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Zorluk: OrtaTriangles: Properties, Perimeter, and Area

In △ABC\triangle ABC, the length of side ABAB is 1313 units, the length of side BCBC is 2121 units, and the area of △ABC\triangle ABC is 126126 square units. If ∠ABC\angle ABC is an acute angle, what is the length of side ACAC?

  1. A
    15
  2. B
    18
  3. 20Cevap
  4. D
    22
  5. E
    25

Cevap

20
The area of triangle ABCABC is 12×21×h=126\frac{1}{2} \times 21 \times h = 126, which yields an altitude AH=12AH = 12 perpendicular to side BCBC. In right triangle ABHABH, the base segment BH=132−122=5BH = \sqrt{13^2 - 12^2} = 5. Because angle ABCABC is acute, HH falls between BB and CC, making HC=21−5=16HC = 21 - 5 = 16. Finally, in right triangle AHCAHC, AC=122+162=400=20AC = \sqrt{12^2 + 16^2} = \sqrt{400} = 20. Therefore, 20 is the correct answer.

Adım Adım Çözüm

1
Calculate the altitude hh from vertex AA to base BCBC.
h=12h = 12 units.
The area formula for a triangle is Area=12×base×height\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}. Substituting BC=21BC = 21 and Area=126\text{Area} = 126 yields 126=12×21×h  ⟹  h=12126 = \frac{1}{2} \times 21 \times h \implies h = 12.
2
Find the length of segment BHBH where HH is the foot of the altitude on BCBC.
BH=5BH = 5 units.
In right triangle △ABH\triangle ABH, AB=13AB = 13 and AH=12AH = 12. By the Pythagorean theorem, BH=132−122=169−144=25=5BH = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5.
3
Determine the length of segment HCHC.
HC=16HC = 16 units.
Since ∠ABC\angle ABC is an acute angle, point HH lies on segment BCBC. Therefore, HC=BC−BH=21−5=16HC = BC - BH = 21 - 5 = 16.
4
Calculate the length of side ACAC.
AC=20AC = 20 units.
In right triangle △AHC\triangle AHC, AH=12AH = 12 and HC=16HC = 16. By the Pythagorean theorem, AC=122+162=144+256=400=20AC = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20.

Anahtar Kavram

Triangles: Altitude, Area, and Pythagorean Theorem
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