Examkin

Question

Difficulty: HardQuadrilaterals and Polygons

In regular octagon ABCDEFGHABCDEFGH with side length 22, diagonals ADAD and BEBE intersect at point PP. What is the area of quadrilateral ABCPABCP?

  1. 2+22 + \sqrt{2}Answer
  2. B
    222\sqrt{2}
  3. C
    44
  4. D
    4+24 + \sqrt{2}
  5. E
    1+221 + 2\sqrt{2}

Answer

2+22 + \sqrt{2}
In regular octagon ABCDEFGHABCDEFGH, each interior angle measures 135∘135^\circ. Segment ABCDABCD forms an isosceles trapezoid where ADAD is parallel to BCBC. The perpendicular distance from ADAD to BCBC is 2sin⁡(45∘)=22 \sin(45^\circ) = \sqrt{2}, and AD=2+2(2cos⁡(45∘))=2+22AD = 2 + 2(2 \cos(45^\circ)) = 2 + 2\sqrt{2}. Diagonal BEBE is parallel to side CDCD, so ∠CBE=45∘\angle CBE = 45^\circ, which implies ∠ABE=135∘−45∘=90∘\angle ABE = 135^\circ - 45^\circ = 90^\circ. Triangle △ABP\triangle ABP is therefore a right isosceles triangle with legs AB=2AB = 2 and BP=2BP = 2, yielding hypotenuse AP=22AP = 2\sqrt{2}. Quadrilateral ABCPABCP is a trapezoid with parallel sides BC=2BC = 2 and AP=22AP = 2\sqrt{2} and height h=2h = \sqrt{2}. Its area is 2+222×2=2+2\frac{2 + 2\sqrt{2}}{2} \times \sqrt{2} = 2 + \sqrt{2}.

Step-by-Step Solution

1
Calculate the interior angle of a regular octagon
Interior angle = (8−2)×180∘8=135∘\frac{(8-2) \times 180^\circ}{8} = 135^\circ
Regular polygon interior angle formula.
2
Determine the orientation and length of diagonal ADAD
AD∥BCAD \parallel BC, height between them is h=2sin⁡(45∘)=2h = 2 \sin(45^\circ) = \sqrt{2}, and base AD=2+2(2cos⁡(45∘))=2+22AD = 2 + 2(2 \cos(45^\circ)) = 2 + 2\sqrt{2}
Quadrilateral ABCDABCD forms an isosceles trapezoid with side length 22 and interior angles 135∘135^\circ.
3
Determine the direction of diagonal BEBE and find intersection point PP
∠CBE=180∘−135∘=45∘\angle CBE = 180^\circ - 135^\circ = 45^\circ, making ∠ABE=135∘−45∘=90∘\angle ABE = 135^\circ - 45^\circ = 90^\circ
Quadrilateral BCDEBCDE is an isosceles trapezoid with BE∥CDBE \parallel CD.
4
Calculate the dimensions of right triangle △ABP\triangle ABP and segment APAP
△ABP\triangle ABP is a right triangle at BB with ∠BAP=45∘\angle BAP = 45^\circ, AB=2AB = 2, BP=2BP = 2, and hypotenuse AP=22AP = 2\sqrt{2}
Since AP∥BCAP \parallel BC, alternate interior angle relationships yield ∠BAP=45∘\angle BAP = 45^\circ.
5
Calculate the area of quadrilateral ABCPABCP
Area(ABCP)=BC+AP2×h=2+222×2=(1+2)2=2+2\text{Area}(ABCP) = \frac{BC + AP}{2} \times h = \frac{2 + 2\sqrt{2}}{2} \times \sqrt{2} = (1 + \sqrt{2})\sqrt{2} = 2 + \sqrt{2}
Quadrilateral ABCPABCP is a trapezoid with parallel bases BC=2BC = 2 and AP=22AP = 2\sqrt{2} and height 2\sqrt{2}.

Key Concept

Properties of regular polygons, isosceles trapezoids, diagonal angles, and area decomposition.
Estimated Time:2m 30s
Rate this question

Topics

Quadrilaterals and Polygons

More questions from this topic