Calculating the area of a right triangle - GMAT Quantitative

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If a right triangle has a hypotenuse that is 10, and one side is 6, find the area of the triangle.

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Becaue this is a right triangle and since the hypotenuse is 10 and one side is 6, the other side will be 8. This can be found using the Pythagorean Theorem, or multiplying a 3-4-5 triangle by 2 to get a 6-8-10 triangle. Since the area of a right triangle is half of the product of the two sides, we have

$A=\frac{1}{2}$LW=\frac{1}{2}$6*8=24.$

A triangle on the coordinate plane has vertices.

Which of the following expressions is equal to the area of the triangle?

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This is a right triangle with legs along the - and -axes, so the area of each can be calculated by taking one-half the product of the two legs.

The vertical leg has length ; the horizontal leg has length .

Now calculate the area:

$A = $\frac{1}{2}$ bh = $\frac{1}{2}$ (c-1)(c+1) = $\frac{1}{2}$ $(c^{2}$-1) = $\frac{1}{2}$ $c^{2}$- $\frac{1}{2}$$

In the above figure, triangle SPQ, SPR and PRQ are right triangles. Given the lengths of SP and PQ are 2cm. What is the area of triangle SPR?

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Since $\triangle SPQ,\triangle SPR,and\ \triangle PRQ$ are right triangles, we know that $\triangle SPQ$ is an isosceles right triangle. So we know that the lengths of $\overline{SP}$ and $\overline{PQ}$ are 2 cm, so we can get the length of $\overline{SQ}$ by using the Pythagorean Theorem:

$SQ $=$\sqrt{2^2$$+2^2$}$=$\sqrt{8}$$

is the midpoint of $\overline{SQ}$ , so the length of $\overline{SR}$ is $\sqrt{2}$ .

Now we can use the Pythagorean Theorem again to solve for $\overline{PR}$ : $$\sqrt{ $(2^2$$-(\sqrt2)^2$)}$=\sqrt2$ .

Finally, we have all the elements needed to solve for the area of $\triangle SPR$ :

$(1/2)\sqrt2\sqrt2=1$

The hypotenuse of a $30 ^{\circ } - $60^{\circ }$ $-90^{\circ }$$ triangle is equal to the sidelength of a square. Give the ratio of the area of the square to that of the triangle.

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Let be the sidelength of the square. Then its area is $s ^{2}$ .

If the hypotenuse of a $30 ^{\circ } - $60^{\circ }$ $-90^{\circ }$$ triangle is , its shorter leg is half that, or $$\frac{1}{2 }$s$ ; its longer leg is $\sqrt{3}$ times the shorter leg, or $\frac{1}{2 }$s $\sqrt{3}$ . The area of the triangle is half the product of the legs, or

$$\frac{1}{2}$ \cdot $\frac{1}{2}$s \cdot$\frac{1}{2 }$s $\sqrt{3}$ =\frac{ \sqrt{3}$}{8} s ^{2}$

The ratio of the area of the square to that of the triangle is

$s ^{2}: $\frac{ \sqrt{3}$}{8} s$ or $1: $\frac{ \sqrt{3}$}{8}$

or $8$\sqrt{3}$:3$

Calculate the area of the following right triangle, leave in terms of .

(Not drawn to scale.)

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The equation for the area of a right triangle is:

$A=\frac{1}{2}$b\times h$

In this case, our values are:

Plugging this into the equation leaves us with:

$A=\frac{1}{2}$(x)(x+1)$

which can be rewritten as $A=\frac{x}{2}$(x+1)$

A right triangle has a hypotenuse of 13 and a base of 12. Calculate the area of the triangle.

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In order to calculate the area of a right triangle, we need to know the height and the base. We are only given the length of the base, so we first need to use the Pythagorean theorem with the given hypotenuse to find the length of the height:

Now that we have the height and the base, we can plug these values into the formula for the area of a right triangle:

$A=\frac{1}{2}$bh=\frac{1}{2}$(12)(5)=\frac{1}{2}$(60)=30$

Find the area of a triangle whose base is and height is .

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To find the area, use the following formula:

$A=\frac{1}{2}$Bh=\frac{1}{2}$(4)(8)=16$

In the above figure, triangle SPQ, SPR and PRQ are right triangles. Given the lengths of SP and PQ are 2cm. What is the area of triangle SPR?

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Since $\triangle SPQ,\triangle SPR,and\ \triangle PRQ$ are right triangles, we know that $\triangle SPQ$ is an isosceles right triangle. So we know that the lengths of $\overline{SP}$ and $\overline{PQ}$ are 2 cm, so we can get the length of $\overline{SQ}$ by using the Pythagorean Theorem:

$SQ $=$\sqrt{2^2$$+2^2$}$=$\sqrt{8}$$

is the midpoint of $\overline{SQ}$ , so the length of $\overline{SR}$ is $\sqrt{2}$ .

Now we can use the Pythagorean Theorem again to solve for $\overline{PR}$ : $$\sqrt{ $(2^2$$-(\sqrt2)^2$)}$=\sqrt2$ .

Finally, we have all the elements needed to solve for the area of $\triangle SPR$ :

$(1/2)\sqrt2\sqrt2=1$

If a right triangle has a hypotenuse that is 10, and one side is 6, find the area of the triangle.

Tap to see back →

Becaue this is a right triangle and since the hypotenuse is 10 and one side is 6, the other side will be 8. This can be found using the Pythagorean Theorem, or multiplying a 3-4-5 triangle by 2 to get a 6-8-10 triangle. Since the area of a right triangle is half of the product of the two sides, we have

$A=\frac{1}{2}$LW=\frac{1}{2}$6*8=24.$

A triangle on the coordinate plane has vertices.

Which of the following expressions is equal to the area of the triangle?

Tap to see back →

This is a right triangle with legs along the - and -axes, so the area of each can be calculated by taking one-half the product of the two legs.

The vertical leg has length ; the horizontal leg has length .

Now calculate the area:

$A = $\frac{1}{2}$ bh = $\frac{1}{2}$ (c-1)(c+1) = $\frac{1}{2}$ $(c^{2}$-1) = $\frac{1}{2}$ $c^{2}$- $\frac{1}{2}$$

The hypotenuse of a $30 ^{\circ } - $60^{\circ }$ $-90^{\circ }$$ triangle is equal to the sidelength of a square. Give the ratio of the area of the square to that of the triangle.

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Let be the sidelength of the square. Then its area is $s ^{2}$ .

If the hypotenuse of a $30 ^{\circ } - $60^{\circ }$ $-90^{\circ }$$ triangle is , its shorter leg is half that, or $$\frac{1}{2 }$s$ ; its longer leg is $\sqrt{3}$ times the shorter leg, or $\frac{1}{2 }$s $\sqrt{3}$ . The area of the triangle is half the product of the legs, or

$$\frac{1}{2}$ \cdot $\frac{1}{2}$s \cdot$\frac{1}{2 }$s $\sqrt{3}$ =\frac{ \sqrt{3}$}{8} s ^{2}$

The ratio of the area of the square to that of the triangle is

$s ^{2}: $\frac{ \sqrt{3}$}{8} s$ or $1: $\frac{ \sqrt{3}$}{8}$

or $8$\sqrt{3}$:3$

Calculate the area of the following right triangle, leave in terms of .

(Not drawn to scale.)

Tap to see back →

The equation for the area of a right triangle is:

$A=\frac{1}{2}$b\times h$

In this case, our values are:

Plugging this into the equation leaves us with:

$A=\frac{1}{2}$(x)(x+1)$

which can be rewritten as $A=\frac{x}{2}$(x+1)$

A right triangle has a hypotenuse of 13 and a base of 12. Calculate the area of the triangle.

Tap to see back →

In order to calculate the area of a right triangle, we need to know the height and the base. We are only given the length of the base, so we first need to use the Pythagorean theorem with the given hypotenuse to find the length of the height:

Now that we have the height and the base, we can plug these values into the formula for the area of a right triangle:

$A=\frac{1}{2}$bh=\frac{1}{2}$(12)(5)=\frac{1}{2}$(60)=30$

Find the area of a triangle whose base is and height is .

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To find the area, use the following formula:

$A=\frac{1}{2}$Bh=\frac{1}{2}$(4)(8)=16$

If a right triangle has a hypotenuse that is 10, and one side is 6, find the area of the triangle.

Tap to see back →

Becaue this is a right triangle and since the hypotenuse is 10 and one side is 6, the other side will be 8. This can be found using the Pythagorean Theorem, or multiplying a 3-4-5 triangle by 2 to get a 6-8-10 triangle. Since the area of a right triangle is half of the product of the two sides, we have

$A=\frac{1}{2}$LW=\frac{1}{2}$6*8=24.$

A triangle on the coordinate plane has vertices.

Which of the following expressions is equal to the area of the triangle?

Tap to see back →

This is a right triangle with legs along the - and -axes, so the area of each can be calculated by taking one-half the product of the two legs.

The vertical leg has length ; the horizontal leg has length .

Now calculate the area:

$A = $\frac{1}{2}$ bh = $\frac{1}{2}$ (c-1)(c+1) = $\frac{1}{2}$ $(c^{2}$-1) = $\frac{1}{2}$ $c^{2}$- $\frac{1}{2}$$

In the above figure, triangle SPQ, SPR and PRQ are right triangles. Given the lengths of SP and PQ are 2cm. What is the area of triangle SPR?

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Since $\triangle SPQ,\triangle SPR,and\ \triangle PRQ$ are right triangles, we know that $\triangle SPQ$ is an isosceles right triangle. So we know that the lengths of $\overline{SP}$ and $\overline{PQ}$ are 2 cm, so we can get the length of $\overline{SQ}$ by using the Pythagorean Theorem:

$SQ $=$\sqrt{2^2$$+2^2$}$=$\sqrt{8}$$

is the midpoint of $\overline{SQ}$ , so the length of $\overline{SR}$ is $\sqrt{2}$ .

Now we can use the Pythagorean Theorem again to solve for $\overline{PR}$ : $$\sqrt{ $(2^2$$-(\sqrt2)^2$)}$=\sqrt2$ .

Finally, we have all the elements needed to solve for the area of $\triangle SPR$ :

$(1/2)\sqrt2\sqrt2=1$

The hypotenuse of a $30 ^{\circ } - $60^{\circ }$ $-90^{\circ }$$ triangle is equal to the sidelength of a square. Give the ratio of the area of the square to that of the triangle.

Tap to see back →

Let be the sidelength of the square. Then its area is $s ^{2}$ .

If the hypotenuse of a $30 ^{\circ } - $60^{\circ }$ $-90^{\circ }$$ triangle is , its shorter leg is half that, or $$\frac{1}{2 }$s$ ; its longer leg is $\sqrt{3}$ times the shorter leg, or $\frac{1}{2 }$s $\sqrt{3}$ . The area of the triangle is half the product of the legs, or

$$\frac{1}{2}$ \cdot $\frac{1}{2}$s \cdot$\frac{1}{2 }$s $\sqrt{3}$ =\frac{ \sqrt{3}$}{8} s ^{2}$

The ratio of the area of the square to that of the triangle is

$s ^{2}: $\frac{ \sqrt{3}$}{8} s$ or $1: $\frac{ \sqrt{3}$}{8}$

or $8$\sqrt{3}$:3$

Calculate the area of the following right triangle, leave in terms of .

(Not drawn to scale.)

Tap to see back →

The equation for the area of a right triangle is:

$A=\frac{1}{2}$b\times h$

In this case, our values are:

Plugging this into the equation leaves us with:

$A=\frac{1}{2}$(x)(x+1)$

which can be rewritten as $A=\frac{x}{2}$(x+1)$

A right triangle has a hypotenuse of 13 and a base of 12. Calculate the area of the triangle.

Tap to see back →

In order to calculate the area of a right triangle, we need to know the height and the base. We are only given the length of the base, so we first need to use the Pythagorean theorem with the given hypotenuse to find the length of the height:

Now that we have the height and the base, we can plug these values into the formula for the area of a right triangle:

$A=\frac{1}{2}$bh=\frac{1}{2}$(12)(5)=\frac{1}{2}$(60)=30$