Line DF bisects angle EDG. Find FG. The diagram is not drawn to scale.
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Answer:
Step-by-step explanation:
10). AD = BC [If ABCD is a parallelogram]
3x + 1 = 6x - 4
6x - 3x = 1 + 4
3x = 5
x =
11). If ABCD is a parallelogram,
AB = CD
5x - 8 = 2x + 7
5x - 2x = 7 + 8
3x = 15
x = 5
12). If ABCD is a parallelogram,
AB = CD
AB║CD and BD is a transversal intersecting these parallel lines.
∠ABD ≅ ∠BDC [Alternate interior angles]
4x - 1 = x + 38
4x - x = 38 + 1
3x = 39
x = 13
See the attached figure.
<span>ad is a diameter of the circle with center p
</span>
∵ pd = radius = 7 ⇒⇒⇒ ∴ ad = 2 * radius = 2 * 7 = 14
∵ ae = 4 ⇒⇒⇒ ∴ ed = ad - ae = 14 - 4 = 10
∵ ad is a diameter
Δ acd is a triangle drawn in a half circle
∴ Δ acd is a right triangle at c
∵ bc ⊥ ad at point e
By applying euclid's theorem inside Δ acd
∴ ce² = ae * ed
∴ ce² = 4 * 10 = 40
∴ ce = √40 = 2√10 ≈ 6.325
<span>ad is a diameter of the circle with center p
</span>
∵ pd = radius = 7 ⇒⇒⇒ ∴ ad = 2 * radius = 2 * 7 = 14
∵ ae = 4 ⇒⇒⇒ ∴ ed = ad - ae = 14 - 4 = 10
∵ ad is a diameter
Δ acd is a triangle drawn in a half circle
∴ Δ acd is a right triangle at c
∵ bc ⊥ ad at point e
By applying euclid's theorem inside Δ acd
∴ ce² = ae * ed
∴ ce² = 4 * 10 = 40
∴ ce = √40 = 2√10 ≈ 6.325