Find the area of a rectangular trapezoid in which the smaller sides are 12 cm each, and the largest angle is 135 °

ABCD – rectangular trapezoid: AB is perpendicular to the bases BC (smaller) and AD (larger). In this trapezoid, the smaller sides are the lateral side AB and the smaller base BC, then:
AB = BC = 12 cm.
The largest angle is ∠BCD = 135 °.
1. Let’s draw the AC diagonal. AC from the entire trapezium cuts off rectangular △ ABC: ∠ABC = 90 °, AB = BC = 12 cm – legs, AC – hypotenuse.
By the Pythagorean theorem:
AC = √ (AB² + BC²) = √ (12² + 12²) = √ (144 + 144) = √ (2 * 144) = 12√2 (cm).
2. Since △ ABC is an isosceles rectangular, then ∠CAB = ∠ACB = (180 ° – ∠ABC) / 2 = (180 ° – 90 °) / 2 = 90 ° / 2 = 45 °.
3. ∠BCD consists of two corners:
∠BCD = ∠ACB + ∠ACD;
45 ° + ∠ACD = 135 °;
∠ACD = 135 ° – 45 °;
∠ACD = 90 °.
4. ∠DAB (aka ∠А) consists of two angles:
∠DAB = ∠CAB + ∠CAD;
45 ° + ∠CAD = 90 °;
∠CAD = 90 ° – 45 °;
∠CAD = 45 °.
5. Consider △ ACD: ∠ACD = 90 °, ∠CAD = 45 °. By the theorem on the sum of the angles of a triangle:
∠ACD + ∠CAD + ∠ADC = 180 °;
∠ADC = 180 ° – 90 ° – 45 °;
∠ADC = 45 °.
Thus, two angles △ ACD are equal to 45 °, then △ ACD is a rectangular isosceles.
AC = CD = 12√2 cm – legs, AD – hypotenuse.
By the Pythagorean theorem:
AD = √ (AC² + CD²) = √ ((12√2) ² + (12√2) ²) = √ (144 * 2 + 144 * 2) = √ (144 * 4) = 12 * 2 = 24 ( cm).
6. The area of ​​the trapezoid ABCD is equal to:
S = (a + b) * h / 2,
where a and b are larger and smaller bases, h is the height of the trapezoid (in this case, its side, which is perpendicular to the bases – side AB).
S = (AD + BC) * AB / 2 = (24 + 12) * 12/2 = 36 * 12/2 = 432/2 = 216 (cm²).
Answer: S = 216 cm².