Give the isosceles trapezoid ABCD
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1. The legs are AB and CD.
2. The bases are BC and AD.
3. The base angles are ∠ABC and ∠ADC.
4. Since ABCD is an isosceles trapezoid, AB = CD. Also, ∠BAC = ∠BCA and ∠CDA = ∠DCA. Therefore,
∠B = 180° - ∠CDA - ∠ABC = 180° - 70° - ∠BCA = 110° - ∠BCA
Since ∠B = ∠BCA, we have ∠B = 110° - ∠B. Solving for ∠B, we get:
2∠B = 110°
∠B = 55°
5. Since ABCD is an isosceles trapezoid, AB = CD. Also, ∠BAC = ∠BCA and ∠CDA = ∠DCA. Therefore,
∠C = 180° - ∠BCA - ∠ADC = 180° - ∠ABC - 105° = ∠ABC - 75°
Since ∠BAC = ∠BCA, we have ∠C = ∠BAC - 75°. Substituting for ∠BAC, we get:
∠C = ∠BCA - 75°
Since ∠C = ∠BCA, we have ∠C = ∠C - 75. Solving for ∠C, we get:
∠C = 75°
6. Since ABCD is an isosceles trapezoid, AB = CD. Also, ∠BAC = ∠BCA and ∠CDA = ∠DCA. Therefore,
∠B = 180° - ∠CDA - ∠ABC = 180° - 82° - ∠BCA = 98° - ∠BCA
Since ∠B = ∠BCA, we have ∠B = 98° - ∠B. Solving for ∠B, we get:
2∠B = 98°
∠B = 49°
Substituting for ∠B in the expression for m B, we get:
2x - 6 = 49°
2x = 55°
x = 27.5°
7. Since ABCD is an isosceles trapezoid, AB = CD. Also, ∠BAC = ∠BCA and ∠CDA = ∠DCA. Therefore,
∠ACD = ∠BCA
Also, ∠CDB = ∠BAC and ∠CDA = ∠BDC.
Applying the sum of angles in a triangle, we have:
∠ACB + ∠BAC + ∠CAB = 180°
∠BCA + ∠BAC + ∠CDA + ∠ADB = 360°
Substituting the known values, we get:
∠ACB + 70° + ∠CAB = 180°
∠BCA + 70° + 4y + 116° = 360°
Solving the first equation for ∠CAB, we get:
∠CAB = 110° - ∠ACB
Substituting this expression for ∠CAB in the second equation, we get:
∠BCA + 70° + 4y + 116° = 360°
∠BCA + 4y = 174°
Substituting the value of ∠BCA from the first equation, we get:
∠ACB + 70° + 110° - ∠ACB + 4y = 174°
4y = 6°
Therefore, y = 1.5°
8. Let h be the altitude of ABCD. Then,
AB + CD = 2h
Since ABCD is an isosceles trapezoid, AB = CD. Also, let a be the length of one of the bases (say, AD) and b be the length of the other base (BC). Then,
AC = BD = √(h² + (b-a)²/4)
But we are given that AC = 56 cm. Substituting this value and the values of a and b, we get:
56 = √(h² + (BC-AD)²/4)
We also know that AB = CD = h(BC-AD)/(b-a). Substituting this value for AB and CD in the equation, we get:
56 = √(h² + h²(BC-AD)²/(b-a)²)
Squaring both sides and simplifying, we get:
h²(BC-AD)²/(b-a)² = 1568
But we also know that BC + AD = b + a. Substituting this value in the equation, we get:
h²(b-a)²/(b-a)² = 392
Therefore, h = 20√2 cm.
Substituting this value in the equation for AB and CD, we get:
AB = CD = 40√2 cm.
9. The problem is not clear. Please provide more information.
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