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Zorluk: OrtaTriangles: Properties, Perimeter, and Area

In triangle PQRPQR, the measure of angle PQRPQR is 90∘90^\circ. Point SS lies on segment PRPR such that line segment QSQS is perpendicular to PRPR. If PS=4PS = 4 and SR=9SR = 9, what is the area of triangle PQRPQR?

  1. A
    1818
  2. 3939Cevap
  3. C
    7878
  4. D
    117117
  5. E
    234234

Cevap

The area of triangle PQRPQR is 39.
By the geometric mean theorem (right triangle altitude theorem), the altitude QSQS satisfies QS2=PS⋅SR=4⋅9=36QS^2 = PS \cdot SR = 4 \cdot 9 = 36, so QS=6QS = 6. The hypotenuse PR=PS+SR=4+9=13PR = PS + SR = 4 + 9 = 13. Substituting base 1313 and height 66 into the triangle area formula 12bh\frac{1}{2}bh gives 12×13×6=39\frac{1}{2} \times 13 \times 6 = 39.

Adım Adım Çözüm

1
Find the length of hypotenuse PRPR
PR=PS+SR=4+9=13PR = PS + SR = 4 + 9 = 13
Point SS lies on segment PRPR, so the total length of the hypotenuse is the sum of its two segments.
2
Calculate altitude QSQS using the geometric mean theorem for right triangles
QS=PS×SR=4×9=36=6QS = \sqrt{PS \times SR} = \sqrt{4 \times 9} = \sqrt{36} = 6
In a right triangle, the altitude to the hypotenuse is the geometric mean of the two segments into which the hypotenuse is divided.
3
Calculate the area of triangle PQRPQR
Area=12×base×height=12×13×6=39\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 13 \times 6 = 39
The area of any triangle is half the product of its base and corresponding height.

Anahtar Kavram

Altitude to the hypotenuse in right triangles and triangle area calculation

Alternatif Yöntem

Alternatively, use similar triangles △PQS∼△QRS\triangle PQS \sim \triangle QRS. The ratio of corresponding sides gives PQPS=PRPQ  ⟹  PQ2=PS⋅PR=4⋅13=52\frac{PQ}{PS} = \frac{PR}{PQ} \implies PQ^2 = PS \cdot PR = 4 \cdot 13 = 52, and QR2=SR⋅PR=9⋅13=117QR^2 = SR \cdot PR = 9 \cdot 13 = 117. Since △PQR\triangle PQR is a right triangle at QQ, its area is 12⋅PQ⋅QR=1252⋅117=126084=12⋅78=39\frac{1}{2} \cdot PQ \cdot QR = \frac{1}{2} \sqrt{52 \cdot 117} = \frac{1}{2} \sqrt{6084} = \frac{1}{2} \cdot 78 = 39.
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Triangles: Properties, Perimeter, and Area

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