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Zorluk: ZorPythagorean Theorem and Special Right Triangles

In the right rectangular solid ABCDEFGHABCDEFGH, the base ABCDABCD is a rectangle with edge lengths AB=6AB = 6 and BC=63BC = 6\sqrt{3}. The vertical edge CG=12CG = 12. Point MM is the midpoint of edge CGCG. What is the perimeter of triangle BDMBDM?

  1. 24+6224 + 6\sqrt{2}Cevap
  2. B
    12+12212 + 12\sqrt{2}
  3. C
    24+6324 + 6\sqrt{3}
  4. D
    18+62+6318 + 6\sqrt{2} + 6\sqrt{3}
  5. E
    12+62+6512 + 6\sqrt{2} + 6\sqrt{5}

Cevap

The perimeter of triangle BDMBDM is 24+6224 + 6\sqrt{2}.
The correct answer is obtained by recognizing three right triangles within the 3D figure: △BCD\triangle BCD has legs 66 and 636\sqrt{3} giving BD=12BD = 12; △BCM\triangle BCM has legs 636\sqrt{3} and 66 giving BM=12BM = 12; and △DCM\triangle DCM has legs 66 and 66 giving DM=62DM = 6\sqrt{2}. Summing the three sides yields a perimeter of 24+6224 + 6\sqrt{2}.

Adım Adım Çözüm

1
Calculate the length of base diagonal BDBD using the right triangle △BCD\triangle BCD.
BD=12BD = 12
In right triangle △BCD\triangle BCD, legs are CD=AB=6CD = AB = 6 and BC=63BC = 6\sqrt{3}. Using the 30∘–60∘–90∘30^\circ\text{--}60^\circ\text{--}90^\circ ratio (1:3:2)(1 : \sqrt{3} : 2), hypotenuse BD=2×6=12BD = 2 \times 6 = 12 (or via Pythagorean theorem: BD=62+(63)2=36+108=144=12BD = \sqrt{6^2 + (6\sqrt{3})^2} = \sqrt{36 + 108} = \sqrt{144} = 12).
2
Calculate the length of segment BMBM using the right triangle △BCM\triangle BCM.
BM=12BM = 12
Since MM is the midpoint of vertical edge CG=12CG = 12, CM=6CM = 6. Vertical edge CGCG is perpendicular to base ABCDABCD, so △BCM\triangle BCM is a right triangle at CC. The legs are CM=6CM = 6 and BC=63BC = 6\sqrt{3}. Applying the 30∘–60∘–90∘30^\circ\text{--}60^\circ\text{--}90^\circ ratio (1:3:2)(1 : \sqrt{3} : 2), hypotenuse BM=2×6=12BM = 2 \times 6 = 12.
3
Calculate the length of segment DMDM using the right triangle △DCM\triangle DCM.
DM=62DM = 6\sqrt{2}
In right triangle △DCM\triangle DCM, legs are CD=6CD = 6 and CM=6CM = 6. Since the legs are equal, △DCM\triangle DCM is a 45∘–45∘–90∘45^\circ\text{--}45^\circ\text{--}90^\circ isosceles right triangle with side ratio 1:1:21 : 1 : \sqrt{2}. Thus, hypotenuse DM=62DM = 6\sqrt{2}.
4
Sum the three side lengths to find the perimeter of △BDM\triangle BDM.
Perimeter = 12+12+62=24+6212 + 12 + 6\sqrt{2} = 24 + 6\sqrt{2}
The perimeter of △BDM\triangle BDM is BD+BM+DMBD + BM + DM.

Anahtar Kavram

Applying special right triangle ratios (30∘–60∘–90∘30^\circ\text{--}60^\circ\text{--}90^\circ and 45∘–45∘–90∘45^\circ\text{--}45^\circ\text{--}90^\circ) to 3D rectangular solids.
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Pythagorean Theorem and Special Right Triangles

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