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Zorluk: OrtaPythagorean Theorem and Special Right Triangles

In triangle ABCABC, the measure of angle AA is 45∘45^\circ and the measure of angle BB is 105∘105^\circ. The length of side ABAB is 66 units. Which of the following statements must be true? Select all that apply.

  1. The length of the altitude from vertex BB to side ACAC is 323\sqrt{2} units.Cevap
  2. The length of side BCBC is 626\sqrt{2} units.Cevap
  3. The area of triangle ABCABC is 9+939 + 9\sqrt{3} square units.Cevap
  4. D
    The length of side ACAC is 62+666\sqrt{2} + 6\sqrt{6} units.
  5. E
    The area of triangle ABCABC is 18+18318 + 18\sqrt{3} square units.

Cevap

The correct statements are: the length of the altitude from vertex BB to side ACAC is 323\sqrt{2} units, the length of side BCBC is 626\sqrt{2} units, and the area of triangle ABCABC is 9+939 + 9\sqrt{3} square units.
By drawing altitude BDBD perpendicular to ACAC, triangle ABCABC decomposes into two special right triangles. In the 45∘−45∘−90∘45^\circ-45^\circ-90^\circ triangle ABDABD, the hypotenuse is 66, yielding leg lengths BD=AD=32BD = AD = 3\sqrt{2}. In the 30∘−60∘−90∘30^\circ-60^\circ-90^\circ triangle BCDBCD, side BD=32BD = 3\sqrt{2} is opposite the 30∘30^\circ angle, making hypotenuse BC=2×32=62BC = 2 \times 3\sqrt{2} = 6\sqrt{2} and long leg CD=36CD = 3\sqrt{6}. Combining ADAD and CDCD gives base AC=32+36AC = 3\sqrt{2} + 3\sqrt{6}, leading to an area of 12(32+36)(32)=9+93\frac{1}{2}(3\sqrt{2} + 3\sqrt{6})(3\sqrt{2}) = 9 + 9\sqrt{3}.

Adım Adım Çözüm

1
Determine the third angle of triangle ABCABC
∠C=180∘−(45∘+105∘)=30∘\angle C = 180^\circ - (45^\circ + 105^\circ) = 30^\circ.
The sum of interior angles in any triangle is 180∘180^\circ.
2
Drop an altitude BDBD perpendicular to side ACAC
Altitude BDBD splits triangle ABCABC into two right triangles: △ABD\triangle ABD (45∘−45∘−90∘45^\circ-45^\circ-90^\circ) and △BCD\triangle BCD (30∘−60∘−90∘30^\circ-60^\circ-90^\circ).
In △ABD\triangle ABD, ∠A=45∘\angle A = 45^\circ and ∠ADB=90∘\angle ADB = 90^\circ, leaving ∠ABD=45∘\angle ABD = 45^\circ. In △BCD\triangle BCD, ∠CBD=105∘−45∘=60∘\angle CBD = 105^\circ - 45^\circ = 60^\circ and ∠C=30∘\angle C = 30^\circ.
3
Calculate side lengths in 45∘−45∘−90∘45^\circ-45^\circ-90^\circ triangle ABDABD
AD=BD=AB2=62=32AD = BD = \frac{AB}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2} units.
The ratio of sides in a 45∘−45∘−90∘45^\circ-45^\circ-90^\circ triangle is 1:1:21:1:\sqrt{2}.
4
Calculate side lengths in 30∘−60∘−90∘30^\circ-60^\circ-90^\circ triangle BCDBCD
Hypotenuse BC=2×BD=62BC = 2 \times BD = 6\sqrt{2} units, and long leg CD=BD×3=32×3=36CD = BD \times \sqrt{3} = 3\sqrt{2} \times \sqrt{3} = 3\sqrt{6} units.
The ratio of sides opposite 30∘:60∘:90∘30^\circ:60^\circ:90^\circ is 1:3:21:\sqrt{3}:2.
5
Calculate total base ACAC and area of triangle ABCABC
AC=AD+CD=32+36AC = AD + CD = 3\sqrt{2} + 3\sqrt{6} units. Area = 12×base×height=12(32+36)(32)=12(18+183)=9+93\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (3\sqrt{2} + 3\sqrt{6})(3\sqrt{2}) = \frac{1}{2} (18 + 18\sqrt{3}) = 9 + 9\sqrt{3} square units.
Standard formula for triangle area is 12bh\frac{1}{2} b h.

Anahtar Kavram

Decomposing an oblique triangle with 45∘45^\circ and 30∘30^\circ angles into 45∘−45∘−90∘45^\circ-45^\circ-90^\circ and 30∘−60∘−90∘30^\circ-60^\circ-90^\circ special right triangles.
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Pythagorean Theorem and Special Right Triangles

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