[1]{2}\times7\sqrt{2}=14\) sq.m

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