Other Quadrilaterals - GMAT Quantitative

Any two consecutive angles of a parallelogram are supplementary, so if one angle has measure , all angles can be proven to have measure . This is the definition of a rectangle. Statement 1 therefore proves the parallelogram to be a rectangle.

Also, a parallelogram is a rectangle if and only if its diagonals are congruent, which is what Statement 2 asserts.

From either statement, it follows that parallelogram is a rectangle.

A circle can be circumscribed about a quadrilateral if and only if both pairs of opposite angles are supplementary. This is not proved or disproved by Statement 1 alone:

Case 1:

This is not a rectangle, and opposite angles are supplementary, so a circle can be constructed to circumscribe the quadrilateral.

Case 2:

This is not a rectangle, and opposite angles are not supplementary, so a circle cannot be constructed to circumscribe the quadrilateral.

From Statement 2, however, it follows that a two opposite angles are not a supplementary pair, so a circle cannot be circumscribed about it.

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. Assume without loss of generality that and are the pairs of base angles.

Then, since base angles are congruent, and . Since the bases of a trapezoid are parallel, from the Same-Side Interior Angles Theorem, and are supplementary, and, subsequently, so are and , as well as and .

If , then and form a supplementary pair, as their measures total ; since the measures of the angles of a quadrilateral total , the measures of and also total , making them supplementary as well.

Therefore, it follows from either statement that both pairs of opposite angles are supplementary, and that a circle can be circumscribed about the quadrilateral.

and , the diagonals of Parallelogram , are perpendicular if and only if Parallelogram is also a rhombus.

Opposite sides of a parallelogram are congruent, so if Statement 1 is assumed, . Parallelogram a rhombus; subsequently, .

The angle measures are irrelevant, so Statement 2 is unhelpful.

For the diagonals of a quadrilateral to be perpendicular, the quadrilateral must be a kite or a rhombus - in either case, there must be two pairs of adjacent congruent sides. Neither statement alone proves this, but both statements together do.

From Statement 1 alone, we can calculate , since two opposite angles of a quadrilateral inscribed inside a circle are supplementary:

From Statement 2 alone, we can calculate , since the degree measure of an inscribed angle of a circle is half that of the arc it intersects:

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

If we assume Statement 1 alone, that , then, since and form a linear pair, is right.

If we assume Statement 2 alone, that , it follows from the converse of the Pythagorean Theorem that is a right triangle with right angle .

Either way, we have proved that the parallelogram is a rectangle.

It is necessary and sufficient to prove that one of the angles of the parallelogram is a right angle.

Assume Statement 1 alone - that . and are supplementary, since they are same-side interior angles of parallel lines. Since , is also supplementary to . But as corresponding angles of parallel lines, . Two angles that are conruent and supplementary are both right angles, so is a right angle.

Assume Statement 2 alone - that and are complementary angles, or, equivalently, . Since the angles of a triangle have measures that add up to , the third angle of , which is , measures , and is a right angle.

Either statement alone proves a right angle and subsequently proves a rectangle.

Assume Statement 1 alone. By congruence, and , making Quadrilateral a parallelogram. However, no clue is given to whether any angles are right or not, so whether the quadrilateral is a rectangle or not remains open.

Assume Statement 2 alone. By congruence, opposite sides , but no clue is provided as to the lengths of opposite sides and . Also, , but no clue is provided as to whether the angles are right. A rectangle would have both characteristics, but so would an isosceles trapezoid with legs and .

Assume both statements are true. Quadrilateral is a parallelogram as a consequence of Statement 1. Since and are consecutive angles of the parallelogram, they are supplementary, but they are also congruent as a consequence of Statement 2. Therefore, they are right angles, and a parallelogram with right angles is a rectangle.

Statement 1 alone is insufficient to answer the question; A quadrilateral in which and are right angles, , and fits the statement, as well as a rectangle, which by defintion has four right angles.

Statement 2 alone is insufficient as well, as a parallelogram with acute and obtuse angles, as well as a rectangle, fits the description.

Assume both statements, and construct diagonal to form two triangles and . By Statement 1, both triangles are right with congruent legs , and congruent hypotenuses, both being the same segment . By the Hypotenuse Leg Theorem, . By congruence, . The quadrilateral, having two sets of congruent opposite sides, is a parallelogram; a parallelogram with right angles is a rectangle.

Each diagonal divides Rhombus into two triangles, both isosceles.

Statement 1 alone establishes the perimeter of one such triangle. However, it does not make it clear what equal side lengths and and diagonal length are. For example, fits the perimeter, but so does .

Statement 2 alone gives no information about the actual lengths of the sides.

Assume both statements are true. Since is equilateral, . It follows that , and . Also, the diagonals of a rhombus bisect their angles and are each other's perpendicular bisectors, so the rhombus, with their diagonals, is given below.

has perimeter , which means that

Since is known to be a triangle, the proportions of the side lengths are known; along with the above equation, , and, subsequently, the perimeter, can be determined.

To find the perimeter of the quadrilateral, we need to know whether it is of a special type of quadrilaterals and we need to know the length of the sides.

Statement 1 tells us only that the quadrilateral is a rhombus. Indeed, a quadrilateral with perpendicular diagonals intersecting at their midpoint must be a rhombus. However we don't know any length of the sides.

Statement 2 says gives us the length us two consecutive sides. It could be tempting to answer that it is sufficient, however, we can't conclude that the quadrilateral has equal lengths. Therefore this statement alone is insufficient.

Both statements together are sufficient since we can conclude that the quadrilateral is a rhombus, and twice will give us the perimeter.

To find perimeter, we need to find the length of all the sides. Recall that rectangles are made up of two pairs of equal sides.

I) Relates one side to another non-equivalent side.

II) Gives us side , which must be equivalent to .

Use II) and I) to find all the side lengths, then add them up. Both are needed.

Recap:

Consider rectangle CONT

I) Side CO is three fourths of side ON

II) Side NT is 15.7 meters long

What is the perimeter of CONT?

Because we are dealing with a rectangle, we know the following:

Find perimeter with:

Use I) and II) to write the following equation:

So:

And finally:

Statement 1): The area of the rectangle is 24.

Write the area for a rectangle and substitute the value of the area.

The length and width of the rectangle are unknown, and each set of dimensions will provide a different perimeter. This statement is insufficient to find the perimeter of the rectangle.

Statement 2): The diagonal of the rectangle is 5.

Given the diagonal of the rectangle, the Pythagorean Theorem can be used to solve for the diagonal. Express the equation in terms of length and width.

Similar to the case in Statement 1), both the length and width are unknown, and the equation by itself is insufficient to solve for the perimeter of the rectangle.

Attempting to use both equations: and to solve for length and width will yield complex numbers as part of the solution.

Therefore:

To find the diagonal of a rectangle, we need the length of two sides.

I) Gives us the length of one side.

II) Lets us find the length of the next side.

Use I) and II) with Pythagorean Theorem to find the diagonal.

Conversely, recognize that we are making a 3/4/5 Pythagorean Triple and see that the last side is 30 fathoms.

Statement 1: The perimeter of the square is known.

A square has four equal sides. Write the perimeter formula for squares.

The side length of a square is a fourth of the perimeter.

Statement 2: A secondary square with a known area encloses the primary square where all 4 corners of the primary square touch each of secondary square's edges.

If the primary square corners touch each of the secondary square edges, they must touch at the midpoint of each edge. Since Statement 2 mentions that the secondary square area is known, it is possible to solve for the edge length and the diagonal of the secondary square. Write the formula for the area of a square.

The diagonal of the secondary square can be solved by using the Pythagorean Theorem.

The side length of the secondary square also must equal the diagonal of the primary square.

Therefore:

Each of the statements, 1 and 2, provide something integral to calculating the length of a diagonal: an angle and the length of the sides connected to its vertex. The diagonal would be the third leg of the resulting triangle, and can be calculated using the law of cosines:

Since this is a parellelogram, knowing one angle allows us to know all the angles.

Area alone is not enough information. Imagine, for instance, a parallelogram with a shorter side of and a longer side of . The diagonal would be well above

However, with the perimeter, the smaller and larger sides must add up to one half of it, .

The longer diagonal reaches its maximum with the larger internal angle widens towards degrees, and the parallelogram flattens into a line. Using the law of cosines, this translates to:

.

At its max, the diagonal could be no greater than

Knowing one pair of sides of a quadrilateral to be congruent is not alone sufficient to prove that figure to be a parallelogram, nor is knowing one pair of sides of a quadrilateral to be parallel. But knowing both about the same pair of sides is sufficient by a theorem of parallelograms.

Consider the area formula for a trapezoid:

is the _midsegment_of the trapezoid, which, by the Trapezoid Midsegment Theorem, has length equal to the mean of the bases - in other words, . The area formula can be expressed, after substitution, as

So, if you know both the area and - but not just one - you can find the height by dividing.