Question 12a

In the figure below, EFGH is a parallelogram and EJK is an isosceles triangle. FG is parallel to KJ. ∠EHG = 70° and JE = JK. Find ∠x. (2 marks)

$Question 12a diagram$

Answer: 40 °

Workings

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∠ x is part of isosceles triangle EJK

∠ x = 180 - ∠ EKJ (purple) x 2

∠ EKJ (purple) → 180 - ∠ FKJ (green)

since FG is parallel to KJ, KF is parallel to HG, this makes KFGA is a parallelogram

∠ FKJ (green) → 180 - 70 = 110

∠ EKJ (purple) → 180 - ∠ FKJ (green) = 180 - 110 = 70

∠ x = 180 - ∠ EKJ (purple) x 2 = 180 - 70 x 2 = 40

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