How to find the area of an equilateral triangle - SSAT Upper Level Quantitative

Card 1 of 28

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

An equilateral triangle is circumscribed about a circle of radius 16. Give the area of the triangle.

Tap to reveal answer

Answer

The circle and triangle referenced are below, along with a radius to one side and a segment to one vertex:

$\Delta ABO$ is a 30-60-90 triangle, so

$AB = OB \cdot $\sqrt{3}$ = 16 $\sqrt{3}$$

$\overline{AB}$ is one-half of a side of the triangle, so the sidelength is $32$\sqrt{3}$$ . The area of the triangle is

$A = $$\frac{s^{2}$$ $\sqrt{3}$}{4} = $$\frac{(32\sqrt{3}$)^{2}$ $\sqrt{3}$}{4} = $\frac{1,024 \cdot 3 \cdot \sqrt{3}$}{4}= 768 $\sqrt{3}$$

← Didn't Know|Knew It →

Question

An equilateral triangle is inscribed inside a circle of radius . Give the area of the triangle.

Tap to reveal answer

Answer

The trick is to know that the circumscribed circle, or the circumcircle, has as its center the intersection of the three altitudes of the triangle, and that this center, or circumcenter, divides each altitude into two segments, one twice the length of the other - the longer one being a radius. Because of this, we can construct the following:

Each of the six smaller triangles is a 30-60-90 triangle, and all six are congruent.

We will find the area of $\Delta COY$ , and multiply it by 6.

By the 30-60-90 Theorem, $CY = OY \cdot $\sqrt{3}$ = 21$\sqrt{3}$$ , so the area of $\Delta COY$ is

$A = $\frac{1}{2}$\cdot OY \cdot CY = $\frac{1}{2}$ \cdot 21\cdot 21$\sqrt{3}$ =\frac{ 441 \sqrt{3}$}{2}$ .

Six times this - $6 \cdot $\frac{ 441 \sqrt{3}$}{2} = 1,323 $\sqrt{3}$$ - is the area of $\Delta ABC$ .

← Didn't Know|Knew It →

Question

In the above diagram, $\Delta ABC$ is equilateral. Give its area.

Tap to reveal answer

Answer

The interior angles of an equilateral triangle all measure 60 degrees, so, by the 30-60-90 Theorem,

$BM = $\frac{AM}{\sqrt{3}$} = $\frac{32}{\sqrt{3}$}$

Also, is the midpoint of $\overline{BC}$ , so $BC = 2 \cdot BM = $\frac{64}{\sqrt{3}$}$ ; this is the base.

The area of this triangle is half the product of the base and the height :

$A= $\frac{1}{2}$ \cdot 32 \cdot $\frac{64}{\sqrt{3}$}= $\frac{1,024}{\sqrt{3}$} = $\frac{1,024\cdot \sqrt{3}$}{$\sqrt{3}$ \cdot $\sqrt{3}$}= $\frac{1,024 \sqrt{3}$}{3}$

This answer is not among the given choices.

← Didn't Know|Knew It →

Question

The perimeter of an equilateral triangle is . Give its area.

Tap to reveal answer

Answer

An equilateral triangle with perimeter 54 has three congruent sides of length

$s = 51 \div 3 = 18$

The area of this triangle is

$A = $$\frac{s^{2}$$$\sqrt{3}$}{4} = $$\frac{18^{2}$$$\sqrt{3}$}{4} = $\frac{324 \sqrt{3}$}{4}= 81 $\sqrt{3}$$

← Didn't Know|Knew It →

Question

The perimeter of an equilateral triangle is $30$\sqrt{6}$$ . Give its area.

Tap to reveal answer

Answer

An equilateral triangle with perimeter $30$\sqrt{6}$$ has three congruent sides of length

$s = 30$\sqrt{6}$\div 3 = 10$\sqrt{6}$$

The area of this triangle is

$A = $$\frac{s^{2}$$$\sqrt{3}$}{4} = $\frac{ \left ( 10\sqrt{6}$ \right $)^{2}$$\sqrt{3}$}{4} = $\frac{100 \cdot 6 \cdot \sqrt{3}$}{4} = 150$\sqrt{3}$$

← Didn't Know|Knew It →

Question

Hexagon is regular and has perimeter 72. $\Delta ACE$ is constructed. What is its area?

Tap to reveal answer

Answer

Since the perimeter of the (six-congruent-sided) regular hexagon is 72, each side has length one sixth this, or 12.

The figure described is given below, with a perpendicular segment drawn from to side $\overline{AC}$ :

Each angle of a regular hexagon measures . Therefore, $m \angle ABM = $60^{\circ }$$ , and $\Delta ABM$ is a 30-60-90 triangle.

By the 30-60-90 Theorem,

$BM = $\frac{1}{2}$AB = $\frac{1}{2}$ \cdot 12 = 6$

and

$AM = BM $\sqrt{3}$ = 6 $\sqrt{3}$$

$AC = 2 \cdot AM = 2 \cdot 6 $\sqrt{3}$ =12 $\sqrt{3}$$ .

$\Delta ACE$ is equilateral, and $12 $\sqrt{3}$$ is its sidelength, making its area

$A = $$\frac{s^{2}$$$\sqrt{3}$}{4} = $\frac{\left (12 \sqrt{3}$ \right $)^{2}$$\sqrt{3}$}{4} = $\frac{144 \cdot 3 \cdot \sqrt{3}$}{4} =108 $\sqrt{3}$$

← Didn't Know|Knew It →

Question

An equilateral triangle has side lengths of . What is the area of this triangle?

Tap to reveal answer

Answer

The area of an equilateral triangle can be found using this formula:

$\frac{\sqrt3}{4}$($\text{side $length}$)^2$

Using $$\text{side length}$=2x$ , we can find the area of the equilateral triangle.

← Didn't Know|Knew It →

Question

An equilateral triangle is circumscribed about a circle of radius 16. Give the area of the triangle.

Tap to reveal answer

Answer

The circle and triangle referenced are below, along with a radius to one side and a segment to one vertex:

$\Delta ABO$ is a 30-60-90 triangle, so

$AB = OB \cdot $\sqrt{3}$ = 16 $\sqrt{3}$$

$\overline{AB}$ is one-half of a side of the triangle, so the sidelength is $32$\sqrt{3}$$ . The area of the triangle is

$A = $$\frac{s^{2}$$ $\sqrt{3}$}{4} = $$\frac{(32\sqrt{3}$)^{2}$ $\sqrt{3}$}{4} = $\frac{1,024 \cdot 3 \cdot \sqrt{3}$}{4}= 768 $\sqrt{3}$$

← Didn't Know|Knew It →

Question

An equilateral triangle is inscribed inside a circle of radius . Give the area of the triangle.

Tap to reveal answer

Answer

The trick is to know that the circumscribed circle, or the circumcircle, has as its center the intersection of the three altitudes of the triangle, and that this center, or circumcenter, divides each altitude into two segments, one twice the length of the other - the longer one being a radius. Because of this, we can construct the following:

Each of the six smaller triangles is a 30-60-90 triangle, and all six are congruent.

We will find the area of $\Delta COY$ , and multiply it by 6.

By the 30-60-90 Theorem, $CY = OY \cdot $\sqrt{3}$ = 21$\sqrt{3}$$ , so the area of $\Delta COY$ is

$A = $\frac{1}{2}$\cdot OY \cdot CY = $\frac{1}{2}$ \cdot 21\cdot 21$\sqrt{3}$ =\frac{ 441 \sqrt{3}$}{2}$ .

Six times this - $6 \cdot $\frac{ 441 \sqrt{3}$}{2} = 1,323 $\sqrt{3}$$ - is the area of $\Delta ABC$ .

← Didn't Know|Knew It →

Question

In the above diagram, $\Delta ABC$ is equilateral. Give its area.

Tap to reveal answer

Answer

The interior angles of an equilateral triangle all measure 60 degrees, so, by the 30-60-90 Theorem,

$BM = $\frac{AM}{\sqrt{3}$} = $\frac{32}{\sqrt{3}$}$

Also, is the midpoint of $\overline{BC}$ , so $BC = 2 \cdot BM = $\frac{64}{\sqrt{3}$}$ ; this is the base.

The area of this triangle is half the product of the base and the height :

$A= $\frac{1}{2}$ \cdot 32 \cdot $\frac{64}{\sqrt{3}$}= $\frac{1,024}{\sqrt{3}$} = $\frac{1,024\cdot \sqrt{3}$}{$\sqrt{3}$ \cdot $\sqrt{3}$}= $\frac{1,024 \sqrt{3}$}{3}$

This answer is not among the given choices.

← Didn't Know|Knew It →

Question

The perimeter of an equilateral triangle is . Give its area.

Tap to reveal answer

Answer

An equilateral triangle with perimeter 54 has three congruent sides of length

$s = 51 \div 3 = 18$

The area of this triangle is

$A = $$\frac{s^{2}$$$\sqrt{3}$}{4} = $$\frac{18^{2}$$$\sqrt{3}$}{4} = $\frac{324 \sqrt{3}$}{4}= 81 $\sqrt{3}$$

← Didn't Know|Knew It →

Question

The perimeter of an equilateral triangle is $30$\sqrt{6}$$ . Give its area.

Tap to reveal answer

Answer

An equilateral triangle with perimeter $30$\sqrt{6}$$ has three congruent sides of length

$s = 30$\sqrt{6}$\div 3 = 10$\sqrt{6}$$

The area of this triangle is

$A = $$\frac{s^{2}$$$\sqrt{3}$}{4} = $\frac{ \left ( 10\sqrt{6}$ \right $)^{2}$$\sqrt{3}$}{4} = $\frac{100 \cdot 6 \cdot \sqrt{3}$}{4} = 150$\sqrt{3}$$

← Didn't Know|Knew It →

Question

Hexagon is regular and has perimeter 72. $\Delta ACE$ is constructed. What is its area?

Tap to reveal answer

Answer

Since the perimeter of the (six-congruent-sided) regular hexagon is 72, each side has length one sixth this, or 12.

The figure described is given below, with a perpendicular segment drawn from to side $\overline{AC}$ :

Each angle of a regular hexagon measures . Therefore, $m \angle ABM = $60^{\circ }$$ , and $\Delta ABM$ is a 30-60-90 triangle.

By the 30-60-90 Theorem,

$BM = $\frac{1}{2}$AB = $\frac{1}{2}$ \cdot 12 = 6$

and

$AM = BM $\sqrt{3}$ = 6 $\sqrt{3}$$

$AC = 2 \cdot AM = 2 \cdot 6 $\sqrt{3}$ =12 $\sqrt{3}$$ .

$\Delta ACE$ is equilateral, and $12 $\sqrt{3}$$ is its sidelength, making its area

$A = $$\frac{s^{2}$$$\sqrt{3}$}{4} = $\frac{\left (12 \sqrt{3}$ \right $)^{2}$$\sqrt{3}$}{4} = $\frac{144 \cdot 3 \cdot \sqrt{3}$}{4} =108 $\sqrt{3}$$

← Didn't Know|Knew It →

Question

An equilateral triangle has side lengths of . What is the area of this triangle?

Tap to reveal answer

Answer

The area of an equilateral triangle can be found using this formula:

$\frac{\sqrt3}{4}$($\text{side $length}$)^2$

Using $$\text{side length}$=2x$ , we can find the area of the equilateral triangle.

← Didn't Know|Knew It →

Question

An equilateral triangle is circumscribed about a circle of radius 16. Give the area of the triangle.

Tap to reveal answer

Answer

The circle and triangle referenced are below, along with a radius to one side and a segment to one vertex:

$\Delta ABO$ is a 30-60-90 triangle, so

$AB = OB \cdot $\sqrt{3}$ = 16 $\sqrt{3}$$

$\overline{AB}$ is one-half of a side of the triangle, so the sidelength is $32$\sqrt{3}$$ . The area of the triangle is

$A = $$\frac{s^{2}$$ $\sqrt{3}$}{4} = $$\frac{(32\sqrt{3}$)^{2}$ $\sqrt{3}$}{4} = $\frac{1,024 \cdot 3 \cdot \sqrt{3}$}{4}= 768 $\sqrt{3}$$

← Didn't Know|Knew It →

Question

An equilateral triangle is inscribed inside a circle of radius . Give the area of the triangle.

Tap to reveal answer

Answer

The trick is to know that the circumscribed circle, or the circumcircle, has as its center the intersection of the three altitudes of the triangle, and that this center, or circumcenter, divides each altitude into two segments, one twice the length of the other - the longer one being a radius. Because of this, we can construct the following:

Each of the six smaller triangles is a 30-60-90 triangle, and all six are congruent.

We will find the area of $\Delta COY$ , and multiply it by 6.

By the 30-60-90 Theorem, $CY = OY \cdot $\sqrt{3}$ = 21$\sqrt{3}$$ , so the area of $\Delta COY$ is

$A = $\frac{1}{2}$\cdot OY \cdot CY = $\frac{1}{2}$ \cdot 21\cdot 21$\sqrt{3}$ =\frac{ 441 \sqrt{3}$}{2}$ .

Six times this - $6 \cdot $\frac{ 441 \sqrt{3}$}{2} = 1,323 $\sqrt{3}$$ - is the area of $\Delta ABC$ .

← Didn't Know|Knew It →

Question

In the above diagram, $\Delta ABC$ is equilateral. Give its area.

Tap to reveal answer

Answer

The interior angles of an equilateral triangle all measure 60 degrees, so, by the 30-60-90 Theorem,

$BM = $\frac{AM}{\sqrt{3}$} = $\frac{32}{\sqrt{3}$}$

Also, is the midpoint of $\overline{BC}$ , so $BC = 2 \cdot BM = $\frac{64}{\sqrt{3}$}$ ; this is the base.

The area of this triangle is half the product of the base and the height :

$A= $\frac{1}{2}$ \cdot 32 \cdot $\frac{64}{\sqrt{3}$}= $\frac{1,024}{\sqrt{3}$} = $\frac{1,024\cdot \sqrt{3}$}{$\sqrt{3}$ \cdot $\sqrt{3}$}= $\frac{1,024 \sqrt{3}$}{3}$

This answer is not among the given choices.

← Didn't Know|Knew It →

Question

The perimeter of an equilateral triangle is . Give its area.

Tap to reveal answer

Answer

An equilateral triangle with perimeter 54 has three congruent sides of length

$s = 51 \div 3 = 18$

The area of this triangle is

$A = $$\frac{s^{2}$$$\sqrt{3}$}{4} = $$\frac{18^{2}$$$\sqrt{3}$}{4} = $\frac{324 \sqrt{3}$}{4}= 81 $\sqrt{3}$$

← Didn't Know|Knew It →

Question

The perimeter of an equilateral triangle is $30$\sqrt{6}$$ . Give its area.

Tap to reveal answer

Answer

An equilateral triangle with perimeter $30$\sqrt{6}$$ has three congruent sides of length

$s = 30$\sqrt{6}$\div 3 = 10$\sqrt{6}$$

The area of this triangle is

$A = $$\frac{s^{2}$$$\sqrt{3}$}{4} = $\frac{ \left ( 10\sqrt{6}$ \right $)^{2}$$\sqrt{3}$}{4} = $\frac{100 \cdot 6 \cdot \sqrt{3}$}{4} = 150$\sqrt{3}$$

← Didn't Know|Knew It →

Question

Hexagon is regular and has perimeter 72. $\Delta ACE$ is constructed. What is its area?

Tap to reveal answer

Answer

Since the perimeter of the (six-congruent-sided) regular hexagon is 72, each side has length one sixth this, or 12.

The figure described is given below, with a perpendicular segment drawn from to side $\overline{AC}$ :

Each angle of a regular hexagon measures . Therefore, $m \angle ABM = $60^{\circ }$$ , and $\Delta ABM$ is a 30-60-90 triangle.

By the 30-60-90 Theorem,

$BM = $\frac{1}{2}$AB = $\frac{1}{2}$ \cdot 12 = 6$

and

$AM = BM $\sqrt{3}$ = 6 $\sqrt{3}$$

$AC = 2 \cdot AM = 2 \cdot 6 $\sqrt{3}$ =12 $\sqrt{3}$$ .

$\Delta ACE$ is equilateral, and $12 $\sqrt{3}$$ is its sidelength, making its area

$A = $$\frac{s^{2}$$$\sqrt{3}$}{4} = $\frac{\left (12 \sqrt{3}$ \right $)^{2}$$\sqrt{3}$}{4} = $\frac{144 \cdot 3 \cdot \sqrt{3}$}{4} =108 $\sqrt{3}$$

← Didn't Know|Knew It →

0

Knew It