Temple geometry
As shown in the figure, four squares of sides shown touch each other at vertices.
If a = 1 3 , b = 9 , c = 1 1 , Find the value of d .
Where a , b , c , d are side lengths of given squares.
The answer is 8.
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2 solutions
Nice way of looking at. +1)
Well Apollonius' is proved using the cosine rule.
I n Δ d b c ∠ b d + i n Δ d b a ∠ b d = 1 8 0 o . ∴ a p p l y i n g C o s L a w t o b o t h Δ s , 1 1 2 = 9 2 + d 2 − 2 ∗ 9 ∗ d ∗ C o s b d . . . . . . . . . . ( 1 ) 1 3 2 = 9 2 + d 2 + 2 ∗ 9 ∗ d ∗ C o s b d . . . . . . . . . . ( 2 ) S i n c e C o s ( π − A ) = − C o s A ( 1 ) + ( 2 ) a n d s i m p l i f y i n g d 2 = 6 4 . S o d = 8
There are two typos (one in (1) and another in (2)), since b=9 and not 3: it should read 2 × 9 × d × cos (bd) instead of 2 × 3 × d × cos (bd)
It is also worth to mention, that we used the following identity:
c o s ( 1 8 0 ° − x ) = − c o s ( x )
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Thank you. I have corrected and added the identity as per your suggestion.
If we take the two white triangles in the center and rotate them until the touch then we have a triangle with a median (see diagram)
It's not hard to prove that the bottom two angles are supplementary, and that the triangles fit together. Now this diagram really should remind you of Apollonius' . Therefore plugging in the numbers we have
⟹ ⟹ 2 d 2 = a 2 + c 2 − 2 b 2 2 d 2 = 1 6 9 + 1 2 1 − 1 6 2 d 2 = 6 4
which gives d = 8