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Difficulty: MediumQuadrilaterals and Polygons

In convex quadrilateral ABCDABCD, ∠ABC=90∘\angle ABC = 90^\circ and ∠ADC=90∘\angle ADC = 90^\circ. If AB=BCAB = BC, AD=6AD = 6, and CD=8CD = 8, what is the area of quadrilateral ABCDABCD?

  1. A
    48
  2. 49Answer
  3. C
    55
  4. D
    73
  5. E
    74

Answer

49
The correct answer is 49. Dividing quadrilateral ABCDABCD along diagonal ACAC creates two right triangles: △ADC\triangle ADC with legs 6 and 8, and △ABC\triangle ABC with hypotenuse ACAC and equal legs. Using the Pythagorean theorem on △ADC\triangle ADC gives hypotenuse AC=62+82=10AC = \sqrt{6^2 + 8^2} = 10, and its area is 12(6)(8)=24\frac{1}{2}(6)(8) = 24. For isosceles right triangle △ABC\triangle ABC, AB2+BC2=102  ⟹  2(AB2)=100  ⟹  AB2=50AB^2 + BC^2 = 10^2 \implies 2(AB^2) = 100 \implies AB^2 = 50, so its area is 12(50)=25\frac{1}{2}(50) = 25. Adding both triangle areas yields 24+25=4924 + 25 = 49.

Step-by-Step Solution

1
Divide the quadrilateral into two right triangles using diagonal ACAC.
Quadrilateral ABCDABCD is split into △ADC\triangle ADC and △ABC\triangle ABC, both of which are right-angled triangles sharing hypotenuse ACAC.
Diagonal ACAC connects the vertices opposite the 90∘90^\circ angles.
2
Calculate the length of diagonal ACAC using △ADC\triangle ADC.
AC2=AD2+CD2=62+82=36+64=100  ⟹  AC=10AC^2 = AD^2 + CD^2 = 6^2 + 8^2 = 36 + 64 = 100 \implies AC = 10.
△ADC\triangle ADC is a right triangle with legs of length 6 and 8.
3
Find the area of △ADC\triangle ADC.
Area(△ADC)=12×AD×CD=12×6×8=24\text{Area}(\triangle ADC) = \frac{1}{2} \times AD \times CD = \frac{1}{2} \times 6 \times 8 = 24.
The area of a right triangle is half the product of its legs.
4
Determine the area of isosceles right triangle △ABC\triangle ABC.
Let AB=BC=sAB = BC = s. Since s2+s2=AC2s^2 + s^2 = AC^2, we have 2s2=100  ⟹  s2=502s^2 = 100 \implies s^2 = 50. Thus, Area(△ABC)=12s2=12×50=25\text{Area}(\triangle ABC) = \frac{1}{2} s^2 = \frac{1}{2} \times 50 = 25.
△ABC\triangle ABC is a right triangle with equal legs ss and hypotenuse AC=10AC = 10.
5
Sum the areas of the two triangles to get the total area.
Total Area=24+25=49\text{Total Area} = 24 + 25 = 49.
The total area of the quadrilateral is the sum of the areas of its non-overlapping constituent triangles.

Key Concept

Polygon area decomposition using diagonal partitioning and the Pythagorean theorem
Estimated Time:1m 30s
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Topics

Quadrilaterals and Polygons

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