Calculating the area of a quadrilateral - GMAT Quantitative

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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?

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Answer

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

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Question

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

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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 ?

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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 ?

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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

Note: Figure NOT drawn to scale

What is the area of Quadrilateral , above?

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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 the quadrilateral on the coordinate plane with vertices ?

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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 .

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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 ?

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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 .

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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 .

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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 .

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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.

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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 .

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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 ?

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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 .

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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.

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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.

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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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Question

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

Tap to reveal answer

Answer

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

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Question

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

Tap to reveal answer

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 ?

Tap to reveal answer

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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