Quadrilaterals

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GMAT Quantitative › Quadrilaterals

Questions 1 - 10

Rectangles

In the above diagram,

$\textup{Rect }ABCF \sim \textup{Rect }CDEF$ .

and . Give the area of $\textup{Rect }ABDE$ .

Explanation

$\textup{Rect }ABCF \sim \textup{Rect }CDEF$ , so

$\frac{AF}{CF} = \frac{CF}{EF}$

$\frac{AF}{24} = \frac{24}{12}$

$AF = \frac{24}{12} \cdot 24 = 48$

The area of the rectangle is

$AE \cdot DE$

$= \left (AF + EF \right ) \cdot CF$

$= \left (48+12 \right ) \cdot 24$

$= 60 \cdot 24$

Parallelogram2

Give the area of the above parallelogram if .

$225\sqrt{3}$

$225\sqrt{2}$

$75\sqrt{2}$

$75\sqrt{3}$

Explanation

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

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

Parallelogram1

Give the area of the above parallelogram if .

$64\sqrt{2}$

$64\sqrt{3}$

$128\sqrt{2}$

Explanation

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,

$A = BD \cdot CD$

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

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

$= 8\cdot 8 \cdot 2$

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

$\dpi{100} \small 63$

$\dpi{100} \small 39$

$\dpi{100} \small 29$

$\dpi{100} \small 43$

$\dpi{100} \small 51$

Explanation

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

Quad

Note: Figure NOT drawn to scale

What is the area of Quadrilateral , above?

$42 \frac{1}{2}$

$38 \frac{1}{2}$

Explanation

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 .

Two angles of a parallelogram measure $\left ( x + 70 \right )^{\circ }$ and $\left ( 2x\right +5) ^{\circ }$ . What are the possible values of ?

$x = 35 \textrm{ or } x = 65$

$x = 35 \textrm{ or } x = 95$

Explanation

Case 1: The two angles are opposite angles of the parallelogram. In this case, they are congruent, and

Case 2: The two angles are consecutive angles of the parallelogram. In this case, they are supplementary, and

$\left ( 2x +5 \right )+\left ( x + 70 \right ) = 180$

$3x \div 3 = 105\div 3$

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A circle is inscribed in a square. The area of the circle is $2\pi$ . What is the area of the square?

$2\sqrt{2}$

$3\sqrt{2}$

Explanation

Since we know the area of the circle, we can tell that: $\pi \cdot r^{2}= 2\pi$ . Where is the radius of the circle.

The length of a side of the square will be since the diameter of the circle is the same length as the side length of the square.

Finally we can calculate the area of the square which will be $(2r)^{2}$ . $r=\sqrt{2}$ so the area will be , which is our final answer.

Rhombus

Note: Figure NOT drawn to scale.

The above figure is of a rhombus and one of its diagonals. What is equal to?

Not enough information is given to answer the question.

Explanation

The four sides of a rhombus are congruent, so a diagonal of the rhombus cuts it into two isosceles triangles. The two angles adjacent to the diagonal are congruent, so the third angle, the one marked, measures:

$\left (180 - 2 \times 24 \right )^{\circ } = \left (180 -48 \right )^{\circ } = 132^{\circ }$