Calculating if two acute / obtuse triangles are similar - GMAT Quantitative

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The triangles are similar. What is the value of x?

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The proportions of corresponding sides of similar triangles must be equal. Therefore, $\frac{8}{12}$ = $\frac{10}{x}$. 8x = 120; x = 15.

$\Delta ABC$ is an equilateral triangle. Points are the midpoints of $\overline{AB}$,$\overline{BC}$,$\overline{AC}$ , respectively. $\Delta DEF$ is constructed.

All of the following are true except:

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The three sides of $\Delta DEF$ are the midsegments of $\Delta A BC$ , so $\Delta DEF$ is similar to $\Delta A BC$ .

By the Triangle Midsegment Theorem, each is parallel to one side of $\Delta A BC$ .

By the same theorem, each has length exactly half of that side, giving $\Delta A BC$ twice the perimeter of $\Delta DEF$ .

But since the sides of $\Delta A BC$ have twice the length of those of $\Delta DEF$ , the area of $\Delta A BC$ , which varies directly as the square of a sidelength, must be four times that of $\Delta DEF$ .

The correct choice is the one that asserts that the area of $\Delta A BC$ is twice that of $\Delta DEF$ .

$\Delta ABC \sim \Delta D EF$

.

Order the angles of $\Delta D EF$ from least to greatest measure.

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In a triangle, the angle of greatest measure is opposite the side of greatest measure, and the angle of least measure is opposite the side of least measure. , so their opposite angles are ranked in this order - that is, $m \angle C < m \angle A < m \angle B$ .

Corresponding angles of similar triangles are congruent, so, since $\Delta ABC \sim \Delta D EF$ , $m \angle A= m \angle D , m \angle B= m \angle E, m \angle C = m \angle F$ .

Therefore, by substitution, $m \angle F < m \angle D < m \angle E$ .

The triangles are similar. What is the value of x?

Tap to see back →

The proportions of corresponding sides of similar triangles must be equal. Therefore, $\frac{8}{12}$ = $\frac{10}{x}$. 8x = 120; x = 15.

$\Delta ABC$ is an equilateral triangle. Points are the midpoints of $\overline{AB}$,$\overline{BC}$,$\overline{AC}$ , respectively. $\Delta DEF$ is constructed.

All of the following are true except:

Tap to see back →

The three sides of $\Delta DEF$ are the midsegments of $\Delta A BC$ , so $\Delta DEF$ is similar to $\Delta A BC$ .

By the Triangle Midsegment Theorem, each is parallel to one side of $\Delta A BC$ .

By the same theorem, each has length exactly half of that side, giving $\Delta A BC$ twice the perimeter of $\Delta DEF$ .

But since the sides of $\Delta A BC$ have twice the length of those of $\Delta DEF$ , the area of $\Delta A BC$ , which varies directly as the square of a sidelength, must be four times that of $\Delta DEF$ .

The correct choice is the one that asserts that the area of $\Delta A BC$ is twice that of $\Delta DEF$ .

$\Delta ABC \sim \Delta D EF$

.

Order the angles of $\Delta D EF$ from least to greatest measure.

Tap to see back →

In a triangle, the angle of greatest measure is opposite the side of greatest measure, and the angle of least measure is opposite the side of least measure. , so their opposite angles are ranked in this order - that is, $m \angle C < m \angle A < m \angle B$ .

Corresponding angles of similar triangles are congruent, so, since $\Delta ABC \sim \Delta D EF$ , $m \angle A= m \angle D , m \angle B= m \angle E, m \angle C = m \angle F$ .

Therefore, by substitution, $m \angle F < m \angle D < m \angle E$ .

The triangles are similar. What is the value of x?

Tap to see back →

The proportions of corresponding sides of similar triangles must be equal. Therefore, $\frac{8}{12}$ = $\frac{10}{x}$. 8x = 120; x = 15.

$\Delta ABC$ is an equilateral triangle. Points are the midpoints of $\overline{AB}$,$\overline{BC}$,$\overline{AC}$ , respectively. $\Delta DEF$ is constructed.

All of the following are true except:

Tap to see back →

The three sides of $\Delta DEF$ are the midsegments of $\Delta A BC$ , so $\Delta DEF$ is similar to $\Delta A BC$ .

By the Triangle Midsegment Theorem, each is parallel to one side of $\Delta A BC$ .

By the same theorem, each has length exactly half of that side, giving $\Delta A BC$ twice the perimeter of $\Delta DEF$ .

But since the sides of $\Delta A BC$ have twice the length of those of $\Delta DEF$ , the area of $\Delta A BC$ , which varies directly as the square of a sidelength, must be four times that of $\Delta DEF$ .

The correct choice is the one that asserts that the area of $\Delta A BC$ is twice that of $\Delta DEF$ .

$\Delta ABC \sim \Delta D EF$

.

Order the angles of $\Delta D EF$ from least to greatest measure.

Tap to see back →

In a triangle, the angle of greatest measure is opposite the side of greatest measure, and the angle of least measure is opposite the side of least measure. , so their opposite angles are ranked in this order - that is, $m \angle C < m \angle A < m \angle B$ .

Corresponding angles of similar triangles are congruent, so, since $\Delta ABC \sim \Delta D EF$ , $m \angle A= m \angle D , m \angle B= m \angle E, m \angle C = m \angle F$ .

Therefore, by substitution, $m \angle F < m \angle D < m \angle E$ .

The triangles are similar. What is the value of x?

Tap to see back →

The proportions of corresponding sides of similar triangles must be equal. Therefore, $\frac{8}{12}$ = $\frac{10}{x}$. 8x = 120; x = 15.

$\Delta ABC$ is an equilateral triangle. Points are the midpoints of $\overline{AB}$,$\overline{BC}$,$\overline{AC}$ , respectively. $\Delta DEF$ is constructed.

All of the following are true except:

Tap to see back →

The three sides of $\Delta DEF$ are the midsegments of $\Delta A BC$ , so $\Delta DEF$ is similar to $\Delta A BC$ .

By the Triangle Midsegment Theorem, each is parallel to one side of $\Delta A BC$ .

By the same theorem, each has length exactly half of that side, giving $\Delta A BC$ twice the perimeter of $\Delta DEF$ .

But since the sides of $\Delta A BC$ have twice the length of those of $\Delta DEF$ , the area of $\Delta A BC$ , which varies directly as the square of a sidelength, must be four times that of $\Delta DEF$ .

The correct choice is the one that asserts that the area of $\Delta A BC$ is twice that of $\Delta DEF$ .

$\Delta ABC \sim \Delta D EF$

.

Order the angles of $\Delta D EF$ from least to greatest measure.

Tap to see back →

In a triangle, the angle of greatest measure is opposite the side of greatest measure, and the angle of least measure is opposite the side of least measure. , so their opposite angles are ranked in this order - that is, $m \angle C < m \angle A < m \angle B$ .

Corresponding angles of similar triangles are congruent, so, since $\Delta ABC \sim \Delta D EF$ , $m \angle A= m \angle D , m \angle B= m \angle E, m \angle C = m \angle F$ .

Therefore, by substitution, $m \angle F < m \angle D < m \angle E$ .

The triangles are similar. What is the value of x?

Tap to see back →

The proportions of corresponding sides of similar triangles must be equal. Therefore, $\frac{8}{12}$ = $\frac{10}{x}$. 8x = 120; x = 15.

$\Delta ABC$ is an equilateral triangle. Points are the midpoints of $\overline{AB}$,$\overline{BC}$,$\overline{AC}$ , respectively. $\Delta DEF$ is constructed.

All of the following are true except:

Tap to see back →

The three sides of $\Delta DEF$ are the midsegments of $\Delta A BC$ , so $\Delta DEF$ is similar to $\Delta A BC$ .

By the Triangle Midsegment Theorem, each is parallel to one side of $\Delta A BC$ .

By the same theorem, each has length exactly half of that side, giving $\Delta A BC$ twice the perimeter of $\Delta DEF$ .

But since the sides of $\Delta A BC$ have twice the length of those of $\Delta DEF$ , the area of $\Delta A BC$ , which varies directly as the square of a sidelength, must be four times that of $\Delta DEF$ .

The correct choice is the one that asserts that the area of $\Delta A BC$ is twice that of $\Delta DEF$ .

$\Delta ABC \sim \Delta D EF$

.

Order the angles of $\Delta D EF$ from least to greatest measure.

Tap to see back →

In a triangle, the angle of greatest measure is opposite the side of greatest measure, and the angle of least measure is opposite the side of least measure. , so their opposite angles are ranked in this order - that is, $m \angle C < m \angle A < m \angle B$ .

Corresponding angles of similar triangles are congruent, so, since $\Delta ABC \sim \Delta D EF$ , $m \angle A= m \angle D , m \angle B= m \angle E, m \angle C = m \angle F$ .

Therefore, by substitution, $m \angle F < m \angle D < m \angle E$ .

The triangles are similar. What is the value of x?

Tap to see back →

The proportions of corresponding sides of similar triangles must be equal. Therefore, $\frac{8}{12}$ = $\frac{10}{x}$. 8x = 120; x = 15.

$\Delta ABC$ is an equilateral triangle. Points are the midpoints of $\overline{AB}$,$\overline{BC}$,$\overline{AC}$ , respectively. $\Delta DEF$ is constructed.

All of the following are true except:

Tap to see back →

The three sides of $\Delta DEF$ are the midsegments of $\Delta A BC$ , so $\Delta DEF$ is similar to $\Delta A BC$ .

By the Triangle Midsegment Theorem, each is parallel to one side of $\Delta A BC$ .

By the same theorem, each has length exactly half of that side, giving $\Delta A BC$ twice the perimeter of $\Delta DEF$ .

But since the sides of $\Delta A BC$ have twice the length of those of $\Delta DEF$ , the area of $\Delta A BC$ , which varies directly as the square of a sidelength, must be four times that of $\Delta DEF$ .

The correct choice is the one that asserts that the area of $\Delta A BC$ is twice that of $\Delta DEF$ .

$\Delta ABC \sim \Delta D EF$

.

Order the angles of $\Delta D EF$ from least to greatest measure.

Tap to see back →

In a triangle, the angle of greatest measure is opposite the side of greatest measure, and the angle of least measure is opposite the side of least measure. , so their opposite angles are ranked in this order - that is, $m \angle C < m \angle A < m \angle B$ .

Corresponding angles of similar triangles are congruent, so, since $\Delta ABC \sim \Delta D EF$ , $m \angle A= m \angle D , m \angle B= m \angle E, m \angle C = m \angle F$ .

Therefore, by substitution, $m \angle F < m \angle D < m \angle E$ .

The triangles are similar. What is the value of x?

Tap to see back →

The proportions of corresponding sides of similar triangles must be equal. Therefore, $\frac{8}{12}$ = $\frac{10}{x}$. 8x = 120; x = 15.

$\Delta ABC$ is an equilateral triangle. Points are the midpoints of $\overline{AB}$,$\overline{BC}$,$\overline{AC}$ , respectively. $\Delta DEF$ is constructed.

All of the following are true except:

Tap to see back →

The three sides of $\Delta DEF$ are the midsegments of $\Delta A BC$ , so $\Delta DEF$ is similar to $\Delta A BC$ .

By the Triangle Midsegment Theorem, each is parallel to one side of $\Delta A BC$ .

By the same theorem, each has length exactly half of that side, giving $\Delta A BC$ twice the perimeter of $\Delta DEF$ .

But since the sides of $\Delta A BC$ have twice the length of those of $\Delta DEF$ , the area of $\Delta A BC$ , which varies directly as the square of a sidelength, must be four times that of $\Delta DEF$ .

The correct choice is the one that asserts that the area of $\Delta A BC$ is twice that of $\Delta DEF$ .