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Zorluk: ZorQuadrilaterals and Polygons

In convex quadrilateral ABCDABCD, the diagonals ACAC and BDBD intersect at point PP at right angles (ACBDAC \perp BD). The length of diagonal ACAC is 1616 and the length of diagonal BDBD is 1212. Points EE, FF, GG, and HH are the midpoints of sides ABAB, BCBC, CDCD, and DADA, respectively. Which of the following statements MUST be true? Select all such statements.

  1. The perimeter of quadrilateral EFGHEFGH is 2828.Cevap
  2. Quadrilateral EFGHEFGH is a rectangle.Cevap
  3. The area of quadrilateral ABCDABCD is 9696.Cevap
  4. D
    The area of quadrilateral EFGHEFGH is 9696.
  5. E
    Quadrilateral EFGHEFGH is a rhombus.

Cevap

The true statements are that the perimeter of quadrilateral EFGHEFGH is 2828, quadrilateral EFGHEFGH is a rectangle, and the area of quadrilateral ABCDABCD is 9696.
Applying the Midpoint Theorem shows that midsegments EFEF and GHGH are parallel to ACAC with length 88, while FGFG and HEHE are parallel to BDBD with length 66. This gives a perimeter of 8+6+8+6=288 + 6 + 8 + 6 = 28. Because ACBDAC \perp BD, the adjacent midsegments meet at 9090^\circ, confirming that quadrilateral EFGHEFGH is a rectangle. Additionally, for any orthodiagonal quadrilateral, the area is 12d1d2=12×16×12=96\frac{1}{2} d_1 d_2 = \frac{1}{2} \times 16 \times 12 = 96.

Adım Adım Çözüm

1
Determine the side lengths of midpoint quadrilateral EFGHEFGH using the Triangle Midpoint Theorem.
EF=GH=12AC=8EF = GH = \frac{1}{2}AC = 8 and FG=HE=12BD=6FG = HE = \frac{1}{2}BD = 6.
In any triangle, the segment connecting the midpoints of two sides is parallel to the third side and half its length.
2
Calculate the perimeter of quadrilateral EFGHEFGH.
Perimeter = EF+FG+GH+HE=8+6+8+6=28EF + FG + GH + HE = 8 + 6 + 8 + 6 = 28.
The perimeter is the sum of all four side lengths of the quadrilateral.
3
Determine the shape classification of quadrilateral EFGHEFGH.
EFGHEFGH is a rectangle.
Since EFACEF \parallel AC and FGBDFG \parallel BD, the angle between EFEF and FGFG equals the angle between diagonals ACAC and BDBD. Given ACBDAC \perp BD, the angle EFG=90\angle EFG = 90^\circ. Opposite sides are equal and parallel with right angles, making EFGHEFGH a rectangle (and not a rhombus, as adjacent sides 868 \neq 6).
4
Calculate the area of quadrilateral ABCDABCD and quadrilateral EFGHEFGH.
Area(ABCDABCD) = 12×AC×BD=12×16×12=96\frac{1}{2} \times AC \times BD = \frac{1}{2} \times 16 \times 12 = 96; Area(EFGHEFGH) = 8×6=488 \times 6 = 48.
The area of a quadrilateral with perpendicular diagonals is half the product of its diagonal lengths. The midpoint quadrilateral has half the area of the outer quadrilateral.

Anahtar Kavram

Midpoint Theorem (Varignon's Theorem) and Area of Orthodiagonal Quadrilaterals
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