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Zorluk: ZorPythagorean Theorem and Special Right Triangles

In right triangle ABCABC, the measure of ∠B\angle B is 90∘90^\circ, the measure of ∠A\angle A is 30∘30^\circ, and hypotenuse AC=12AC = 12. Point DD lies on hypotenuse ACAC such that BDBD is perpendicular to ACAC, and point EE lies on leg BCBC such that DEDE is perpendicular to BCBC. What is the area of quadrilateral ABDEABDE?

  1. 13538\frac{135\sqrt{3}}{8}Cevap
  2. B
    11738\frac{117\sqrt{3}}{8}
  3. C
    2732\frac{27\sqrt{3}}{2}
  4. D
    6334\frac{63\sqrt{3}}{4}
  5. E
    1443−98\frac{144\sqrt{3} - 9}{8}

Cevap

The area of quadrilateral ABDEABDE is 13538\frac{135\sqrt{3}}{8}.
The area of quadrilateral ABDEABDE is obtained by subtracting the area of the smaller right triangle △DEC\triangle DEC from the area of the outer right triangle △ABC\triangle ABC. Since both triangles share the 60∘60^\circ angle at vertex CC, repeated application of the 30∘−60∘−90∘30^\circ-60^\circ-90^\circ ratio (1:3:21:\sqrt{3}:2) gives Area(△ABC)=183\text{Area}(\triangle ABC) = 18\sqrt{3} and Area(△DEC)=938\text{Area}(\triangle DEC) = \frac{9\sqrt{3}}{8}. Subtraction yields 13538\frac{135\sqrt{3}}{8}.

Adım Adım Çözüm

1
Calculate side lengths and area of △ABC\triangle ABC.
BC=6BC = 6, AB=63AB = 6\sqrt{3}, and Area(△ABC)=183\text{Area}(\triangle ABC) = 18\sqrt{3}.
In 30∘−60∘−90∘30^\circ-60^\circ-90^\circ triangle ABCABC with hypotenuse AC=12AC = 12, the side opposite 30∘30^\circ (BCBC) is half the hypotenuse (BC=6BC = 6), and the side opposite 60∘60^\circ (ABAB) is 636\sqrt{3}. The area is 12×6×63=183\frac{1}{2} \times 6 \times 6\sqrt{3} = 18\sqrt{3}.
2
Find the hypotenuse DCDC of △BDC\triangle BDC.
DC=3DC = 3.
Since BD⊥ACBD \perp AC, △BDC\triangle BDC is a right triangle with ∠BDC=90∘\angle BDC = 90^\circ and ∠C=60∘\angle C = 60^\circ, making ∠DBC=30∘\angle DBC = 30^\circ. Its hypotenuse is BC=6BC = 6. The side opposite 30∘30^\circ is DC=12BC=3DC = \frac{1}{2} BC = 3.
3
Determine the sides and area of △DEC\triangle DEC.
EC=32EC = \frac{3}{2}, DE=332DE = \frac{3\sqrt{3}}{2}, and Area(△DEC)=938\text{Area}(\triangle DEC) = \frac{9\sqrt{3}}{8}.
Since DE⊥BCDE \perp BC, △DEC\triangle DEC is another 30∘−60∘−90∘30^\circ-60^\circ-90^\circ right triangle with hypotenuse DC=3DC = 3. Thus EC=32EC = \frac{3}{2} and DE=332DE = \frac{3\sqrt{3}}{2}. Its area is 12×32×332=938\frac{1}{2} \times \frac{3}{2} \times \frac{3\sqrt{3}}{2} = \frac{9\sqrt{3}}{8}.
4
Subtract Area(△DEC)\text{Area}(\triangle DEC) from Area(△ABC)\text{Area}(\triangle ABC) to get the area of quadrilateral ABDEABDE.
Area(ABDE)=183−938=13538\text{Area}(ABDE) = 18\sqrt{3} - \frac{9\sqrt{3}}{8} = \frac{135\sqrt{3}}{8}.
Quadrilateral ABDEABDE is formed by removing △DEC\triangle DEC from △ABC\triangle ABC.

Anahtar Kavram

Iterative application of 30∘−60∘−90∘30^\circ-60^\circ-90^\circ special right triangle side ratios (1:3:21:\sqrt{3}:2) along dropped altitudes.
Tahmini Süre:2m 30s
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Pythagorean Theorem and Special Right Triangles

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