Quadrilaterals

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GMAT Quantitative › Quadrilaterals

Questions 1 - 10

Rhombus_1

The above figure shows a rhombus . Give its area.

$40 \sqrt{11}$

$40\sqrt{61}$

Explanation

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

Rhombus_1

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

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

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

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

Parallelogram2

Give the area of the above parallelogram if .

$25\sqrt{3}$

$25\sqrt{2}$

$50\sqrt{2}$

Explanation

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

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

Export-png

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:

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

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

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

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

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

$AE \cdot DE$

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

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

$= 60 \cdot 24$

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

Explanation

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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