Calculating the length of the diagonal of a quadrilateral - 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

Given the area of a square is , what is the diagonal?

Answer

Write the formula for finding the area of a square given the diagonal.

$A=\frac{d^2}{2}$

Rearrange the equation so that the diagonal term is isolated.

$d=\sqrt{2A}$

Substitute the known area and simplify.

$d=\sqrt{2(9x^2)} = \sqrt{18x^2} = 3x\sqrt2$

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Question

What is the length of the diagonal for a square with a side length of ?

Answer

The diagonal of a square is simply the hypotenuse of a right triangle whose other two sides are the length and width of the square. Because all sides of a square are equal in length, this means the length and width are both , which gives us a right triangle with a base of and a height of , for which the hypotenuse is the diagonal of the square. Applying the Pythagorean Theorem to find the length of the diagonal, we have:

$c^2=32\rightarrow c=\sqrt{32}=\sqrt{2*16}=4\sqrt{2}$

So the length of the diagonal for a square with a side length of is $4\sqrt{2}$ . In general, we could check the length of the diagonal for any square with side length , and we would see that the diagonal length is always $a\sqrt{2}$ .

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Question

Calculate the length of the diagonal for a rectangle with a length of and a width of .

Answer

The diagonal of a rectangle can be thought of as the hypotenuse of a right triangle whose base and height are the length and width of the rectangle, respectively. This means we can use the Pythagorean Theorem to calculate the length of the diagonal for a rectangle:

$d^2=90\rightarrow d=\sqrt{90}=\sqrt{9\cdot10}=3\sqrt{10}$

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Question

Rhombus has area 72. The lengths of $\overline{RO}$ and $\overline{HM}$ are both whole numbers, and $\overline{RO}$ is the longer diagonal. Which of the following could be the length of $\overline{HM}$ ?

Answer

The area of a rhombus is half the product of the lengths of its diagonals, which here are $\overline{RO}$ and $\overline{HM}$ . This means

$\frac{1}{2} \cdot RO \cdot HM = 72$

$2 \cdot \frac{1}{2} \cdot RO \cdot HM =2 \cdot 72$

$RO \cdot HM =144$

Since both diagonals have whole numbers as their lengths, and , we are looking for to be a whole number that can be divided into 144 to yield a quotient less than . Equivalently,

The quotient is $HM< 144 \div HM$

Since we can multiply both sides by to yield the inequality

$(HM)^{2}< 144$ ,

we know that

,

so we can eliminate 12 and 16.

Also, since $144 \div 10 = 14 .4$ , 10 is not correct, as would not be a whole number.

If , then $RO = 144 \div 9 = 16$ . Both diagonals have lengths that are whole numbers, and $\overline{RO}$ is the longer diagonal. 9 is the correct choice.

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Question

Rhombus has perimeter 80; $m \angle R = 90 ^{\circ }$ . What is the length of $\overline{RO}$ ?

Answer

A rhombus has four sides of equal length. Since Rhombus has a right angle $\angle R$ , it follows that, the rhombus being a parallelogram, all four angles are right angles, and, by definition, Rhombus is a square. The length of a diagonal of a square is $\sqrt{2}$ times the length of a side; since the rhombus has perimeter 80, each side measures one fourth of this, or 20, and diagonal $\overline{RO}$ has length $20 \sqrt{2}$ .

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Question

Rhombus has perimeter 48; $m \angle R = 60 ^{\circ }$ . What is the length of $\overline{RO}$ ?

Answer

The referenced rhombus, along with diagonals $\overline{RO}$ and $\overline{HM}$ , is below.

The four sides of a rhombus have equal measure, so each side has measure one fourth of the perimeter of 48, which is 12.

Since consecutive angles of a rhombus, as with any other parallelogram, are suplementary, $\angle RHO$ and $\angle RMO$ have measure $120^{\circ }$ ; the diagonals bisect $\angle MRH$ and $\angle RHO$ into $30 ^{\circ }$ and $60 ^{\circ }$ angles, respectively, to form four 30-60-90 triangles. $\bigtriangleup HXR$ is one of them; by the 30-60-90 Triangle Theorem, $HX = \frac{1}{2} \cdot RH = \frac{1}{2} \cdot 12 =6$ ,

and

$RX = HX \cdot \sqrt{3} = 6 \sqrt{3}$ .

Since the diagonals of a rhombus bisect each other, $RO = 2 \cdot RX = 2 \cdot 6 \sqrt{3} = 12\sqrt{3}$ .

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Question

Rhombus has perimeter 64; $m \angle R = 60 ^{\circ }$ . What is the length of $\overline{HM}$ ?

Answer

The sides of a rhombus are all congruent; since the perimeter of Rhombus is 64, each side measures one fourth of this, or 16.

The referenced rhombus, along with diagonal $\overline{HM}$ , is below:

Since consecutive angles of a rhombus, as with any other parallelogram, are supplementary, $\angle RHO$ and $\angle RMO$ have measure $120^{\circ }$ ; $\overline{HM}$ bisects both into $60 ^{\circ }$ angles, making $\bigtriangleup RHM$ equilangular and, as a consequence, equilateral. Therefore, .

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Question

Rhombus has area 56.

Which of the following could be true about the values of and ?

Answer

The area of a rhombus is half the product of the lengths of its diagonals, which here are $\overline{RO}$ and $\overline{HM}$ . This means

$\frac{1}{2} \cdot RO \cdot HM = 56$

Therefore, we need to test each of the choices to find the pair of diagonal lengths for which this holds.

:

Area: $\frac{1}{2} \cdot RO \cdot HM = \frac{1}{2} \cdot 28 \times 28 = 392$

Area: $\frac{1}{2} \cdot RO \cdot HM = \frac{1}{2} \cdot 14 \times 14 =98$

Area: $\frac{1}{2} \cdot RO \cdot HM = \frac{1}{2} \cdot 4 \times 14 = 28$

Area: $\frac{1}{2} \cdot RO \cdot HM = \frac{1}{2} \cdot16 \times 7 =56$

is the correct choice.

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Question

Given: Quadrilateral such that $\overline{KI} \cong \overline{KE}$ , $\overline{TI} \cong \overline{TE}$ , $m \angle K = 60^{\circ }$ , $\angle T$ is a right angle, and diagonal $\overline{I E}$ has length 24.

Give the length of diagonal $\overline{KT}$ .

Answer

The Quadrilateral is shown below with its diagonals $\overline{I E}$ and $\overline{KT}$ .

. We call the point of intersection :

The diagonals of a quadrilateral with two pairs of adjacent congruent sides - a kite - are perpendicular; also, $\overline{KT}$ bisects the $60 ^{\circ }$ and $90^{\circ }$ angles of the kite. Consequently, $\bigtriangleup KXI$ is a 30-60-90 triangle and $\bigtriangleup TXI$ is a 45-45-90 triangle. Also, the diagonal that connects the common vertices of the pairs of adjacent sides bisects the other diagonal, making the midpoint of $\overline{IE}$ . Therefore,

$IX = \frac{1}{2} \cdot IE = \frac{1}{2} \cdot 24 = 12$ .

By the 30-60-90 Theorem, since $\overline{IX}$ and $\overline{KX}$ are the short and long legs of $\bigtriangleup KXI$ ,

$KX = IX \cdot \sqrt{3}= 12 \sqrt{3}$

By the 45-45-90 Theorem, since $\overline{IX}$ and $\overline{XT}$ are the legs of a 45-45-90 Theorem,

.

The diagonal $\overline{KT}$ has length

$KT =XT + KX= 12+ 12\sqrt{3}$ .

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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

Compare your answer with the correct one above

Question

Given the area of a square is , what is the diagonal?

Answer

Write the formula for finding the area of a square given the diagonal.

$A=\frac{d^2}{2}$

Rearrange the equation so that the diagonal term is isolated.

$d=\sqrt{2A}$

Substitute the known area and simplify.

$d=\sqrt{2(9x^2)} = \sqrt{18x^2} = 3x\sqrt2$

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Question

What is the length of the diagonal for a square with a side length of ?

Answer

The diagonal of a square is simply the hypotenuse of a right triangle whose other two sides are the length and width of the square. Because all sides of a square are equal in length, this means the length and width are both , which gives us a right triangle with a base of and a height of , for which the hypotenuse is the diagonal of the square. Applying the Pythagorean Theorem to find the length of the diagonal, we have:

$c^2=32\rightarrow c=\sqrt{32}=\sqrt{2*16}=4\sqrt{2}$

So the length of the diagonal for a square with a side length of is $4\sqrt{2}$ . In general, we could check the length of the diagonal for any square with side length , and we would see that the diagonal length is always $a\sqrt{2}$ .

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Question

Calculate the length of the diagonal for a rectangle with a length of and a width of .

Answer

The diagonal of a rectangle can be thought of as the hypotenuse of a right triangle whose base and height are the length and width of the rectangle, respectively. This means we can use the Pythagorean Theorem to calculate the length of the diagonal for a rectangle:

$d^2=90\rightarrow d=\sqrt{90}=\sqrt{9\cdot10}=3\sqrt{10}$

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Question

Rhombus has area 72. The lengths of $\overline{RO}$ and $\overline{HM}$ are both whole numbers, and $\overline{RO}$ is the longer diagonal. Which of the following could be the length of $\overline{HM}$ ?

Answer

The area of a rhombus is half the product of the lengths of its diagonals, which here are $\overline{RO}$ and $\overline{HM}$ . This means

$\frac{1}{2} \cdot RO \cdot HM = 72$

$2 \cdot \frac{1}{2} \cdot RO \cdot HM =2 \cdot 72$

$RO \cdot HM =144$

Since both diagonals have whole numbers as their lengths, and , we are looking for to be a whole number that can be divided into 144 to yield a quotient less than . Equivalently,

The quotient is $HM< 144 \div HM$

Since we can multiply both sides by to yield the inequality

$(HM)^{2}< 144$ ,

we know that

,

so we can eliminate 12 and 16.

Also, since $144 \div 10 = 14 .4$ , 10 is not correct, as would not be a whole number.

If , then $RO = 144 \div 9 = 16$ . Both diagonals have lengths that are whole numbers, and $\overline{RO}$ is the longer diagonal. 9 is the correct choice.

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Question

Rhombus has perimeter 80; $m \angle R = 90 ^{\circ }$ . What is the length of $\overline{RO}$ ?

Answer

A rhombus has four sides of equal length. Since Rhombus has a right angle $\angle R$ , it follows that, the rhombus being a parallelogram, all four angles are right angles, and, by definition, Rhombus is a square. The length of a diagonal of a square is $\sqrt{2}$ times the length of a side; since the rhombus has perimeter 80, each side measures one fourth of this, or 20, and diagonal $\overline{RO}$ has length $20 \sqrt{2}$ .

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Question

Rhombus has perimeter 48; $m \angle R = 60 ^{\circ }$ . What is the length of $\overline{RO}$ ?

Answer

The referenced rhombus, along with diagonals $\overline{RO}$ and $\overline{HM}$ , is below.

The four sides of a rhombus have equal measure, so each side has measure one fourth of the perimeter of 48, which is 12.

Since consecutive angles of a rhombus, as with any other parallelogram, are suplementary, $\angle RHO$ and $\angle RMO$ have measure $120^{\circ }$ ; the diagonals bisect $\angle MRH$ and $\angle RHO$ into $30 ^{\circ }$ and $60 ^{\circ }$ angles, respectively, to form four 30-60-90 triangles. $\bigtriangleup HXR$ is one of them; by the 30-60-90 Triangle Theorem, $HX = \frac{1}{2} \cdot RH = \frac{1}{2} \cdot 12 =6$ ,

and

$RX = HX \cdot \sqrt{3} = 6 \sqrt{3}$ .

Since the diagonals of a rhombus bisect each other, $RO = 2 \cdot RX = 2 \cdot 6 \sqrt{3} = 12\sqrt{3}$ .

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Question

Rhombus has perimeter 64; $m \angle R = 60 ^{\circ }$ . What is the length of $\overline{HM}$ ?

Answer

The sides of a rhombus are all congruent; since the perimeter of Rhombus is 64, each side measures one fourth of this, or 16.

The referenced rhombus, along with diagonal $\overline{HM}$ , is below:

Since consecutive angles of a rhombus, as with any other parallelogram, are supplementary, $\angle RHO$ and $\angle RMO$ have measure $120^{\circ }$ ; $\overline{HM}$ bisects both into $60 ^{\circ }$ angles, making $\bigtriangleup RHM$ equilangular and, as a consequence, equilateral. Therefore, .

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Question

Rhombus has area 56.

Which of the following could be true about the values of and ?

Answer

The area of a rhombus is half the product of the lengths of its diagonals, which here are $\overline{RO}$ and $\overline{HM}$ . This means

$\frac{1}{2} \cdot RO \cdot HM = 56$

Therefore, we need to test each of the choices to find the pair of diagonal lengths for which this holds.

:

Area: $\frac{1}{2} \cdot RO \cdot HM = \frac{1}{2} \cdot 28 \times 28 = 392$

Area: $\frac{1}{2} \cdot RO \cdot HM = \frac{1}{2} \cdot 14 \times 14 =98$

Area: $\frac{1}{2} \cdot RO \cdot HM = \frac{1}{2} \cdot 4 \times 14 = 28$

Area: $\frac{1}{2} \cdot RO \cdot HM = \frac{1}{2} \cdot16 \times 7 =56$

is the correct choice.

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Question

Given: Quadrilateral such that $\overline{KI} \cong \overline{KE}$ , $\overline{TI} \cong \overline{TE}$ , $m \angle K = 60^{\circ }$ , $\angle T$ is a right angle, and diagonal $\overline{I E}$ has length 24.

Give the length of diagonal $\overline{KT}$ .

Answer

The Quadrilateral is shown below with its diagonals $\overline{I E}$ and $\overline{KT}$ .

. We call the point of intersection :

The diagonals of a quadrilateral with two pairs of adjacent congruent sides - a kite - are perpendicular; also, $\overline{KT}$ bisects the $60 ^{\circ }$ and $90^{\circ }$ angles of the kite. Consequently, $\bigtriangleup KXI$ is a 30-60-90 triangle and $\bigtriangleup TXI$ is a 45-45-90 triangle. Also, the diagonal that connects the common vertices of the pairs of adjacent sides bisects the other diagonal, making the midpoint of $\overline{IE}$ . Therefore,

$IX = \frac{1}{2} \cdot IE = \frac{1}{2} \cdot 24 = 12$ .

By the 30-60-90 Theorem, since $\overline{IX}$ and $\overline{KX}$ are the short and long legs of $\bigtriangleup KXI$ ,

$KX = IX \cdot \sqrt{3}= 12 \sqrt{3}$

By the 45-45-90 Theorem, since $\overline{IX}$ and $\overline{XT}$ are the legs of a 45-45-90 Theorem,

.

The diagonal $\overline{KT}$ has length

$KT =XT + KX= 12+ 12\sqrt{3}$ .

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