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Zorluk: OrtaPythagorean Theorem and Special Right Triangles

In quadrilateral ABCDABCD, diagonal ACAC divides the figure into two right triangles, △ABC\triangle ABC and △ACD\triangle ACD. It is given that ∠ABC=90∘\angle ABC = 90^\circ, ∠ACD=90∘\angle ACD = 90^\circ, ∠BAC=30∘\angle BAC = 30^\circ, and ∠CAD=45∘\angle CAD = 45^\circ. If AB=63AB = 6\sqrt{3}, what is the perimeter of quadrilateral ABCDABCD?

  1. A
    18+63+6218 + 6\sqrt{3} + 6\sqrt{2}
  2. 18+63+12218 + 6\sqrt{3} + 12\sqrt{2}Cevap
  3. C
    24+63+12224 + 6\sqrt{3} + 12\sqrt{2}
  4. D
    30+63+12230 + 6\sqrt{3} + 12\sqrt{2}
  5. E
    18+183+12618 + 18\sqrt{3} + 12\sqrt{6}

Cevap

18+63+12218 + 6\sqrt{3} + 12\sqrt{2}
The correct answer is derived by sequentially calculating the side lengths of the two special right triangles that share segment ACAC. In △ABC\triangle ABC, the given side AB=63AB = 6\sqrt{3} is adjacent to the 30∘30^\circ angle, making it the side opposite 60∘60^\circ. Using the 1:3:21 : \sqrt{3} : 2 ratio yields BC=6BC = 6 and hypotenuse AC=12AC = 12. In △ACD\triangle ACD, leg AC=12AC = 12 is adjacent to a 45∘45^\circ angle, making △ACD\triangle ACD a 45∘−45∘−90∘45^\circ-45^\circ-90^\circ triangle with equal leg CD=12CD = 12 and hypotenuse AD=122AD = 12\sqrt{2}. Summing the outer edges AB+BC+CD+AD=63+6+12+122=18+63+122AB + BC + CD + AD = 6\sqrt{3} + 6 + 12 + 12\sqrt{2} = 18 + 6\sqrt{3} + 12\sqrt{2}.

Adım Adım Çözüm

1
Analyze right triangle △ABC\triangle ABC using 30∘−60∘−90∘30^\circ-60^\circ-90^\circ special right triangle ratios.
BC=6BC = 6 and AC=12AC = 12
In a 30∘−60∘−90∘30^\circ-60^\circ-90^\circ triangle, the ratio of sides opposite to 30∘:60∘:90∘30^\circ : 60^\circ : 90^\circ is 1:3:21 : \sqrt{3} : 2. Since AB=63AB = 6\sqrt{3} is opposite 60∘60^\circ, the shorter leg BC=633=6BC = \frac{6\sqrt{3}}{\sqrt{3}} = 6. The hypotenuse AC=2×BC=12AC = 2 \times BC = 12.
2
Analyze right triangle △ACD\triangle ACD using 45∘−45∘−90∘45^\circ-45^\circ-90^\circ special right triangle ratios.
CD=12CD = 12 and AD=122AD = 12\sqrt{2}
In right triangle △ACD\triangle ACD with ∠ACD=90∘\angle ACD = 90^\circ and ∠CAD=45∘\angle CAD = 45^\circ, △ACD\triangle ACD is an isosceles right triangle with side ratio 1:1:21 : 1 : \sqrt{2}. Since leg AC=12AC = 12, leg CD=12CD = 12 and hypotenuse AD=122AD = 12\sqrt{2}.
3
Sum the lengths of the four outer boundary segments to compute the perimeter of quadrilateral ABCDABCD.
Perimeter =18+63+122= 18 + 6\sqrt{3} + 12\sqrt{2}
Perimeter =AB+BC+CD+AD=63+6+12+122=18+63+122= AB + BC + CD + AD = 6\sqrt{3} + 6 + 12 + 12\sqrt{2} = 18 + 6\sqrt{3} + 12\sqrt{2}.

Anahtar Kavram

Special Right Triangles (30∘−60∘−90∘30^\circ-60^\circ-90^\circ and 45∘−45∘−90∘45^\circ-45^\circ-90^\circ side ratios)
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Pythagorean Theorem and Special Right Triangles

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