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15 Jun 2018, 02:24
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
55% (02:28) correct
45%
(02:01)
wrong
based on 51
sessions
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The area of a square is equal to the area of an isosceles triangle and its altitude equal to the side of the square s (see diagram). In terms of s, what defines the perimeter of the isosceles triangle?
A. \(2 s (\sqrt{2} + 2 )\)
B. \(2 + \sqrt{2} s\)
C. \(s (\sqrt{2} + 2)\)
D. \(2 s (\sqrt{2} + 1 )\)
E. None of the above
A. \(2 s (\sqrt{2} + 2 )\)
B. \(2 + \sqrt{2} s\)
C. \(s (\sqrt{2} + 2)\)
D. \(2 s (\sqrt{2} + 1 )\)
E. None of the above
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15 Jun 2018, 04:30
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blueviper
Area of square =\(s^2\)...
Area of isosceles triangle= \(\frac{1}{2}*b*s=s^2\)...
b=2s...
Take one sub triangle....
Base = 2s/2=s...... altitude=s, so third side=s√2 as triangle is right angle triangle..
Thus three sides are 2s,s√2 and s√2..
Therefore perimeter=s2√2+2s=2s(√2+1)
D
15 Jun 2018, 07:08
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blueviper
Attachment:
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Triangle \(h = s\)
• Triangle area = Area of square: find triangle base
\((\frac{1}{2}*b*s)=s^2\)
\(b*s=2s^2\)
\(b= 2s\)
• Two congruent right triangles: find BC
The altitude of an isosceles triangle to the unequal side
is a perpendicular bisector:
base is bisected at a right angle, and vertex angle is bisected
Result: two congruent right triangles
(SAS: AB = BC, bisected vertex creates equal included angles, BD is shared)
From above \(b = 2s\)
Base is bisected: \(\frac{1}{2}\) AC = \(s\)
So the legs of one right triangle both = \(s\)
∆ BCD is an isosceles right triangle
with sides in ratio \(x : x: x\sqrt{2}\)
\(s\) corresponds with \(x\)
Side BC = \(s\sqrt{2}\)
• Perimeter
Side BC = side AB = \(s\sqrt{2}\)
Perimeter, \(P=(2s + s\sqrt{2}+s\sqrt{2})\)
\(P=2s\sqrt{2}+2s\)
\(P=2s(\sqrt{2} +1)\)
Answer D
*Or, Pythagorean theorem
\(s^2+s^2=BC^2\)
\(2s^2=BC^2\)
\(\sqrt{2s^2}=\sqrt{BC^2}\)
\(\sqrt{2}*\sqrt{s^2}=\sqrt{BC^2}\)
\(BC= s\sqrt{2}\)
