Area of a right quadrilateral
$begingroup$
Quadrilateral $ABCD$ has right angles only at vertices $A$ and $D$. The numbers show the areas of two of the triangles. What is the area of $ABCD$?
The rectangle $DABC'$ will have an area of $30$. I am not sure how to proceed after that. Can anyone help?
quadrilateral
edited Dec 7 '18 at 22:50
Lucky
13815
asked Dec 7 '18 at 21:51
HariHari
7818
$endgroup$
add a comment |
$begingroup$
Quadrilateral $ABCD$ has right angles only at vertices $A$ and $D$. The numbers show the areas of two of the triangles. What is the area of $ABCD$?
The rectangle $DABC'$ will have an area of $30$. I am not sure how to proceed after that. Can anyone help?
quadrilateral
edited Dec 7 '18 at 22:50
Lucky
13815
asked Dec 7 '18 at 21:51
HariHari
7818
$endgroup$
add a comment |
$begingroup$
Quadrilateral $ABCD$ has right angles only at vertices $A$ and $D$. The numbers show the areas of two of the triangles. What is the area of $ABCD$?
The rectangle $DABC'$ will have an area of $30$. I am not sure how to proceed after that. Can anyone help?
quadrilateral
edited Dec 7 '18 at 22:50
Lucky
13815
asked Dec 7 '18 at 21:51
HariHari
7818
$endgroup$
Quadrilateral $ABCD$ has right angles only at vertices $A$ and $D$. The numbers show the areas of two of the triangles. What is the area of $ABCD$?
The rectangle $DABC'$ will have an area of $30$. I am not sure how to proceed after that. Can anyone help?
quadrilateral
quadrilateral
edited Dec 7 '18 at 22:50
Lucky
13815
asked Dec 7 '18 at 21:51
HariHari
7818
edited Dec 7 '18 at 22:50
Lucky
13815
asked Dec 7 '18 at 21:51
HariHari
7818
edited Dec 7 '18 at 22:50
Lucky
13815
edited Dec 7 '18 at 22:50
Lucky
13815
edited Dec 7 '18 at 22:50
Lucky
13815
13815
asked Dec 7 '18 at 21:51
HariHari
7818
asked Dec 7 '18 at 21:51
HariHari
7818
asked Dec 7 '18 at 21:51
HariHari
7818
7818
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Let $E$ be the intersection of $AC$ and $BD$. Observe that $Delta ABE$ and $Delta CDE$ are similar (alternate interior angles). You can also figure out the ratio of $overline{DE}$ to $overline{BE}$ by comparing areas. Namely, $frac{overline{DB}}{overline{BE}} = frac{A(Delta ADB)}{A(Delta ABE)} = frac{15}{5} = 3$, so $overline{DE} = 2 overline{BE}$. Then using the similarity observed above, we have $overline{CD} = 2overline{AB}$. In particular, $A(Delta BCD) = 2 A(Delta ABD) = 30$; summing these two triangles gives a total are of $45$.
answered Dec 7 '18 at 22:15
plattyplatty
3,360320
$endgroup$
$begingroup$
Thank You for the help, but can you clarify getting DE in terms of BE. As ΔADB is not comparable to ΔABE because they are not similar triangles.
$endgroup$
– Hari
Dec 8 '18 at 2:04
$begingroup$
Right, I kind of skipped half a step there. Really, you can construct the altitude from $E$ to $AB$, to show that $AD$ is three times as long as this. Then this altitude gives you a similar triangle to work with.
$endgroup$
– platty
Dec 8 '18 at 3:17
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $E$ be the intersection of $AC$ and $BD$. Observe that $Delta ABE$ and $Delta CDE$ are similar (alternate interior angles). You can also figure out the ratio of $overline{DE}$ to $overline{BE}$ by comparing areas. Namely, $frac{overline{DB}}{overline{BE}} = frac{A(Delta ADB)}{A(Delta ABE)} = frac{15}{5} = 3$, so $overline{DE} = 2 overline{BE}$. Then using the similarity observed above, we have $overline{CD} = 2overline{AB}$. In particular, $A(Delta BCD) = 2 A(Delta ABD) = 30$; summing these two triangles gives a total are of $45$.
answered Dec 7 '18 at 22:15
plattyplatty
3,360320
$endgroup$
$begingroup$
Thank You for the help, but can you clarify getting DE in terms of BE. As ΔADB is not comparable to ΔABE because they are not similar triangles.
$endgroup$
– Hari
Dec 8 '18 at 2:04
$begingroup$
Right, I kind of skipped half a step there. Really, you can construct the altitude from $E$ to $AB$, to show that $AD$ is three times as long as this. Then this altitude gives you a similar triangle to work with.
$endgroup$
– platty
Dec 8 '18 at 3:17
add a comment |
$begingroup$
Let $E$ be the intersection of $AC$ and $BD$. Observe that $Delta ABE$ and $Delta CDE$ are similar (alternate interior angles). You can also figure out the ratio of $overline{DE}$ to $overline{BE}$ by comparing areas. Namely, $frac{overline{DB}}{overline{BE}} = frac{A(Delta ADB)}{A(Delta ABE)} = frac{15}{5} = 3$, so $overline{DE} = 2 overline{BE}$. Then using the similarity observed above, we have $overline{CD} = 2overline{AB}$. In particular, $A(Delta BCD) = 2 A(Delta ABD) = 30$; summing these two triangles gives a total are of $45$.
answered Dec 7 '18 at 22:15
plattyplatty
3,360320
$endgroup$
$begingroup$
Thank You for the help, but can you clarify getting DE in terms of BE. As ΔADB is not comparable to ΔABE because they are not similar triangles.
$endgroup$
– Hari
Dec 8 '18 at 2:04
$begingroup$
Right, I kind of skipped half a step there. Really, you can construct the altitude from $E$ to $AB$, to show that $AD$ is three times as long as this. Then this altitude gives you a similar triangle to work with.
$endgroup$
– platty
Dec 8 '18 at 3:17
add a comment |
$begingroup$
Let $E$ be the intersection of $AC$ and $BD$. Observe that $Delta ABE$ and $Delta CDE$ are similar (alternate interior angles). You can also figure out the ratio of $overline{DE}$ to $overline{BE}$ by comparing areas. Namely, $frac{overline{DB}}{overline{BE}} = frac{A(Delta ADB)}{A(Delta ABE)} = frac{15}{5} = 3$, so $overline{DE} = 2 overline{BE}$. Then using the similarity observed above, we have $overline{CD} = 2overline{AB}$. In particular, $A(Delta BCD) = 2 A(Delta ABD) = 30$; summing these two triangles gives a total are of $45$.
answered Dec 7 '18 at 22:15
plattyplatty
3,360320
$endgroup$
Let $E$ be the intersection of $AC$ and $BD$. Observe that $Delta ABE$ and $Delta CDE$ are similar (alternate interior angles). You can also figure out the ratio of $overline{DE}$ to $overline{BE}$ by comparing areas. Namely, $frac{overline{DB}}{overline{BE}} = frac{A(Delta ADB)}{A(Delta ABE)} = frac{15}{5} = 3$, so $overline{DE} = 2 overline{BE}$. Then using the similarity observed above, we have $overline{CD} = 2overline{AB}$. In particular, $A(Delta BCD) = 2 A(Delta ABD) = 30$; summing these two triangles gives a total are of $45$.
answered Dec 7 '18 at 22:15
plattyplatty
3,360320
answered Dec 7 '18 at 22:15
plattyplatty
3,360320
answered Dec 7 '18 at 22:15
plattyplatty
3,360320
answered Dec 7 '18 at 22:15
plattyplatty
3,360320
3,360320
$begingroup$
Thank You for the help, but can you clarify getting DE in terms of BE. As ΔADB is not comparable to ΔABE because they are not similar triangles.
$endgroup$
– Hari
Dec 8 '18 at 2:04
$begingroup$
Right, I kind of skipped half a step there. Really, you can construct the altitude from $E$ to $AB$, to show that $AD$ is three times as long as this. Then this altitude gives you a similar triangle to work with.
$endgroup$
– platty
Dec 8 '18 at 3:17
add a comment |
$begingroup$
Thank You for the help, but can you clarify getting DE in terms of BE. As ΔADB is not comparable to ΔABE because they are not similar triangles.
$endgroup$
– Hari
Dec 8 '18 at 2:04
$begingroup$
Right, I kind of skipped half a step there. Really, you can construct the altitude from $E$ to $AB$, to show that $AD$ is three times as long as this. Then this altitude gives you a similar triangle to work with.
$endgroup$
– platty
Dec 8 '18 at 3:17
$begingroup$
Thank You for the help, but can you clarify getting DE in terms of BE. As ΔADB is not comparable to ΔABE because they are not similar triangles.
$endgroup$
– Hari
Dec 8 '18 at 2:04
$begingroup$
Thank You for the help, but can you clarify getting DE in terms of BE. As ΔADB is not comparable to ΔABE because they are not similar triangles.
$endgroup$
– Hari
Dec 8 '18 at 2:04
$begingroup$
Right, I kind of skipped half a step there. Really, you can construct the altitude from $E$ to $AB$, to show that $AD$ is three times as long as this. Then this altitude gives you a similar triangle to work with.
$endgroup$
– platty
Dec 8 '18 at 3:17
$begingroup$
Right, I kind of skipped half a step there. Really, you can construct the altitude from $E$ to $AB$, to show that $AD$ is three times as long as this. Then this altitude gives you a similar triangle to work with.
$endgroup$
– platty
Dec 8 '18 at 3:17
add a comment |
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