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In trapezoidal ABCD, ad ∥ BC, ab = 4, BC = 9, CD = 5, Da = 6. (1) prove ab ⊥ BC; (2) find the area of trapezoidal ABCD
The parallel line of AB passing through point d intersects point E
Then CE = BC-AD = 9-6 = 3
DE=AB=4,
And CD = 5
We know 3 ^ 2 + 4 ^ 2 = 5 ^ 2 by Pythagorean theorem
Let de and EC be right angles
So ab ⊥ BC
The trapezoid area is (upper bottom + lower bottom) × height △ 2 = (6 + 9) × 4 △ 2 = 30
In trapezoidal ABCD, ab ∥ CD, ad ⊥ BD, DC = 3, BC = 7, Da = 8, then ab=____ ,BD=_____ ,AC=_____
The answer is ab = 12, BD = 4, radical 5, AC = radical 105
Please help me write the correct process,
Let BD = x, then by using the cosine relation, X / sqrt (x ^ 2 + 64) = (x ^ 2-40) / 6x, we can solve x = 4sqrt5, thus AB = 12, we can also solve the cosine value of abd from above, ADC cosine value is its opposite number, we can solve AC by using cosine theorem
It is known that in the trapezoidal ABCD, ad ∥ BC, ab = 4, BC = 9, CD = 5, Da = 6 (1) explains: ab ⊥ BC (2) calculates the area of the trapezoidal ABCD
Trapezoidal formula = [(up + down) * high] / 2
=[(6+9)*4]/2
=30
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