In the figure, EFGH is a parallelogram with the length of EH 2 times the length of EF. EHJ is an equilateral triangle. K is a point on EJ such that EK = KJ. ∠FGH is 110°. Find ∠KFG .
4 m
In the figure, EFGH is a parallelogram with the length of EH 2 times the length of EF. EHJ is an equilateral triangle. K is a point on EJ such that EK = KJ. ∠FGH is 110°. Find ∠KFG .
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Ischool Pte Ltd
Date:25 Apr 2024
Topic:
ANGLES
Question Code:
P6.AN.NK.17051716
Student:
Class:
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In the figure, EFGH is a parallelogram with the length of EH 2 times the length of EF. EHJ is an equilateral triangle. K is a point on EJ such that EK = KJ. ∠FGH is 110°. Find ∠KFG .
(4 m)
Ischool Pte Ltd
Date:
25 Apr 2024
Topic:
ANGLES
Question Code:
P6.AN.NK.17051716
Student:
Class:
Level:
School:
In the figure, EFGH is a parallelogram with the length of EH 2 times the length of EF. EHJ is an equilateral triangle. K is a point on EJ such that EK = KJ. ∠FGH is 110°. Find ∠KFG .
In the figure, EFGH is a parallelogram with the length of EH 2 times the length of EF. EHJ is an equilateral triangle. K is a point on EJ such that EK = KJ. ∠FGH is 110°. Find ∠KFG .
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ISchool Pte Ltd
Date:25 Apr 2024
Topic:ANGLES
Question Number:P6.AN.NK.17051716-A1
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