MHB Solving the Tangram Puzzle Geometrically (No Constructions)

[squigley5]

I have looked at the threads posted by others who were working on what seems to be the same task. I have correctly determined all the lengths and angles as the task requires, however am struggling to justify the lengths geometrically, without constructing any midpoints or additional constructions. I was told by the graders that I could not use any constructions, it has to be done with logic. What am I missing or doing wrong?

Here is the task:

Given:
● AFJ and JFK are large, congruent right isosceles triangles.
● EFH and ABD are small, congruent right isosceles triangles.
● BCG is a right isosceles triangle.
● BEFD is a square.
● EGKH is a parallelogram.
● ACKJ is a square with dimensions of 1 unit by 1 unit (i.e., the entire area of the square is 1 unit2).

Note: The right angle for each triangle can be determined by inspection. All line segments that appear straight are straight (e.g., JE ̅̅̅ is straight, with no bend at F).

There are no gaps or overlapping figures.

Requirements:
A. Determine the dimensions and area of each of the seven individual pieces from the square arrangement in Figure 1. (rearranging the pieces is not allowed).
1. Explain with full geometric justification, how you determined the dimensions of each piece.

Note: You cannot make midpoint assumptions (e.g., B is the midpoint between A and C).

B. Determine the angle measures of each of the seven individual pieces from the square arrangement in Figure 1. (rearranging the pieces is not allowed).
1. Explain with full geometric justification, how you determined the angle measures of each piece.

View attachment 5569

 

[squigley5]

Here is the work I have completed, the cells that mention the midpoint construction are the ones that are causing the problem I believe, but I cannot figure a different way to justify the lengths.

For▲AFJand ▲JFK

Statement

[TD="width: 50%"] Reason

[/TD]

[TD="width: 50%"] ▲ AFJ and ▲ JFK are congruent right isosceles triangles, AJ and JK have a length of 1 unit, m∠AFJ, m∠JFK and m∠AJK = 90°

[/TD]
[TD="width: 50%"] Given and visual observation

[/TD]

[TR]
[TD="width: 50%"] m∠AJF + m∠FAJ + m∠AFJ = m∠FJK + m∠FKJ +m∠JFK = 180°

[/TD]
[TD="width: 50%"] Sum of internal angles in triangle is 180°

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m∠AJF + m∠FAJ = m∠FJK + m∠FKJ = 90°

[/TD]
[TD="width: 50%"] Subtraction

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m∠AJF = m∠FAJ = 45°

[/TD]
[TD="width: 50%"] Division, Isosceles triangle definition

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m∠AJF = m∠FAJ = m∠FJK = m∠FKJ = 45°

[/TD]
[TD="width: 50%"] CPCTC

[/TD]
[/TR]
[TR]
[TD="width: 50%"] ▲AJK is an isosceles right triangle

[/TD]
[TD="width: 50%"] SAS Postulate, Complementary angles definition

[/TD]
[/TR]
[TR]
[TD="width: 50%"] ▲AJK is similar to ▲AFJ and ▲ JFK

[/TD]
[TD="width: 50%"] Definition of right isosceles triangle

[/TD]
[/TR]
[TR]
[TD="width: 50%"] mAJ² + m JK² = m AK² , 1² + 1² = 2 , therefore AK has a length of √2

[/TD]
[TD="width: 50%"] Given Information and Pythagorean Theorem

[/TD]
[/TR]
[TR]
[TD="width: 50%"] ▲AFJ and ▲ JFK are congruent right isosceles triangles, m∠KFJ = m∠AFJ = 90°, and AF = FK, therefore FJ is a perpendicular bisector of AK and also the median of ▲AJK

[/TD]
[TD="width: 50%"] Given Information, Observation, CPCTC and Definitions of Right Isosceles Triangles, Median of a Triangle and Perpendicular Bisectors


[/TD]
[/TR]
[TR]
[TD="width: 50%"] AF and FK have a length of √2/2

[/TD]
[TD="width: 50%"] Definition and Division

[/TD]
[/TR]
[TR]
[TD="width: 50%"] FJ = AF = FK

[/TD]
[TD="width: 50%"] Congruency, Isosceles triangle definition

[/TD]
[/TR]
[TR]
[TD="width: 50%"] ▲AJK has an area of ½ Units²

[/TD]
[TD="width: 50%"] A=1/2 bh

[/TD]
[/TR]
[TR]
[TD="width: 50%"] ▲AFJ and ▲ JFK each have an area of ¼ units²

[/TD]
[TD="width: 50%"] Given they are congruent and by definition each is ½ of ▲AJK

[/TD]
[/TR]

For▲ABDand ▲HEF

Statement

[TD="width: 50%"] Reason

[/TD]

[TD="width: 50%"] ▲ ABD and ▲ HEF are congruent right isosceles triangles, m∠ADB and m∠EFH = 90°

[/TD]
[TD="width: 50%"] Given and visual observation

[/TD]

[TR]
[TD="width: 50%"] 90° + m∠DAB + m∠ABD = m∠FHE + m∠HEF + 90° = 180°

[/TD]
[TD="width: 50%"] Sum of internal angles in triangle is 180°

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m∠DAB + m∠ABD = m∠FHE + m∠HEF = 90°

[/TD]
[TD="width: 50%"] Subtraction

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m∠DAB = m∠ABD = 45°

[/TD]
[TD="width: 50%"] Division, Isosceles triangle definition

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m∠DAB = m∠ABD = m∠FHE = m∠HEF = 45°

[/TD]
[TD="width: 50%"] CPCTC

[/TD]
[/TR]
[TR]
[TD="width: 50%"] EF, FH, BD and AD are Congruent

[/TD]
[TD="width: 50%"] CPCTC

[/TD]
[/TR]
[TR]
[TD="width: 50%"] HE and AB are congruent

[/TD]
[TD="width: 50%"] CPCTC

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m FH = √2/4 Units


[/TD]
[TD="width: 50%"] By completing a midpoint construction, the mid point of FK was determined to be point H, therefore FH and HK are are congruent and ½ the length of FK. Length of FK was determined in a previous proof. (see Figure 1 below for Midpoint Construction proof)


[/TD]
[/TR]
[TR]
[TD="width: 50%"] m FH, EF, BD and AD = √2/4 Units

[/TD]
[TD="width: 50%"] Definition of Isosceles triangle and CPCTC

[/TD]
[/TR]
[TR]
[TD="width: 50%"] mFH² + m FE² = m EH², √2/4² + √2/4² = 4/16, therefore m EH = √4/16 = ½ Unit

[/TD]
[TD="width: 50%"] Pythagorean Theorem

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m EH = m AB = ½ Unit

[/TD]
[TD="width: 50%"] CPCTC

[/TD]
[/TR]

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[squigley5]

Work Continued

For▲BCG

Statement

[TD="width: 50%"] Reason

[/TD]

[TD="width: 50%"] ▲ BCG is a right isosceles triangle, m∠BCG = 90°

[/TD]
[TD="width: 50%"] Given and inspection

[/TD]

[TR]
[TD="width: 50%"] m∠CGB + m∠CBG + 90° = 180°

[/TD]
[TD="width: 50%"] Sum of internal angles in triangle is 180°

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m∠CGB + m∠CBG = 90°

[/TD]
[TD="width: 50%"] Subtraction

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m∠CGB = m∠CBG = 45°

[/TD]
[TD="width: 50%"] Division, Isosceles triangle definition

[/TD]
[/TR]
[TR]
[TD="width: 50%"] BC and CG are congruent

[/TD]
[TD="width: 50%"] Definition of Isosceles triangle

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m CG + m GK = 1 Unit

[/TD]
[TD="width: 50%"] Given length of CK is 1 Unit

[/TD]
[/TR]
[TR]
[TD="width: 50%"] 1 Unit - m GK = m CG = ½ Unit

[/TD]
[TD="width: 50%"] Subtraction, proof for parallelogram

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m CG = m BC = ½ Unit

[/TD]
[TD="width: 50%"] Congruent sides of an isosceles triangle

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m BG = m BE + m EG

[/TD]
[TD="width: 50%"] Two segments added together make a whole

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m BG = √2/2 units

[/TD]
[TD="width: 50%"] Addition

[/TD]
[/TR]

For■BEFD

Statement

[TD="width: 50%"] Reason

[/TD]

[TD="width: 50%"] BEFD is a square, all four sides and angles are congruent, all angles are 90°

[/TD]
[TD="width: 50%"] Definition and given information

[/TD]

[TR]
[TD="width: 50%"] Length of DB = √2/4 Units

[/TD]
[TD="width: 50%"] Line segments are congruent with themselves, the length of DB was found during exercise for ▲ABD

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m DB, FD, EF and BE = √2/4 Units

[/TD]
[TD="width: 50%"] Definition of a square

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m∠BEF, m∠EFD, m∠FDB, m∠DBE = 90°

[/TD]
[TD="width: 50%"] Definition of a square

[/TD]
[/TR]

 

[squigley5]

Work Continued

ForParallelogram EGKH

Statement

[TD="width: 50%"] Reason

[/TD]

[TD="width: 50%"] EGKH is a parallelogram

[/TD]
[TD="width: 50%"] Given

[/TD]

[TR]
[TD="width: 50%"] m HK = √2/4 Units

[/TD]
[TD="width: 50%"] By completing a midpoint construction, the mid point of FK was determined to be point H, therefore FH and HK are are congruent and ½ the length of FK , as found in proof for ▲ HEF

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m HK = m EG

[/TD]
[TD="width: 50%"] Opposite sides of a parallelogram are congruent

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m HE = ½ Unit

[/TD]
[TD="width: 50%"] Line segments are congruent with themselves, the length of HE was found during exercise for ▲HEF

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m GK = m HE = ½ Unit

[/TD]
[TD="width: 50%"] Opposite sides of a parallelogram are congruent

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m∠EHK = 135°

[/TD]
[TD="width: 50%"] ∠EHK and ∠EHF are supplementary therefore by subtraction 180° - (m∠EHF)45° = 135°, Previous proof

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m∠EHK = m∠EGK = 135°

[/TD]
[TD="width: 50%"] Opposite angles of parallelograms are congruent

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m∠JKF = 45°

[/TD]
[TD="width: 50%"] Previous proof

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m∠JKF + m∠GKH = 90°

[/TD]
[TD="width: 50%"] Complementary angle definition. Given ACKJ is a Square therefore all angles are 90°

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m∠GKH = 45°

[/TD]
[TD="width: 50%"] Complementary angle definition

[/TD]
[/TR]
[TR]
[TD="width: 50%"] m∠GKH = m∠HEG = 45°

[/TD]
[TD="width: 50%"] Opposite angles of parallelograms are congruent

[/TD]
[/TR]

 

[Greg]

Hi squigley5 and welcome to MHB! :D

Given:
● and are large, congruent right isosceles triangles.
● and are small, congruent right isosceles triangles.
● is a right isosceles triangle.
● is a square.
● is a parallelogram.
● is a square with dimensions of 1 unit by 1 unit (i.e., the entire area of the square is 1 unit2).

The above list is incomplete. Can you please clarify?

 

[squigley5]

Thanks for the heads up. Took so long to get it all posted I missed the dropped characters. Fixed it.

 

[squigley5]

Solved: Since BEFD is a square and ABD is a right isosceles triangle, segments DB, BE, EF and FD are congruent. Since ABD is a right isosceles triangle AD is congruent with BD, therefore AD is congruent with FD and their sum is equal to AF. Since we now can mathematically calculate FD and AD we can solve the rest of the lengths without constructions.