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(i drew this diagram on paper so its easier to see. great title btw)
We know that AB=AH, which makes triangle ABH isosceles with side length BH being the side that isn’t equal to the others. We also know that HK is the same length as the other two which means that triangle HCK is also isosceles (HC is the same as CK because a rhombus has all equal sides and H, K are midpoints of BC and CD respectively). KA is also the same length as the other three, which would mean that triangle AHK is equilateral. This makes all of these angle inside of triangle AHK 60 degrees each. We need to give angle ABH a value, so we can solve.
Let x represent the value of angle ABH. Since angle ABH and angle AHB are the same, we can label them both as x. This would mean that angle ABH is (180-2x). By using parallel lines and their rules, we find that angle ADK and AKD are both x, angle KAD is also (180 - 2x), and angle HCK is (180-x).
Since we know that BC is parallel to AD, using the C-rule (what I would call TPT-C), we can conclude that…..
Angle BAD = 180- angle ABC (180-2x)+(60)+(180-2x) = (180-x) 420 - 3x= 180-x 240 = 3x x = 80 degrees Angle ABH is 80 degrees. But since we are looking for angle ABD, we can… (180-x) = 180-80 = 100
Grade 7/8 POTW Triangle ABCD is a rhombus. H is in between B and C while K is in between C and D. We also know that AB = BC = CD = DA = AH = HK = KA. This means that ABH is an isosceles triangle as AB = AH. AKD is a congruent isosceles triangle as well as AD = AK. AHK is an equilateral triangle as AH = HK = KA. We can now use this information to try to solve for angle BAD. Firstly, each angle of AHK must be 60 degrees as it is equilateral meaning all angles are the same and 60 degrees (180/3 = 60). We can also let angle ABH = AHB = ADK = AKD = x degrees. We know all angles add up to 180 degrees in a triangle, so for both congruent isosceles triangles, the remaining angles would be BAH = KAD = 180 - 2x degrees. Finally, using interior angles adding up to 180 degrees, we know that BAD = BCD = 180 - x degrees. We can now use this information to solve for angle BAD. 180 - Angle ABC = Angle BAD 180 - x = (180 - 2x) + (180 - 2x) + 60 180 - x = 180 - 2x + 180 - 2x + 60 180 - x = 420 - 4x 180 - x + 4x - 180 = 420 - 4x + 4x - 180 3x = 240 3x/3 = 240/3 x = 80 Angle BAD = 180 - x = 180 - 80 = 100 degrees. Therefore, angle BAD is equal to 100 degrees.
Grade 7/8 POTW:
ReplyDelete(i drew this diagram on paper so its easier to see. great title btw)
We know that AB=AH, which makes triangle ABH isosceles with side length BH being the side that isn’t equal to the others. We also know that HK is the same length as the other two which means that triangle HCK is also isosceles (HC is the same as CK because a rhombus has all equal sides and H, K are midpoints of BC and CD respectively). KA is also the same length as the other three, which would mean that triangle AHK is equilateral. This makes all of these angle inside of triangle AHK 60 degrees each. We need to give angle ABH a value, so we can solve.
Let x represent the value of angle ABH. Since angle ABH and angle AHB are the same, we can label them both as x. This would mean that angle ABH is (180-2x). By using parallel lines and their rules, we find that angle ADK and AKD are both x, angle KAD is also (180 - 2x), and angle HCK is (180-x).
Since we know that BC is parallel to AD, using the C-rule (what I would call TPT-C), we can conclude that…..
Angle BAD = 180- angle ABC
(180-2x)+(60)+(180-2x) = (180-x)
420 - 3x= 180-x
240 = 3x
x = 80 degrees
Angle ABH is 80 degrees. But since we are looking for angle ABD, we can…
(180-x)
= 180-80
= 100
Angle BAD is 100 degrees
Grade 7/8 POTW
ReplyDeleteTriangle ABCD is a rhombus. H is in between B and C while K is in between C and D. We also know that AB = BC = CD = DA = AH = HK = KA. This means that ABH is an isosceles triangle as AB = AH. AKD is a congruent isosceles triangle as well as AD = AK. AHK is an equilateral triangle as AH = HK = KA. We can now use this information to try to solve for angle BAD. Firstly, each angle of AHK must be 60 degrees as it is equilateral meaning all angles are the same and 60 degrees (180/3 = 60). We can also let angle ABH = AHB = ADK = AKD = x degrees. We know all angles add up to 180 degrees in a triangle, so for both congruent isosceles triangles, the remaining angles would be BAH = KAD = 180 - 2x degrees. Finally, using interior angles adding up to 180 degrees, we know that BAD = BCD = 180 - x degrees. We can now use this information to solve for angle BAD.
180 - Angle ABC = Angle BAD
180 - x = (180 - 2x) + (180 - 2x) + 60
180 - x = 180 - 2x + 180 - 2x + 60
180 - x = 420 - 4x
180 - x + 4x - 180 = 420 - 4x + 4x - 180
3x = 240
3x/3 = 240/3
x = 80
Angle BAD = 180 - x = 180 - 80 = 100 degrees.
Therefore, angle BAD is equal to 100 degrees.
POTW:
ReplyDeleteMy POTW was done on paper.
The answer was: The measure of angle BAD is 100 degrees.