An isosceles triangle has two sides of length 10 units each and a third side of integer le… — GRE General Test

An isosceles triangle has two sides of length $10$ units each and a third side of integer length $x$ units. If the area of the triangle is strictly greater than $24$ square units and less than or equal to $48$ square units, which of the following could be the value of $x$ ? Select all such values.

A
4
6Cevap
C
14
16Cevap
E
20

Cevap

The values 6 and 16 are the valid side lengths.

The values 6 and 16 produce valid areas of approximately 28.62 and exactly 48 square units respectively, both of which satisfy the given condition that the area must be strictly greater than 24 and less than or equal to 48.

Adım Adım Çözüm

Express the height and area of the isosceles triangle in terms of

x

Height

h = \sqrt{10^2 - (x/2)^2} = \sqrt{100 - \frac{x^2}{4}}

, so Area

A = \frac{1}{2} x \sqrt{100 - \frac{x^2}{4}} = \frac{1}{4} x \sqrt{400 - x^2}

The altitude to the base of an isosceles triangle bisects the base into two equal segments of length

\frac{x}{2}

Set up the inequality for the area constraints

24 < A \le 48

24 < \frac{1}{4} x \sqrt{400 - x^2} \le 48 \implies 96 < x \sqrt{400 - x^2} \le 192

Multiplying all parts by

4

isolates the radical expression.

Square all terms to analyze the function

f(x^2) = x^2(400 - x^2)

9216 < x^2(400 - x^2) \le 36864

Squaring positive quantities preserves the inequality direction.

Evaluate the area function for each given option choice.

For

x=4

A \approx 19.6

(too small). For

x=6

A \approx 28.6

(valid). For

x=14

A \approx 50.0

(too large). For

x=16

A = 48

(valid). For

x=20

: degenerate triangle with

A = 0

(invalid).

Direct evaluation identifies which integer choices satisfy

24 < A \le 48

Anahtar Kavram

Properties of isosceles triangles, Pythagorean theorem for altitude, area bounds, and triangle inequality.

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

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