Area of a right quadrilateral
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 at 22:50
Lucky
14015
asked Dec 7 at 21:51
Hari
7818
add a comment |
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 at 22:50
Lucky
14015
asked Dec 7 at 21:51
Hari
7818
add a comment |
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 at 22:50
Lucky
14015
asked Dec 7 at 21:51
Hari
7818
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 at 22:50
Lucky
14015
asked Dec 7 at 21:51
Hari
7818
edited Dec 7 at 22:50
Lucky
14015
asked Dec 7 at 21:51
Hari
7818
edited Dec 7 at 22:50
Lucky
14015
edited Dec 7 at 22:50
Lucky
14015
edited Dec 7 at 22:50
Lucky
14015
14015
asked Dec 7 at 21:51
Hari
7818
asked Dec 7 at 21:51
Hari
7818
asked Dec 7 at 21:51
Hari
7818
7818
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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 at 22:15
platty
3,360320
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.
– Hari
Dec 8 at 2:04
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.
– platty
Dec 8 at 3:17
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 at 22:15
platty
3,360320
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.
– Hari
Dec 8 at 2:04
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.
– platty
Dec 8 at 3:17
add a comment |
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 at 22:15
platty
3,360320
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.
– Hari
Dec 8 at 2:04
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.
– platty
Dec 8 at 3:17
add a comment |
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 at 22:15
platty
3,360320
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 at 22:15
platty
3,360320
answered Dec 7 at 22:15
platty
3,360320
answered Dec 7 at 22:15
platty
3,360320
answered Dec 7 at 22:15
platty
3,360320
3,360320
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.
– Hari
Dec 8 at 2:04
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.
– platty
Dec 8 at 3:17
add a comment |
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.
– Hari
Dec 8 at 2:04
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.
– platty
Dec 8 at 3:17
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.
– Hari
Dec 8 at 2:04
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.
– Hari
Dec 8 at 2:04
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.
– platty
Dec 8 at 3:17
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.
– platty
Dec 8 at 3:17
add a comment |
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