Calculating the area of a quadrilateral - GMAT Quantitative Reasoning

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Question

Note: Figure NOT drawn to scale

What is the area of Quadrilateral , above?

Answer

Quadrilateral is a composite of two right triangles, $\Delta ADU$ and $\Delta DUQ$ , so we find the area of each and add the areas. First, we need to find and , since the area of a right triangle is half the product of the lengths of its legs.

By the Pythagorean Theorem:

$\left (AD \right )^{2} +\left ( AU\right ) ^{2} =\left ( DU\right ) ^{2}$

$\left (AD \right )^{2} +4 ^{2} =5 ^{2}$

$\left (AD \right )^{2} +16 =25$

$\left (AD \right )^{2} +16 -16 =25 -16$

$\left (AD \right )^{2} =9$

Also by the Pythagorean Theorem:

$\left ( QU\right ) ^{2} + \left ( DU\right ) ^{2}=\left ( DQ\right ) ^{2}$

$\left ( QU\right ) ^{2} + 5^{2}=13^{2}$

$\left ( QU\right ) ^{2} + 25 =169$

$\left ( QU\right ) ^{2} + 25 -25=169 -25$

$\left ( QU\right ) ^{2} =144$

The area of $\Delta ADU$ is $\frac{1}{2} \cdot AD \cdot AU = \frac{1}{2} \cdot 3 \cdot 4 = 6$ .

The area of $\Delta DUQ$ is $\frac{1}{2} \cdot DU \cdot QU = \frac{1}{2} \cdot 5 \cdot 12 = 30$ .

Add the areas to get , the area of Quadrilateral .

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Question

What is the area of a trapezoid with a height of 7, a base of 5, and another base of 13?

Answer

$area = \frac{(b_{1}+ b_{2}\cdot h)}{2} = \frac{(5 + 13)\cdot 7}{2} = \frac{18\cdot 7}{2} = \frac{126}{2} = 63$

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Question

A circle can be circumscribed about each of the following figures except:

Answer

A circle can be circumscribed about any triangle regardless of its sidelengths or angle measures, so we can eliminate the two triangle choices.

A circle can be circumscribed about any regular polygon, so we can eliminate those two choices as well.

The correct choice is that each figure can have a circle circumscribed about it.

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Question

What is the area of a quadrilateral on the coordinate plane with vertices ?

Answer

As can be seen from this diagram, this is a parallelogram with base 8 and height 4:

The area of this parallelogram is the product of its base and its height:

$A = bh = 8 \cdot 4 = 32$

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Question

What is the area of a quadrilateral on the coordinate plane with vertices ?

Answer

As can be seen in this diagram, this is a trapezoid with bases 10 and 5 and height 8.

Setting in the following formula, we can calculate the area of the trapezoid:

$A = \frac{1}{2} (B + b) h = \frac{1}{2} \cdot (10 + 5)\cdot 8 = 60$

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Question

What is the area of the quadrilateral on the coordinate plane with vertices ?

Answer

The quadrilateral formed is a trapezoid with two horizontal bases. One base connects (0,0) and (9,0) and therefore has length ; the other connects (4,7) and (7,7) and has length . The height is the vertical distance between the two bases, which is the difference of the coorindates: . Therefore, the area of the trapezoid is

$A = \frac{1}{2} (B + b ) h = \frac{1}{2} \cdot (9+3 ) \cdot 7 = 42$

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Question

What is the area of the quadrilateral on the coordinate plane with vertices .

Answer

The quadrilateral is a trapezoid with horizontal bases; one connects and and has length , and the other connects and and has length . The height is the vertical distance between the bases, which is the difference of the -coordinates; this is . Substitute in the formula for the area of a trapezoid:

$A = \frac{1}{2} (B + b)h = \frac{1}{2}\cdot (10+6) \cdot 6 = 48$

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Question

What is the area of the quadrilateral on the coordinate plane with vertices ?

Answer

The quadrilateral is a parallelogram with two vertical bases, each with length . Its height is the distance between the bases, which is the difference of the -coordinates: . The area of the parallelogram is the product of its base and its height:

$7 \times 5 = 35$

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Question

Give the area of the above parallelogram if .

Answer

Multiply height by base to get the area.

By the 30-60-90 Theorem:

$CD= \frac{BC}{2} = \frac{10}{2}} = 5$ .

$BD =CD \sqrt{3} = 5\sqrt{3}$

The area is therefore

$BD \cdot CD = 5\sqrt{3} \cdot 5= 25\sqrt{3}$

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Question

Give the area of the above parallelogram if .

Answer

Multiply height by base to get the area.

By the 45-45-90 Theorem,

$BD=CD = BC \div \sqrt{2} = \frac{16}{\sqrt{2}} = \frac{16\cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{16 \sqrt{2}}{2} = 8 \sqrt{2}$ .

Since the product of the height and the base of a parallelogram is its area,

$A = BD \cdot CD$

$= 8 \sqrt{2} \cdot 8 \sqrt{2}$

$= 8\cdot 8 \cdot \sqrt{2} \cdot \sqrt{2}$

$= 8\cdot 8 \cdot 2$

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Question

Give the area of the above parallelogram if .

Answer

Multiply height by base to get the area.

By the 30-60-90 Theorem:

$BD= \frac{BC}{2} = \frac{30}{2}} = 15$ .

$CD =BD \sqrt{3} = 15\sqrt{3}$

The area is therefore

$BD \cdot CD =15 \cdot 15\sqrt{3} = 225\sqrt{3}$

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Question

The above figure shows a rhombus . Give its area.

Answer

Construct the other diagonal of the rhombus, which, along with the first one, form a pair of mutual perpendicular bisectors.

By the Pythagorean Theorem,

$n = \sqrt{12^{2}- 10^{2}} =\sqrt{144- 100} = \sqrt{44}= \sqrt{4}\cdot \sqrt{11} = 2 \sqrt{11}$

The rhombus can be seen as the composite of four congruent right triangles, each with legs 10 and $2 \sqrt{11}$ , so the area of the rhombus is

$A = 4 \cdot \frac{1}{2}\cdot 10 \cdot 2\sqrt{11} =40 \sqrt{11}$ .

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. Give the area of Quadrilateral .

Answer

Apply the Pythagorean Theorem twice here.

$BD = \sqrt{6^{2}+ 8^{2}} = \sqrt{36+64 } = \sqrt{100}= 10$

$BC = \sqrt{26^{2}-10^{2}} = \sqrt{676-100} = \sqrt{576} = 24$

The quadrilateral is a composite of two right triangles, each of whose area is half the product of its legs:

Area of $\Delta ABD$ : $\frac{1}{2} \times 6 \times 8 = 24$

Area of $\Delta BCD$ : $\frac{1}{2} \times 10 \times 24= 120$

Add:

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Question

Rhombus has perimeter 48; . What is the area of Rhombus ?

Answer

Each side of a rhombus is congruent, so if it has perimeter 48, it has sidelength 12. Also, the diagonals of a rhombus are each other's perpendicular bisectors, so if they are both constructed, and their point of intersection is called , then . The following figure is formed by the rhombus and its diagonals.

$\bigtriangleup AMB$ is a right triangle with its short leg half the length of its hypotenuse, so it is a 30-60-90 triangle, and its long leg measures $6\sqrt{3}$ by the 30-60-90 Theorem. Therefore, $BD = 12\sqrt{3}$ . The area of a rhombus is half the product of the lengths of its diagonals:

$\frac{1}{2} \times 12 \times 12 \sqrt{3} = 72 \sqrt{3}$

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Question

Note: figure NOT drawn to scale.

The above figure depicts a rhombus, .

$HM = 2^{X}$

Give the area of Rhombus .

Answer

The area of a rhombus is half the product of the lengths of its diagonals, so

$A = \frac{1}{2} \cdot RO \cdot HM$

$= \frac{1}{2} \cdot 2 (HM)\cdot HM$

$= HM \cdot HM$

$= 2 ^{x} \cdot 2 ^{x}$

$= 2^{x+x}$

$= 2^{2x}$

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Question

Note: figure NOT drawn to scale.

Give the area of the above trapezoid.

Answer

The area of a trapezoid with height and bases of length and is

$A = \frac{1}{2} (B + b)h$ .

Setting $h= b = 2^{x}, B = 2^{x+1}$ :

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

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

$= 2^{-1}\cdot 2^{x} \cdot ( 2 \cdot 2^{x } + 2^{x})$

$= 2^{x-1} \cdot ( 2 \cdot 2^{x } + 2^{x})$

$= 2^{x-1} \cdot 3 \cdot 2^{x }$

$= 3 \cdot 2^{x+x-1 }$

$= 3 \cdot 2^{2x-1 }$

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Question

Note: figure NOT drawn to scale.

Give the area of the above trapezoid.

Answer

The area of a trapezoid with height and bases of length and is

$A = \frac{1}{2} (B + b)h$ .

Setting $h= b = 2^{x}, B = 3 \cdot 2^{x}$ :

$A = \frac{1}{2} (3 \cdot 2^{x} + 2^{x}) \cdot 2^{x}$

$= \frac{1}{2} \cdot 4 \cdot 2^{x} \cdot 2^{x}$

$= 2 \cdot 2^{x} \cdot 2^{x}$

$= 2^{1} \cdot 2^{x} \cdot 2^{x}$

$= 2^{1+x+x}$

$= 2^{2x+1}$

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