Can someone explain me how to solve this one? Thanks
What is the perimeter of ABCDE?
Perimeter calculation
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- steven_ghoos
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Hi Steven,
We're given three segments already, so we just need to solve for segments AE and ED.
The clue here is in the 30-degree angles we're given. Imagine that AED were a triangle, and then locate the angle of DAE based on the fact that DAB is 90 degrees and BAE is 30. 90 - 30 = 60. And the same goes for angle ADE. Which makes this an equilateral triangle.
Since we know all sides of this square are the same, that means imaginary segment AD would also be 12. And in an equilateral triangle, all sides are the same. So we know this five-sided figure has a perimeter of 12 + 12 + 12 + 12 + 12 = 60.
We're given three segments already, so we just need to solve for segments AE and ED.
The clue here is in the 30-degree angles we're given. Imagine that AED were a triangle, and then locate the angle of DAE based on the fact that DAB is 90 degrees and BAE is 30. 90 - 30 = 60. And the same goes for angle ADE. Which makes this an equilateral triangle.
Since we know all sides of this square are the same, that means imaginary segment AD would also be 12. And in an equilateral triangle, all sides are the same. So we know this five-sided figure has a perimeter of 12 + 12 + 12 + 12 + 12 = 60.
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Draw an Imaginary line to join the points A and D. Now AD must be parallel to BC (since angles ABC and DCB are both 90 deg and both BA and CD measure 12).
Also, angle BAD = angle CDA = 90 degrees.
Angle EAD = 90-30 = 60
Angle ADE = 90-30 = 60
Triangle ADE is an equilateral triangle. Thus AE = ED = AD = 12
Perimeter = 12*5 = 60
- ankur.agrawal
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Hey Aleph. Some Doubts:aleph777 wrote:Hi Steven,
We're given three segments already, so we just need to solve for segments AE and ED.
The clue here is in the 30-degree angles we're given. Imagine that AED were a triangle, and then locate the angle of DAE based on the fact that DAB is 90 degrees and BAE is 30. 90 - 30 = 60. And the same goes for angle ADE. Which makes this an equilateral triangle.
Since we know all sides of this square are the same, that means imaginary segment AD would also be 12. And in an equilateral triangle, all sides are the same. So we know this five-sided figure has a perimeter of 12 + 12 + 12 + 12 + 12 = 60.
Angle DAB =90 . How?
Why would AD be 12. CAn we imagine this figure to be a square?
Thanx.
- ankur.agrawal
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How can we say that AD is parallel to BC. Ur reasoning: (since angles ABC and DCB are both 90 deg and both BA and CD measure 12). Is this sum kind of theorem. Kindly explain.stormier wrote:
Draw an Imaginary line to join the points A and D. Now AD must be parallel to BC (since angles ABC and DCB are both 90 deg and both BA and CD measure 12).
Also, angle BAD = angle CDA = 90 degrees.
Angle EAD = 90-30 = 60
Angle ADE = 90-30 = 60
Triangle ADE is an equilateral triangle. Thus AE = ED = AD = 12
Perimeter = 12*5 = 60
- GMATGuruNY
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Here is one way to prove that ABCD is square:ankur.agrawal wrote:How can we say that AD is parallel to BC. Ur reasoning: (since angles ABC and DCB are both 90 deg and both BA and CD measure 12). Is this sum kind of theorem. Kindly explain.stormier wrote:
Draw an Imaginary line to join the points A and D. Now AD must be parallel to BC (since angles ABC and DCB are both 90 deg and both BA and CD measure 12).
Also, angle BAD = angle CDA = 90 degrees.
Angle EAD = 90-30 = 60
Angle ADE = 90-30 = 60
Triangle ADE is an equilateral triangle. Thus AE = ED = AD = 12
Perimeter = 12*5 = 60
Draw diagonals AC and BD.
Since AB = BC = 12 and angle ABC is a right angle, diagonal AC = 12√2, forming a 45:45:90 triangle.
Since BC = CD = 12 and angle BCD is a right angle, diagonal BD = 12√2, forming another 45:45:90 triangle.
The result is that diagonals AC and BD are both equal and perpendicular, proving that ABCD is a square. A square is the only quadrilateral in which the diagonals are equal and perpendicular.
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- ankur.agrawal
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Why will they be perpendicular? Plz clear my doubtGMATGuruNY wrote:Here is one way to prove that ABCD is square:ankur.agrawal wrote:How can we say that AD is parallel to BC. Ur reasoning: (since angles ABC and DCB are both 90 deg and both BA and CD measure 12). Is this sum kind of theorem. Kindly explain.stormier wrote:
Draw an Imaginary line to join the points A and D. Now AD must be parallel to BC (since angles ABC and DCB are both 90 deg and both BA and CD measure 12).
Also, angle BAD = angle CDA = 90 degrees.
Angle EAD = 90-30 = 60
Angle ADE = 90-30 = 60
Triangle ADE is an equilateral triangle. Thus AE = ED = AD = 12
Perimeter = 12*5 = 60
Draw diagonals AC and BD.
Since AB = BC = 12 and angle ABC is a right angle, diagonal AC = 12√2, forming a 45:45:90 triangle.
Since BC = CD = 12 and angle BCD is a right angle, diagonal BD = 12√2, forming another 45:45:90 triangle.
The result is that diagonals AC and BD are both equal and perpendicular, proving that ABCD is a square. A square is the only quadrilateral in which the diagonals are equal and perpendicular.
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ankur.agrawal wrote:Why will they be perpendicular? Plz clear my doubtGMATGuruNY wrote:Here is one way to prove that ABCD is square:ankur.agrawal wrote:How can we say that AD is parallel to BC. Ur reasoning: (since angles ABC and DCB are both 90 deg and both BA and CD measure 12). Is this sum kind of theorem. Kindly explain.stormier wrote:
Draw an Imaginary line to join the points A and D. Now AD must be parallel to BC (since angles ABC and DCB are both 90 deg and both BA and CD measure 12).
Also, angle BAD = angle CDA = 90 degrees.
Angle EAD = 90-30 = 60
Angle ADE = 90-30 = 60
Triangle ADE is an equilateral triangle. Thus AE = ED = AD = 12
Perimeter = 12*5 = 60
Draw diagonals AC and BD.
Since AB = BC = 12 and angle ABC is a right angle, diagonal AC = 12√2, forming a 45:45:90 triangle.
Since BC = CD = 12 and angle BCD is a right angle, diagonal BD = 12√2, forming another 45:45:90 triangle.
The result is that diagonals AC and BD are both equal and perpendicular, proving that ABCD is a square. A square is the only quadrilateral in which the diagonals are equal and perpendicular.
The drawing above shows the 45:45:90 triangles that are formed when the diagonals AC and BD are drawn. It illustrates why the diagonals both equal and perpendicular, proving that ABCD is a square. Please refer to the explanation in my earlier post.
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As a tutor, I don't simply teach you how I would approach problems.
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- ankur.agrawal
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Thanks for the explaination. Apology for asking small doubts.GMATGuruNY wrote:ankur.agrawal wrote:Why will they be perpendicular? Plz clear my doubtGMATGuruNY wrote:Here is one way to prove that ABCD is square:ankur.agrawal wrote:How can we say that AD is parallel to BC. Ur reasoning: (since angles ABC and DCB are both 90 deg and both BA and CD measure 12). Is this sum kind of theorem. Kindly explain.stormier wrote:
Draw an Imaginary line to join the points A and D. Now AD must be parallel to BC (since angles ABC and DCB are both 90 deg and both BA and CD measure 12).
Also, angle BAD = angle CDA = 90 degrees.
Angle EAD = 90-30 = 60
Angle ADE = 90-30 = 60
Triangle ADE is an equilateral triangle. Thus AE = ED = AD = 12
Perimeter = 12*5 = 60
Draw diagonals AC and BD.
Since AB = BC = 12 and angle ABC is a right angle, diagonal AC = 12√2, forming a 45:45:90 triangle.
Since BC = CD = 12 and angle BCD is a right angle, diagonal BD = 12√2, forming another 45:45:90 triangle.
The result is that diagonals AC and BD are both equal and perpendicular, proving that ABCD is a square. A square is the only quadrilateral in which the diagonals are equal and perpendicular.
The drawing above shows the 45:45:90 triangles that are formed when the diagonals AC and BD are drawn. It illustrates why the diagonals both equal and perpendicular, proving that ABCD is a square. Please refer to the explanation in my earlier post.