Other Quadrilaterals - GMAT Quantitative

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Question

While walking with some friends on campus, you come across an open grassy rectangular area. You shout "Pythagoras!" and run across the rectangular area from one corner to another. If the rectangle measures meters by meters, what distance did you cover?

Answer

This problem can be solved in a variety of ways, however you do it, it may be worth eliminating any options shorter than either side of the quad. The diagonal distance is the hypotenuse of a triangle, so it must be longer than 25 or 60 meters.

Then, we can either find our hypotenuse via Pythagorean Theorem, or our knowledge of Pythagorean Triples.

Using Pythagorean Theorem:

$c=\sqrt{25^2+60^2}=\sqrt{4225}=65$

So 65 meters.

Alternatively, recognize the triangle as a 5x/12x/13x triangle

so

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Question

Note: Diagram is NOT drawn to scale.

Refer to the above diagram.

Any of the following facts alone would be enough to prove that $\textrm{Quad} ; ABCD$ is not a parallelogram, EXCEPT:

Answer

Opposite sides of a parallelogram are congruent; if , then , violating this condition.

Consecutive angles of a parallelogram are supplementary; if $m \angle A = 114^{\circ}$ , then $m \angle A + m \angle B= 114 + 76 = 190$ , violating this condition.

Opposite angles of a parallelogram are congruent; if $m \angle D <76^{\circ}$ , then $m \angle D eq m \angle B$ , violating this condition.

Adjacent sides of a parallelogram, however, may or may not be congruent, so the condition that would not by itself prove that the quadrilateral is not a parallelogram.

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Question

Which of the following can not be the measures of the four interior angles of a quadrilateral?

Answer

The four interior angles of a quadrilateral measure a total of $360^{\circ }$ , so we test each group of numbers to see if they have this sum.

This last group does not have the correct sum, so it is the correct choice.

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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 a quadrilateral if and only if both pairs of its opposite angles are supplementary. An isosceles trapezoid has this characteristic; this can be proved by the fact that base angles are congruent, and by the Same-Side Interior Angles Statement. For a parallelogram to have this characteristic, since opposite angles are congruent also, all angles must measure $90^{\circ }$ ; the rectangle fits this description, but the rhombus does not.

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Question

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

Answer

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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Question

Rhombus has two diagonals that intersect at point ; $\angle RHO = 80 ^{\circ }$ .

What is $m \angle RPH$ ?

Answer

The diagonals of a rhombus always intersect at right angles, so $m \angle HPO = 90 ^{\circ }$ . The measures of the interior angles of the rhombus are irrelevant.

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Question

Quadrilateral is inscribed in circle $\bigodot$ . $m \angle A = 100 ^{\circ }$ . What is $m \angle Q$ ?

Answer

Two opposite angles of a quadrilateral inscribed inside a circle are supplementary, so

$m \angle Q = 180^{\circ } - m \angle A = 180^{\circ } - 100^{\circ } = 80^{\circ }$

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Question

Note: Figure NOT drawn to scale.

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

Answer

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:

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

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Question

Refer to the above figure. You are given that Polygon is a parallelogram, but NOT that it is a rectangle.

Which of the following statements is not enough to prove that Polygon is also a rectangle?

Answer

To prove that Polygon is also a rectangle, we need to prove that any one of its angles is a right angle.

If $\overline{BC} \perp \overline{DC}$ , then by definition of perpendicular lines, $\angle BCD$ is right.

If $m \angle 1 = 90^{\circ }$ , then, since $\angle 1$ and $\angle BCD$ form a linear pair, $\angle BCD$ is right.

If $(AD)^{2} + (DC)^{2} = (AC)^{2}$ , then, by the Converse of the Pythagorean Theorem, $\Delta ADC$ is a right triangle with right angle $\angle D$ .

If $\angle BAC$ and $\angle CAD$ are complementary angles, then, since $m \angle BAC + m \angle CAD =m \angle BA D$

$m \angle BAD = m \angle BAC + m \angle CAD = 90 ^{\circ }$ , making $m \angle BAD$ right.

However, since, by definition of a parallelogram, $\overline{AB } \parallel \overline{ CD }$ , by the Alternate Interior Angles Theorem, $\angle BAC \cong \angle ACD$ regardless of whether the parallelogram is a rectangle or not.

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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

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 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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