The bases of the trapezoid are 2cm and 7cm, and its diagonals are 10 and 17cm. Find the area.
Let ABCD be a trapezoid: AD = 7 cm and BC = 2 cm – bases, AC = 10 cm and BD = 17 cm – diagonals.
1. Draw the height BH from the vertex B.
In △ BHD, by the Pythagorean theorem, we express BH:
BH = √ (BD² – DH²).
Since AD is divided by point H into two segments AH and DH, then:
DH = AD – AH.
Thus:
BH = √ (BD² – (AD – AH) ²) = √ (17² – (7 – AH) ²) = √ (289 – (7 – AH) ²) = √ (289 – (49 – 14AH + AH²)) = √ (289 – 49 + 14AH – AH²) = √ (240 + 14AH – AH²).
2. Draw the height CK from the vertex C.
In △ CKA, by the Pythagorean theorem, we express CK:
CK = √ (AC² – AK²).
Since AD is divided by point K into two segments AK and DK, then:
AK = AD – DK.
Thus:
CK = √ (AC² – (AD – DK) ²) = √ (10² – (7 – DK) ²) = √ (100 – (49 – 14DK + DK²)) = √ (100 – 49 + 14DK – DK²)) = √ (51 + 14DK – DK²).
3. AD by points H and K is divided into three segments AH, HK, DK:
AD = AH + HK + DK.
Since BH and CK are perpendicular to parallel straight lines, they are parallel to each other. The quadrilateral HBCK is a rectangle with sides BH = CK and BC = HK.
Since BC = HK, then HK = 2 cm.
Means:
AH + 2 + DK = 7;
AH + DK = 7-2;
AH + DK = 5.
Since BH = CK, then:
√ (240 + 14AH – AH²) = √ (51 + 14DK – DK²).
4. We have obtained a system of equations with two unknowns:
√ (240 + 14AH – AH²) = √ (51 + 14DK – DK²);
AH + DK = 5.
In the second equation of the system, we express AH in terms of DK:
AH = 5 – DK.
We substitute the resulting expression into the first equation of the system and solve the equation with one unknown:
√ (240 + 14 (5 – DK) – (5 – DK) ²) = √ (51 + 14DK – DK²) (square both sides of the equation);
(√ (240 + 70 – 14DK – 25 + 10DK – DK²) ² = (√ (51 + 14DK – DK²)) ²;
– DK² – 4 DK + 285 = – DK² + 14DK + 51 (we transfer all terms from the unknown to the left side of the equation, and all natural terms – to the right);
– DK² + DK² – 4 DK – 14DK + 285 – 51 = 51 – 285 (we give similar terms);
– 18DK = – 234;
DK = (-234) / (- 18);
DK = 13.
5. Find the length CK:
CK = √ (51 + 14 * 13 – 13²) = √ (51 + 182 – 169) = √ 64 = 8 (cm).
6. The area of the trapezoid is found by the formula:
S = (a + b) * h / 2,
where a and b are the bases of the trapezoid, h is the height of the trapezoid.
S = (7 + 2) * 8/2 = 9 * 8/2 = 9 * 4 = 36 (cm²).
Answer: S = 36 cm².
