Calculating whether quadrilaterals are similar - GMAT Quantitative

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Question

Which of the following rectangles is similar to one with a length of and a width of ?

Answer

In order for two rectangles to be similar, the ratio of their dimensions must be equal. We can check which dimensions are those of a rectangle similar to the given one by first calculating the ratio of the length to the width for the given rectangle, and then doing the same for each of the answer choices until we find which has an equal ratio between its dimensions:

$Given:\frac{L}{W}=\frac{16}{10}=\frac{8}{5}$

So in order for a rectangle to be similar to the given rectangle, this must be the ratio of its length to its width. Now we check the answer choices, in no particular order, for one with this ratio:

$Option1:\frac{L}{W}=\frac{4}{3}$

$Option2:\frac{L}{W}=\frac{18}{6}=3$

$Option3:\frac{L}{W}=\frac{20}{12}=\frac{5}{3}$

$Option4:\frac{L}{W}=\frac{6}{4}=\frac{3}{2}$

$Option5:\frac{L}{W}=\frac{32}{20}=\frac{8}{5}$

We can see that only the rectangle with a length of and a width of has the same ratio as the given rectangle, so this is the similar one.

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Question

Which of the following dimensions would a rectangle need to have in order to be similar to one with a length of and a width of ?

Answer

In order for two rectangles to be similar, the ratio of their dimensions must be equal. We can calculate the ratio of length to width for the given rectangle, and then check the answer choices for the rectangle whose dimensions have the same ratio:

$Given:\frac{L}{W}=\frac{21}{9}=\frac{7}{3}$

Now we check the answer choices, in no particular order, and the dimensions with the same ratio are those of the rectangle that is similar:

$Option1:\frac{L}{W}=\frac{10}{4}=\frac{5}{2}$

$Option2:\frac{L}{W}=\frac{15}{8}$

$Option3:\frac{L}{W}=\frac{16}{9}$

$Option4:\frac{L}{W}=\frac{18}{10}=\frac{9}{5}$

$Option5:\frac{L}{W}=\frac{14}{6}=\frac{7}{3}$

We can see that a rectangle with a length of and a width of has the same ratio of dimensions as the given rectangle, so this is the one that is similar.

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Question

Refer to the above Trapezoid . There exists Trapezoid such that

Trapezoid $ABCD \sim$ Trapezoid , and $\overline{EH}$ has length 60.

Give the length of $\overline{EF}$ .

Answer

Construct the perpendicular segment from to $\overline{CD}$ and let be its point of intersection with $\overline{CD}$ . By construction, the trapezoid is divided into Rectangle and right triangle $\bigtriangleup AZD$ . Since opposite sides of a rectangle are congruent, and ; as a consequence of the latter, . By the Pythagrean Theorem, the length of the hypotenuse $\overline{AD}$ of right triangle $\bigtriangleup AZD$ can be calculated from the length of legs $\overline{AZ}$ and $\overline{ZD}$ :

$AD = \sqrt{(AZ)^{2}+(DZ)^{2}} = \sqrt{3^{2}+4^{2}} =\sqrt{9+16} = \sqrt{25} = 5$

The figure, with the segment and the calculated measurements, is below.

Since Trapezoid $ABCD \sim$ Trapezoid , by proportionality of corresponding sides of similar figures:

$\frac{EF}{AB}=\frac{EH}{AD}$

$\frac{EF}{12}=\frac{60}{5} =12$

$\frac{EF}{12} \cdot 12 =12 \cdot 12$

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Question

Refer to the above Trapezoid . There exists Trapezoid such that

Trapezoid $ABCD \sim$ Trapezoid , and the length of the midsegment of Trapezoid is 91.

Give the length of $\overline{FG}$ .

Answer

The length of the midsegment of a trapezoid - the segment that has as its endpoints the midpoints of its legs - is half the sum of the lengths of its legs. Therefore, Trapezoid has as the length of its midsegment

$\frac{1}{2} (AB + CD) = \frac{1}{2} (12+16) = \frac{1}{2} \cdot 28 = 14$ .

Sidelengths of similar figures are in proportion. If the similarity ratio is , then the bases of Trapezoid have length and , so their midsegment will have length

$\frac{1}{2} (EF+ GH) = \frac{1}{2} (12R+16R) = \frac{1}{2} \cdot 28R = 14R$ ,

meaning that the ratio of the lengths of the midsegments will be the same as the similarity ratio. Since the length of the midsegment of Trapezoid is 91, this similarity ratio is

$\frac{91}{14}= \frac{13}{2}$ .

The ratio of the length of $\overline{FG}$ to that of corresponding side $\overline{BC}$ is therefore $\frac{13}{2}$ , so

$\frac{FG}{BC} = \frac{13}{2}$

$\frac{FG}{3} = \frac{13}{2}$

$\frac{FG}{3} \cdot 3= \frac{13}{2} \cdot 3$

$FG= \frac{39}{2} = 19 \frac{1}{2}$

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Question

Refer to the above Trapezoid . There exists Trapezoid such that

Trapezoid $ABCD \sim$ Trapezoid , and $\overline{EF}$ has length 66.

To the nearest whole, give the area of Trapezoid .

Answer

The similarity ratio of the trapezoids is the ratio of the length of one side to that of the corresponding side of the other. For these trapezoids, we take the ratio of the lengths of corresponding sides $\overline{EF}$ and $\overline{AB}$ :

$\frac{EF}{AB} = \frac{66}{12} = \frac{11}{2}$ .

The ratio of the areas is the square of this, or

$\left ( \frac{11}{2} \right )^{2} = \frac{121}{4}$ .

The area of Trapezoid is one half the product of the height and the sum of bases and :

$A _{1} = \frac{1}{2} \cdot BC \cdot (AB + CD)$

$= \frac{1}{2} \cdot 3 \cdot (12+ 16)$

$= \frac{1}{2} \cdot 3 \cdot 28$

Multiply this by the area ratio:

$A _{2} = A _{1} \cdot \frac{121}{4}$

$A _{2} = 42 \cdot \frac{121}{4} = 1,270\frac{1}{2}$ .

The correct response is 1,271.

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Question

Rhombus $RHOM \sim$ Rhombus . Rhombus has perimeter 80; Rhombus has perimeter 180; .

Find the length of diagonal $\overline{SP}$ .

Answer

A rhombus has four sides the same length, so each side of Rhombus has length one fourth of 80, or 20; each side of Rhombus has length one fourth of 180, or 45.

The diagonals of a rhombus are each other's perpendicular bisectors, so, if we let be the point of intersection of the diagonals of Rhombus , $\overline{RO}$ and $\overline{HM}$ , we form four congruent right triangles.

We will examine $\bigtriangleup RXM$ . ; and, $XM = \frac{1}{2} \cdot HM = \frac{1}{2} \cdot 32 = 16$ .

By the Pythagorean Theorem,

$RX = \sqrt{(RM)^{2}-(RX)^{2}}$

$= \sqrt{20^{2}-16^{2}}$

$= \sqrt{400-256}$

$= \sqrt{144}$

and $RO = 2 \cdot 12 = 24$

Since corresponding sides of similar figures are in proportion, so are corresponding diagonals. Therefore,

$\frac{SP}{RO} = \frac{SN}{RM}$

$\frac{SP}{24} = \frac{45}{20}$

$\frac{SP}{24} \cdot 24= \frac{45}{20} \cdot 24$

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Question

Rhombus $RHOM \sim$ Rhombus . $m \angle R = 60^{\circ }$ ; ; .

Give the ratio of the area of Rhombus to that of Rhombus .

Answer

The angles of the rhombus measure:

$m \angle R = 60 ^{\circ }$ ;

$m \angle O = 60 ^{\circ }$ , since opposite angles of a rhombus, as in any other parallelogram, are congruent;

$m \angle H = 120^{\circ }$ and $m \angle M = 120^{\circ }$ , since consecutive angles of a rhombus are supplementary (the sum of their degree measures is 180).

The diagonals of a rhombus are each other's perpendicular bisectors as well as the bisectors of the angles, so $\overline{HM}$ and $\overline{RO}$ , whose point of intersection we will call , divide Rhombus into four 30-60-90 triangles. If we examine $\bigtriangleup RXH$ , we see that its short leg $\overline{HX}$ has length half that of $\overline{HM}$ ; , so . By the 30-60-90 Triangle Theorem, long leg $\overline{RX}$ has length $\sqrt{3}$ this, or $6 \sqrt{3}$ , and the diagonal $\overline{RO}$ has measure twice this, or $12\sqrt{3}$ .

The ratio of the lengths of corresponding diagonals of the rhombuses is the same as the similarity ratios of the sides, so the similarity ratio of Rhombus to Rhombus is

$\frac{SP }{RO} = \frac{72}{12\sqrt{3}}= \frac{6}{\sqrt{3}}$

The ratio of the areas is the square of the similarity ratio, which is

$\left ( \frac{6}{\sqrt{3}} \right )^{2}= \frac{36}{3} = \frac{12}{1}$

That is, 12 to 1.

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Question

In Quadrilateral , $\overline{KI} \cong \overline{KE}$ , $\overline{TI} \cong \overline{TE}$ , $m \angle K = 60^{\circ }$ , and $\angle T$ is a right angle; .

There exists Quadrilateral such that Quadrilateral $KITE \sim$ Quadrilateral , and .

Give the length of $\overline{LF}$ .

Answer

The Quadrilateral with its diagonals is shown below. 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, two congruent 30-60-90 triangles, $\bigtriangleup KXE$ and $\bigtriangleup KXI$ , and two congruent 45-45-90 triangles , $\bigtriangleup TXE$ and $\bigtriangleup TXI$ , are formed; also, $\bigtriangleup TIE$ , being an isosceles right triangle, is a 45-45-90 triangle. $\overline{KE}$ , the hypotenuse of $\bigtriangleup KXE$ , has length 16, so by the 30-60-90 Triangle Theorem, its shorter leg $\overline{XE}$ has length half this, or 8. Also, $\overline{XE}$ is a leg of $\bigtriangleup TXE$ , so by the 45-45-90 Theorem, the hypotenuse $\overline{TE}$ has length $\sqrt{2}$ times this, or $8\sqrt{2}$ .

Corresponding sides of similar quadrilaterals are in proportion, so

$\frac{LF}{KE} = \frac{UF}{TE}$

$\frac{LF}{16} = \frac{24}{8 \sqrt{2}}$

$\frac{LF}{16} \cdot 16 = \frac{24}{8 \sqrt{2}} \cdot 16$

$LF= \frac{48}{ \sqrt{2}} = \frac{48 \cdot \sqrt{2}}{ \sqrt{2} \cdot \sqrt{2}}= \frac{48 \sqrt{2}}{2} = 24 \sqrt{2}$

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Question

Rhombus $RHOM \sim$ Rhombus . ; Rhombus has area 90; Rhombus has area 360. What is the length of diagonal $\overline{IN}$ ?

Answer

The area of a rhombus is half the product of the lengths of its diagonals, so we can take advantage of this to find $\overline{HM}$ from the known measurements of Rhombus :

$A_{1} = \frac{1}{2} (RO)(HM)$

$90= \frac{1}{2} \cdot 15 \cdot HM$

$90= \frac{15}{2} \cdot HM$

$\frac{2} {15}\cdot 90= \frac{2} {15}\cdot \frac{15}{2} \cdot HM$

The ratio of the area of Rhombus to that of Rhombus is 4, so their similarity ratio is the square root of this, or 2. We can calculate now:

$\frac{IN}{HM} = 2$

$\frac{IN}{12} = 2$

$\frac{IN}{12}\cdot 12 = 2 \cdot 12$

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Question

In Quadrilateral , $\overline{KI} \cong \overline{KE}$ , $\overline{TI} \cong \overline{TE}$ , $m \angle K = 60^{\circ }$ , $\angle T$ is a right angle.

There exists Quadrilateral such that Quadrilateral $KITE \sim$ Quadrilateral , and .

Which of the following is true about the areas of the two quadrilaterals?

Answer

We will assume that $\overline{KI}$ and $\overline{JU}$ have common measure 1 for the sake of simplcity; this reasoning is independent of the actual measure of $\overline{KI}$ .

The Quadrilateral with its diagonals is shown below. 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. By the 30-60-90 Theorem, since $\overline{ IX}$ and $\overline{KI}$ are the short leg and hypotenuse of $\bigtriangleup KXI$ ,

$IX = \frac{1}{2} \cdot KI = \frac{1}{2} \cdot 1 = \frac{1}{2}$ .

By the 45-45-90 Theorem, since $\overline{ IX}$ and $\overline{IT}$ are a leg and the hypotenuse of $\bigtriangleup TXI$ ,

$IT = IX \cdot \sqrt{2} =\frac{1}{2} \sqrt{2} = \frac{ \sqrt{2} }{2}$

The similarity ratio of Quadrilateral to Quadrilateral can be found by finding the ratio of the length of side $\overline{JU}$ to corresponding side $\overline{IT}$ :

$\frac{JU}{IT}= \frac{1}{ \frac{\sqrt{2}}{2} } = \frac {2}{\sqrt{2}}$

The ratio of the areas is the square of the similarity ratio:

$\left ( \frac {2}{\sqrt{2}} \right )^{2}= \frac{2^{2}}{\left ( \sqrt{2} \right )^{2}} = \frac{4}{2}=2$

The correct choice is that Quadrilateral has area twice that of Quadrilateral .

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Question

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

There exists Quadrilateral such that Quadrilateral $KITE \sim$ Quadrilateral , and .

Give the length of $\overline{LJ}$ .

Answer

The Quadrilateral with its diagonals is shown below. We call the point of intersection :

The diagonals of a quadrilateral with two pairs of adjacent congruent sides - a kite - are perpendicular. $\overline{KT}$ , the diagonal that connects the common vertices of the pairs of adjacent sides, bisects the other diagonal, making the midpoint of $\overline{IE}$ . Therefore, $IX= \frac{1}{2} \cdot IE = \frac{1}{2} \cdot 24 = 12$ .

$\overline{KT}$ also bisects the $60 ^{\circ }$ and $90^{\circ }$ angles of the kite, so the result is that two 30-60-90 triangles, $\bigtriangleup KXE$ and $\bigtriangleup KXI$ , and two 45-45-90 triangles , $\bigtriangleup TXE$ and $\bigtriangleup TXI$ , are formed; also, $\bigtriangleup TIE$ , being isosceles, is a 45-45-90 triangle.

Examine $\bigtriangleup KXI$ . Since its short leg $\overline{IX}$ has length 12, by the 30-60-90 Theorem, its hypotenuse, $\overline{KI}$ , has twice this length, or 24.

Examine $\bigtriangleup TXI$ . Since a leg $\overline{IX}$ has length 12, by the 45-45-90 Theorem, its hypotenuse, $\overline{IT}$ , has length $\sqrt{2}$ times this, or $12\sqrt{2}$ .

Since by similarity, corresponding sides are in proportion,

$\frac{LJ}{KI}= \frac{JU}{IT}$

$\frac{LJ}{24}= \frac{36}{12\sqrt{2}}$

$\frac{LJ}{24} \cdot 24 = \frac{36}{12\sqrt{2}} \cdot 24$

$LJ = \frac{72}{ \sqrt{2}} = \frac{72\cdot \sqrt{2}}{ \sqrt{2}\cdot \sqrt{2}} = \frac{72 \sqrt{2}}{2}= 36 \sqrt{2}$

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Question

Which of the following rectangles is similar to one with a length of and a width of ?

Answer

In order for two rectangles to be similar, the ratio of their dimensions must be equal. We can check which dimensions are those of a rectangle similar to the given one by first calculating the ratio of the length to the width for the given rectangle, and then doing the same for each of the answer choices until we find which has an equal ratio between its dimensions:

$Given:\frac{L}{W}=\frac{16}{10}=\frac{8}{5}$

So in order for a rectangle to be similar to the given rectangle, this must be the ratio of its length to its width. Now we check the answer choices, in no particular order, for one with this ratio:

$Option1:\frac{L}{W}=\frac{4}{3}$

$Option2:\frac{L}{W}=\frac{18}{6}=3$

$Option3:\frac{L}{W}=\frac{20}{12}=\frac{5}{3}$

$Option4:\frac{L}{W}=\frac{6}{4}=\frac{3}{2}$

$Option5:\frac{L}{W}=\frac{32}{20}=\frac{8}{5}$

We can see that only the rectangle with a length of and a width of has the same ratio as the given rectangle, so this is the similar one.

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Question

Which of the following dimensions would a rectangle need to have in order to be similar to one with a length of and a width of ?

Answer

In order for two rectangles to be similar, the ratio of their dimensions must be equal. We can calculate the ratio of length to width for the given rectangle, and then check the answer choices for the rectangle whose dimensions have the same ratio:

$Given:\frac{L}{W}=\frac{21}{9}=\frac{7}{3}$

Now we check the answer choices, in no particular order, and the dimensions with the same ratio are those of the rectangle that is similar:

$Option1:\frac{L}{W}=\frac{10}{4}=\frac{5}{2}$

$Option2:\frac{L}{W}=\frac{15}{8}$

$Option3:\frac{L}{W}=\frac{16}{9}$

$Option4:\frac{L}{W}=\frac{18}{10}=\frac{9}{5}$

$Option5:\frac{L}{W}=\frac{14}{6}=\frac{7}{3}$

We can see that a rectangle with a length of and a width of has the same ratio of dimensions as the given rectangle, so this is the one that is similar.

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Question

Refer to the above Trapezoid . There exists Trapezoid such that

Trapezoid $ABCD \sim$ Trapezoid , and $\overline{EH}$ has length 60.

Give the length of $\overline{EF}$ .

Answer

Construct the perpendicular segment from to $\overline{CD}$ and let be its point of intersection with $\overline{CD}$ . By construction, the trapezoid is divided into Rectangle and right triangle $\bigtriangleup AZD$ . Since opposite sides of a rectangle are congruent, and ; as a consequence of the latter, . By the Pythagrean Theorem, the length of the hypotenuse $\overline{AD}$ of right triangle $\bigtriangleup AZD$ can be calculated from the length of legs $\overline{AZ}$ and $\overline{ZD}$ :

$AD = \sqrt{(AZ)^{2}+(DZ)^{2}} = \sqrt{3^{2}+4^{2}} =\sqrt{9+16} = \sqrt{25} = 5$

The figure, with the segment and the calculated measurements, is below.

Since Trapezoid $ABCD \sim$ Trapezoid , by proportionality of corresponding sides of similar figures:

$\frac{EF}{AB}=\frac{EH}{AD}$

$\frac{EF}{12}=\frac{60}{5} =12$

$\frac{EF}{12} \cdot 12 =12 \cdot 12$

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Question

Refer to the above Trapezoid . There exists Trapezoid such that

Trapezoid $ABCD \sim$ Trapezoid , and the length of the midsegment of Trapezoid is 91.

Give the length of $\overline{FG}$ .

Answer

The length of the midsegment of a trapezoid - the segment that has as its endpoints the midpoints of its legs - is half the sum of the lengths of its legs. Therefore, Trapezoid has as the length of its midsegment

$\frac{1}{2} (AB + CD) = \frac{1}{2} (12+16) = \frac{1}{2} \cdot 28 = 14$ .

Sidelengths of similar figures are in proportion. If the similarity ratio is , then the bases of Trapezoid have length and , so their midsegment will have length

$\frac{1}{2} (EF+ GH) = \frac{1}{2} (12R+16R) = \frac{1}{2} \cdot 28R = 14R$ ,

meaning that the ratio of the lengths of the midsegments will be the same as the similarity ratio. Since the length of the midsegment of Trapezoid is 91, this similarity ratio is

$\frac{91}{14}= \frac{13}{2}$ .

The ratio of the length of $\overline{FG}$ to that of corresponding side $\overline{BC}$ is therefore $\frac{13}{2}$ , so

$\frac{FG}{BC} = \frac{13}{2}$

$\frac{FG}{3} = \frac{13}{2}$

$\frac{FG}{3} \cdot 3= \frac{13}{2} \cdot 3$

$FG= \frac{39}{2} = 19 \frac{1}{2}$

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Question

Refer to the above Trapezoid . There exists Trapezoid such that

Trapezoid $ABCD \sim$ Trapezoid , and $\overline{EF}$ has length 66.

To the nearest whole, give the area of Trapezoid .

Answer

The similarity ratio of the trapezoids is the ratio of the length of one side to that of the corresponding side of the other. For these trapezoids, we take the ratio of the lengths of corresponding sides $\overline{EF}$ and $\overline{AB}$ :

$\frac{EF}{AB} = \frac{66}{12} = \frac{11}{2}$ .

The ratio of the areas is the square of this, or

$\left ( \frac{11}{2} \right )^{2} = \frac{121}{4}$ .

The area of Trapezoid is one half the product of the height and the sum of bases and :

$A _{1} = \frac{1}{2} \cdot BC \cdot (AB + CD)$

$= \frac{1}{2} \cdot 3 \cdot (12+ 16)$

$= \frac{1}{2} \cdot 3 \cdot 28$

Multiply this by the area ratio:

$A _{2} = A _{1} \cdot \frac{121}{4}$

$A _{2} = 42 \cdot \frac{121}{4} = 1,270\frac{1}{2}$ .

The correct response is 1,271.

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Question

Rhombus $RHOM \sim$ Rhombus . Rhombus has perimeter 80; Rhombus has perimeter 180; .

Find the length of diagonal $\overline{SP}$ .

Answer

A rhombus has four sides the same length, so each side of Rhombus has length one fourth of 80, or 20; each side of Rhombus has length one fourth of 180, or 45.

The diagonals of a rhombus are each other's perpendicular bisectors, so, if we let be the point of intersection of the diagonals of Rhombus , $\overline{RO}$ and $\overline{HM}$ , we form four congruent right triangles.

We will examine $\bigtriangleup RXM$ . ; and, $XM = \frac{1}{2} \cdot HM = \frac{1}{2} \cdot 32 = 16$ .

By the Pythagorean Theorem,

$RX = \sqrt{(RM)^{2}-(RX)^{2}}$

$= \sqrt{20^{2}-16^{2}}$

$= \sqrt{400-256}$

$= \sqrt{144}$

and $RO = 2 \cdot 12 = 24$

Since corresponding sides of similar figures are in proportion, so are corresponding diagonals. Therefore,

$\frac{SP}{RO} = \frac{SN}{RM}$

$\frac{SP}{24} = \frac{45}{20}$

$\frac{SP}{24} \cdot 24= \frac{45}{20} \cdot 24$

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Question

Rhombus $RHOM \sim$ Rhombus . $m \angle R = 60^{\circ }$ ; ; .

Give the ratio of the area of Rhombus to that of Rhombus .

Answer

The angles of the rhombus measure:

$m \angle R = 60 ^{\circ }$ ;

$m \angle O = 60 ^{\circ }$ , since opposite angles of a rhombus, as in any other parallelogram, are congruent;

$m \angle H = 120^{\circ }$ and $m \angle M = 120^{\circ }$ , since consecutive angles of a rhombus are supplementary (the sum of their degree measures is 180).

The diagonals of a rhombus are each other's perpendicular bisectors as well as the bisectors of the angles, so $\overline{HM}$ and $\overline{RO}$ , whose point of intersection we will call , divide Rhombus into four 30-60-90 triangles. If we examine $\bigtriangleup RXH$ , we see that its short leg $\overline{HX}$ has length half that of $\overline{HM}$ ; , so . By the 30-60-90 Triangle Theorem, long leg $\overline{RX}$ has length $\sqrt{3}$ this, or $6 \sqrt{3}$ , and the diagonal $\overline{RO}$ has measure twice this, or $12\sqrt{3}$ .

The ratio of the lengths of corresponding diagonals of the rhombuses is the same as the similarity ratios of the sides, so the similarity ratio of Rhombus to Rhombus is

$\frac{SP }{RO} = \frac{72}{12\sqrt{3}}= \frac{6}{\sqrt{3}}$

The ratio of the areas is the square of the similarity ratio, which is

$\left ( \frac{6}{\sqrt{3}} \right )^{2}= \frac{36}{3} = \frac{12}{1}$

That is, 12 to 1.

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Question

In Quadrilateral , $\overline{KI} \cong \overline{KE}$ , $\overline{TI} \cong \overline{TE}$ , $m \angle K = 60^{\circ }$ , and $\angle T$ is a right angle; .

There exists Quadrilateral such that Quadrilateral $KITE \sim$ Quadrilateral , and .

Give the length of $\overline{LF}$ .

Answer

The Quadrilateral with its diagonals is shown below. 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, two congruent 30-60-90 triangles, $\bigtriangleup KXE$ and $\bigtriangleup KXI$ , and two congruent 45-45-90 triangles , $\bigtriangleup TXE$ and $\bigtriangleup TXI$ , are formed; also, $\bigtriangleup TIE$ , being an isosceles right triangle, is a 45-45-90 triangle. $\overline{KE}$ , the hypotenuse of $\bigtriangleup KXE$ , has length 16, so by the 30-60-90 Triangle Theorem, its shorter leg $\overline{XE}$ has length half this, or 8. Also, $\overline{XE}$ is a leg of $\bigtriangleup TXE$ , so by the 45-45-90 Theorem, the hypotenuse $\overline{TE}$ has length $\sqrt{2}$ times this, or $8\sqrt{2}$ .

Corresponding sides of similar quadrilaterals are in proportion, so

$\frac{LF}{KE} = \frac{UF}{TE}$

$\frac{LF}{16} = \frac{24}{8 \sqrt{2}}$

$\frac{LF}{16} \cdot 16 = \frac{24}{8 \sqrt{2}} \cdot 16$

$LF= \frac{48}{ \sqrt{2}} = \frac{48 \cdot \sqrt{2}}{ \sqrt{2} \cdot \sqrt{2}}= \frac{48 \sqrt{2}}{2} = 24 \sqrt{2}$

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Question

Rhombus $RHOM \sim$ Rhombus . ; Rhombus has area 90; Rhombus has area 360. What is the length of diagonal $\overline{IN}$ ?

Answer

The area of a rhombus is half the product of the lengths of its diagonals, so we can take advantage of this to find $\overline{HM}$ from the known measurements of Rhombus :

$A_{1} = \frac{1}{2} (RO)(HM)$

$90= \frac{1}{2} \cdot 15 \cdot HM$

$90= \frac{15}{2} \cdot HM$

$\frac{2} {15}\cdot 90= \frac{2} {15}\cdot \frac{15}{2} \cdot HM$

The ratio of the area of Rhombus to that of Rhombus is 4, so their similarity ratio is the square root of this, or 2. We can calculate now:

$\frac{IN}{HM} = 2$

$\frac{IN}{12} = 2$

$\frac{IN}{12}\cdot 12 = 2 \cdot 12$

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