In the coordinate plane, triangle JKL has vertices J(0,0), K(14,0), and L(x, 12), where x … — GRE General Test

In the coordinate plane, triangle $JKL$ has vertices $J(0,0)$ , $K(14,0)$ , and $L(x, 12)$ , where $x > 0$ . If the perimeter of triangle $JKL$ is $42$ units and the length of side $JL$ is less than the length of side $KL$ , what is the value of $x$ ?

A
4
5Cevap
C
7
D
9
E
12

Cevap

The correct answer is 5. Using the distance formula, the base length

JK = 14

. Expressing the side lengths as

JL = \sqrt{x^2 + 144}

and

KL = \sqrt{(14-x)^2 + 144}

, setting the perimeter

JK + JL + KL = 42

leads to the quadratic equation

x^2 - 14x + 45 = 0

. This yields

x = 5

x = 9

. Evaluating the sides for

x = 5

gives

JL = 13

and

KL = 15

, which satisfies the problem condition

JL < KL

Adım Adım Çözüm

Calculate the length of base

JK

using the distance formula.

The distance between

J(0,0)

and

K(14,0)

14 - 0 = 14

units.

Base

JK

lies along the horizontal

x

-axis.

Express side lengths

JL

and

KL

in terms of

x

JL = \sqrt{(x-0)^2 + (12-0)^2} = \sqrt{x^2 + 144}

and

KL = \sqrt{(14-x)^2 + (12-0)^2} = \sqrt{(14-x)^2 + 144}

Apply the distance formula between coordinates

L(x,12)

and vertices

J

and

K

Set up and solve the perimeter equation.

14 + \sqrt{x^2 + 144} + \sqrt{(14-x)^2 + 144} = 42 \implies \sqrt{x^2 + 144} + \sqrt{(14-x)^2 + 144} = 28

. Squaring both sides systematically yields

x^2 - 14x + 45 = 0

, giving roots

x = 5

and

x = 9

The total perimeter is given as 42 units.

Apply the constraint

JL < KL

to choose the valid root.

For

x = 5

JL = \sqrt{25 + 144} = 13

and

KL = \sqrt{81 + 144} = 15

, satisfying

JL < KL

For

x = 9

JL = 15

and

KL = 13

, which violates

JL < KL

Anahtar Kavram

Coordinate Geometry and Triangle Side Length Constraints

Tahmini Süre:1m 30s

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