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Zorluk: ZorTriangles: Properties, Perimeter, and Area

In △ABC\triangle ABC, the lengths of sides ABAB, BCBC, and ACAC are 1313, 1414, and 1515, respectively. A line segment DEDE is drawn parallel to side BCBC, with point DD lying on side ABAB and point EE lying on side ACAC. If the perimeter of △ADE\triangle ADE is equal to the perimeter of quadrilateral DBCEDBCE, what is the area of △ADE\triangle ADE?

Cevap: 47.25

Cevap

47.25
The area of the original triangle △ABC\triangle ABC is computed as 8484 using Heron's formula. By defining the linear scale factor kk between △ADE\triangle ADE and △ABC\triangle ABC, the perimeters of △ADE\triangle ADE and quadrilateral DBCEDBCE are expressed as 42k42k and 42−14k42 - 14k, respectively. Setting these equal yields k=0.75k = 0.75. The area of △ADE\triangle ADE is then k2×84=0.5625×84=47.25k^2 \times 84 = 0.5625 \times 84 = 47.25.

Adım Adım Çözüm

1
Calculate the perimeter and area of the main triangle △ABC\triangle ABC.
The perimeter of △ABC\triangle ABC is 13+14+15=4213 + 14 + 15 = 42. Using Heron's formula with semi-perimeter s=21s = 21, Area(△ABC)=21(21−13)(21−14)(21−15)=21×8×7×6=84\text{Area}(\triangle ABC) = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = 84.
Finding the area and perimeter of the full triangle sets the required baseline for proportional scaling.
2
Set up expressions for the perimeters of △ADE\triangle ADE and quadrilateral DBCEDBCE using a scale factor kk.
Because DE∥BCDE \parallel BC, △ADE∼△ABC\triangle ADE \sim \triangle ABC with scale factor k=ADAB=AEAC=DEBCk = \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}. Thus, Perimeter(△ADE)=13k+15k+14k=42k\text{Perimeter}(\triangle ADE) = 13k + 15k + 14k = 42k. The segments DB=13(1−k)DB = 13(1-k) and EC=15(1−k)EC = 15(1-k), so Perimeter(DBCE)=13(1−k)+14+15(1−k)+14k=42−14k\text{Perimeter}(DBCE) = 13(1-k) + 14 + 15(1-k) + 14k = 42 - 14k.
Parallel lines create similar triangles, which allows all perimeter segment lengths to be represented in terms of one variable kk.
3
Solve for the scale factor kk by equating the two perimeters.
42k=42−14k  ⟹  56k=42  ⟹  k=4256=34=0.7542k = 42 - 14k \implies 56k = 42 \implies k = \frac{42}{56} = \frac{3}{4} = 0.75.
Equating the perimeters satisfies the condition specified in the question stem.
4
Calculate the area of △ADE\triangle ADE using the square of the linear scale factor.
Area(△ADE)=k2×Area(△ABC)=(34)2×84=916×84=1894=47.25\text{Area}(\triangle ADE) = k^2 \times \text{Area}(\triangle ABC) = \left(\frac{3}{4}\right)^2 \times 84 = \frac{9}{16} \times 84 = \frac{189}{4} = 47.25.
The area ratio of similar geometric figures is proportional to the square of their linear scale factor.

Anahtar Kavram

Properties of Similar Triangles, Area via Heron's Formula, and Perimeter Scaling
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