GMAT CLUB OLYMPICS: In two congruent squares ABCD and AEFG shown above

Question

https://gmatclub.com/forum/download/file.php?id=62920 In two congruent squares ABCD and AEFG shown above, points C, E, H and G are collinear. What is the measure of yellow angle ? A. 95° B. 100° C. 105° D. 110° E. 115° M37-31 Untitled10.png

Bunuel · Accepted Answer

Official Solution:

https://gmatclub.com/tests-beta/resources/import/img/M37-31.png 
 
In two congruent squares ABCD and AEFG shown above, points C, E, H and G are collinear. What is the measure of yellow angle ?

A.  95° 
B.  100° 
C.  105° 
D.  110° 
E.  115°

Say the length of the side of the square is  x . Consider triangle AOC below:
 
  https://gmatclub.com/tests-beta/resources/import/img/M37-31-2.png 
 
AC is the diagonal of the square and thus will equal to  x\sqrt{2} 
 
AO is half of the diagonal of the square and thus will equal to  \frac{x\sqrt{2}}{2} 
 
So, the ratio of the hypotenuse to one of the legs is 2:1. This means that we have a 30°-60°-90° triangle. In such triangles, the sides are in the ratio  1:√3:2 . From this it follows that the angle opposite the smallest side (AO) would be 30°. This makes angle HCD equal to  45° - 30°=15° , angle DHC equal to  90° - 15°=75°  and finally angle OHD equal to  180° - 75°=105°

Answer: C

rajatchopra1994 · Answer

IMO-C

As to get answer figure required modifications, I solved this question with the help of pen & paper. Please find the attached image of the explanation. Please check.

Lenel · Answer

Let the measure of the yellow angle be  x^{\circ}  
Let  a  be the side of the square 
Its diagonal will be  a\sqrt{2} 
We notice that the line GHEC and AF (diagonal) are perpendicular to each other
We get a right angle triangle as shown in the image attached., hypotenuse =  a\sqrt{2} 
One side of this right angle triangle will be half the measure of the diagonal AF ie =  \frac{a\sqrt{2}}{2}

from the measures of sides of the right angle triangle we notice that the The right angle triangle is of the form ,  30^{\circ}, 60^{\circ}, 90^{\circ} 
△HDC is also a right angle triangle, Hence its angles are  75^{\circ}, 90^{\circ}, 15^{\circ}

Also  75^{\circ} + x^{\circ}  =  180^{\circ} 
  x^{\circ}  =  180^{\circ} - 75^{\circ} 
  x^{\circ} = 105^{\circ}

Answer : C

chetan2u · Answer

Draw the diameter AF intersecting the second diagonal at O. => so  \angle AOC=90

Thus, AOC is a right angled triangle......The grey shaded triangle
AC is the diagonal, say X, and AO is diagonal/2 or X/2
The third side OC= \sqrt{X^2-(X/2)^2}=\frac{\sqrt{3}X}{2} 
Ratio of sides are  \frac{X}{2}:\frac{\sqrt{3}X}{2}:X=1:\sqrt{3}:2 
So AOC is 30:60:90 triangle, where  \angle OCA=30

Take  	riangle CDH 
 \angle CDH=90 
 \angle DCH=45-30=15 
So,  \angle DHC=180-90-15=75

Thus, the yellow shaded angle = 180-75=105

C

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GMAT CLUB OLYMPICS: In two congruent squares ABCD and AEFG shown above

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GMAT CLUB OLYMPICS: In two congruent squares ABCD and AEFG shown above

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