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Ratio of areas determined by a square inscribed in the corner of a right triangle
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I’m having trouble working out how to algebraically get to the answer of this question. (See original image below.)
A square is drawn in the corner of a right-angled triangle with side lengths $a$, $b$, and [hypotenuse] $c$, as shown.
Which expression gives the ratio of the unshaded area [inside the triangle, but outside the square] to the shaded area [of the square] in all cases?
- (A) $1:1$
- (B) $c:(a+b)$
- (C) $a b: c^2$
- (D) $( a + b )^2 : 2 c^2$
- (E) $c^2 : 2 a b$
Apparently the answer is $c^2 : 2 a b$ (choice E), but how?
Your help is greatly appreciated! Thank you in advance.
(Please ignore the pen marks! They are incorrect assumptions a friend made on the diagram.)
algebra-precalculus geometry triangle ratio
edited yesterday
Blue
43.6k868141
asked yesterday
Zelda_01
82
 |Â
show 8 more comments
up vote
1
down vote
favorite
I’m having trouble working out how to algebraically get to the answer of this question. (See original image below.)
A square is drawn in the corner of a right-angled triangle with side lengths $a$, $b$, and [hypotenuse] $c$, as shown.
Which expression gives the ratio of the unshaded area [inside the triangle, but outside the square] to the shaded area [of the square] in all cases?
- (A) $1:1$
- (B) $c:(a+b)$
- (C) $a b: c^2$
- (D) $( a + b )^2 : 2 c^2$
- (E) $c^2 : 2 a b$
Apparently the answer is $c^2 : 2 a b$ (choice E), but how?
Your help is greatly appreciated! Thank you in advance.
(Please ignore the pen marks! They are incorrect assumptions a friend made on the diagram.)
algebra-precalculus geometry triangle ratio
edited yesterday
Blue
43.6k868141
asked yesterday
Zelda_01
82
Do you see any similar triangles in the figure, and can you see how to use them to determine the length of the side of the square?
– Blue
yesterday
Hi, yeah all three triangles in the diagram are similar, but I don't know how to use them to determine the length of the side of the square. I ended up splitting a and b and calling them a1/a2 and b1/b2, but that made it confusing.
– Zelda_01
yesterday
Suppose the side of the square is $s$. What are the sides of the two smaller triangles?
– Blue
yesterday
(a-s) and (b-s)? Or am I missing something easier?
– Zelda_01
yesterday
Good! Specifically, $a-s$ and $b-s$ are two of the legs. (I meant "legs", not "sides".) The two other legs are ... what? And what proportion links all four of these values?
– Blue
yesterday
 |Â
show 8 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I’m having trouble working out how to algebraically get to the answer of this question. (See original image below.)
A square is drawn in the corner of a right-angled triangle with side lengths $a$, $b$, and [hypotenuse] $c$, as shown.
Which expression gives the ratio of the unshaded area [inside the triangle, but outside the square] to the shaded area [of the square] in all cases?
- (A) $1:1$
- (B) $c:(a+b)$
- (C) $a b: c^2$
- (D) $( a + b )^2 : 2 c^2$
- (E) $c^2 : 2 a b$
Apparently the answer is $c^2 : 2 a b$ (choice E), but how?
Your help is greatly appreciated! Thank you in advance.
(Please ignore the pen marks! They are incorrect assumptions a friend made on the diagram.)
algebra-precalculus geometry triangle ratio
edited yesterday
Blue
43.6k868141
asked yesterday
Zelda_01
82
I’m having trouble working out how to algebraically get to the answer of this question. (See original image below.)
A square is drawn in the corner of a right-angled triangle with side lengths $a$, $b$, and [hypotenuse] $c$, as shown.
Which expression gives the ratio of the unshaded area [inside the triangle, but outside the square] to the shaded area [of the square] in all cases?
- (A) $1:1$
- (B) $c:(a+b)$
- (C) $a b: c^2$
- (D) $( a + b )^2 : 2 c^2$
- (E) $c^2 : 2 a b$
Apparently the answer is $c^2 : 2 a b$ (choice E), but how?
Your help is greatly appreciated! Thank you in advance.
(Please ignore the pen marks! They are incorrect assumptions a friend made on the diagram.)
algebra-precalculus geometry triangle ratio
edited yesterday
Blue
43.6k868141
asked yesterday
Zelda_01
82
edited yesterday
Blue
43.6k868141
edited yesterday
Blue
43.6k868141
edited yesterday
Blue
43.6k868141
43.6k868141
asked yesterday
Zelda_01
82
asked yesterday
Zelda_01
82
asked yesterday
Zelda_01
82
82
Do you see any similar triangles in the figure, and can you see how to use them to determine the length of the side of the square?
– Blue
yesterday
Hi, yeah all three triangles in the diagram are similar, but I don't know how to use them to determine the length of the side of the square. I ended up splitting a and b and calling them a1/a2 and b1/b2, but that made it confusing.
– Zelda_01
yesterday
Suppose the side of the square is $s$. What are the sides of the two smaller triangles?
– Blue
yesterday
(a-s) and (b-s)? Or am I missing something easier?
– Zelda_01
yesterday
Good! Specifically, $a-s$ and $b-s$ are two of the legs. (I meant "legs", not "sides".) The two other legs are ... what? And what proportion links all four of these values?
– Blue
yesterday
 |Â
show 8 more comments
Do you see any similar triangles in the figure, and can you see how to use them to determine the length of the side of the square?
– Blue
yesterday
Hi, yeah all three triangles in the diagram are similar, but I don't know how to use them to determine the length of the side of the square. I ended up splitting a and b and calling them a1/a2 and b1/b2, but that made it confusing.
– Zelda_01
yesterday
Suppose the side of the square is $s$. What are the sides of the two smaller triangles?
– Blue
yesterday
(a-s) and (b-s)? Or am I missing something easier?
– Zelda_01
yesterday
Good! Specifically, $a-s$ and $b-s$ are two of the legs. (I meant "legs", not "sides".) The two other legs are ... what? And what proportion links all four of these values?
– Blue
yesterday
Do you see any similar triangles in the figure, and can you see how to use them to determine the length of the side of the square?
– Blue
yesterday
Do you see any similar triangles in the figure, and can you see how to use them to determine the length of the side of the square?
– Blue
yesterday
Hi, yeah all three triangles in the diagram are similar, but I don't know how to use them to determine the length of the side of the square. I ended up splitting a and b and calling them a1/a2 and b1/b2, but that made it confusing.
– Zelda_01
yesterday
Hi, yeah all three triangles in the diagram are similar, but I don't know how to use them to determine the length of the side of the square. I ended up splitting a and b and calling them a1/a2 and b1/b2, but that made it confusing.
– Zelda_01
yesterday
Suppose the side of the square is $s$. What are the sides of the two smaller triangles?
– Blue
yesterday
Suppose the side of the square is $s$. What are the sides of the two smaller triangles?
– Blue
yesterday
(a-s) and (b-s)? Or am I missing something easier?
– Zelda_01
yesterday
(a-s) and (b-s)? Or am I missing something easier?
– Zelda_01
yesterday
Good! Specifically, $a-s$ and $b-s$ are two of the legs. (I meant "legs", not "sides".) The two other legs are ... what? And what proportion links all four of these values?
– Blue
yesterday
Good! Specifically, $a-s$ and $b-s$ are two of the legs. (I meant "legs", not "sides".) The two other legs are ... what? And what proportion links all four of these values?
– Blue
yesterday
 |Â
show 8 more comments
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
Let the side of the square is $x$. Then, using similarity of triangles, $fracb-xx=fracxa-x$ or $(b-x)(a-x)=x^2$; $ab-ax-bx+x^2=x^2$; $ab=x(a+b)$ or $x=fracaba+b$. Thus, the shaded area is $frac(ab)^2(a+b)^2$.
The unshaded area is $$0.5x(b-x)+0.5x(a-x)=0.5left(fracaba+b(b-fracaba+b)+fracaba+b(a-fracaba+bright)=fracab(b^2+a^2)2(a+b)^2=fracabc^22(a+b)^2$$ so the final ratio is $$frac2abc^2$$
answered yesterday
Vasya
2,4601413
Thanks for this, was just wondering how you went from working out both areas to then finding the final ratio? I haven't done ratios in a while sorry!
– Zelda_01
yesterday
Wait, nevermind I worked it out, thanks!
– Zelda_01
yesterday
add a comment |Â
up vote
0
down vote
Let $x$ be the side length of the square.
By similarity,
- The hypotenuse of the lower small right triangle is $bigl(largefraccbbigr)x$.$\[4pt]$
- The hypotenuse of the upper small right triangle is $bigl(largefraccabigr)x$.
Hence we get
$$left(fraccbright)x+left(fraccaright)x=c$$
which yields
$$x=fracaba+b$$
If $S,U$ are the respective areas of the shaded and unshaded regions, then
- $S=x^2$$\[4pt]$
- $U=bigl(largefrac12bigr)ab-x^2$
hence, the required ratio can be expressed as
beginalign*
fracUS
&=fracleft(frac12right)ab-x^2x^2\[4pt]
&=fracleft(frac12right)abx^2-1\[4pt]
&=left(
left(smallfrac12right)ab
right)
left(smallfrac1xright)^2
-1
\[4pt]
&=left(
left(smallfrac12right)ab
right)
left(smallfraca+babright)^2
-1
\[4pt]
&=frac(a+b)^22ab-1\[4pt]
&=frac(a+b)^2-2ab2ab\[4pt]
&=fraca^2+b^22ab\[4pt]
&=fracc^22ab\[4pt]
endalign*
answered yesterday
quasi
32.9k22258
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Let the side of the square is $x$. Then, using similarity of triangles, $fracb-xx=fracxa-x$ or $(b-x)(a-x)=x^2$; $ab-ax-bx+x^2=x^2$; $ab=x(a+b)$ or $x=fracaba+b$. Thus, the shaded area is $frac(ab)^2(a+b)^2$.
The unshaded area is $$0.5x(b-x)+0.5x(a-x)=0.5left(fracaba+b(b-fracaba+b)+fracaba+b(a-fracaba+bright)=fracab(b^2+a^2)2(a+b)^2=fracabc^22(a+b)^2$$ so the final ratio is $$frac2abc^2$$
answered yesterday
Vasya
2,4601413
Thanks for this, was just wondering how you went from working out both areas to then finding the final ratio? I haven't done ratios in a while sorry!
– Zelda_01
yesterday
Wait, nevermind I worked it out, thanks!
– Zelda_01
yesterday
add a comment |Â
up vote
0
down vote
accepted
Let the side of the square is $x$. Then, using similarity of triangles, $fracb-xx=fracxa-x$ or $(b-x)(a-x)=x^2$; $ab-ax-bx+x^2=x^2$; $ab=x(a+b)$ or $x=fracaba+b$. Thus, the shaded area is $frac(ab)^2(a+b)^2$.
The unshaded area is $$0.5x(b-x)+0.5x(a-x)=0.5left(fracaba+b(b-fracaba+b)+fracaba+b(a-fracaba+bright)=fracab(b^2+a^2)2(a+b)^2=fracabc^22(a+b)^2$$ so the final ratio is $$frac2abc^2$$
answered yesterday
Vasya
2,4601413
Thanks for this, was just wondering how you went from working out both areas to then finding the final ratio? I haven't done ratios in a while sorry!
– Zelda_01
yesterday
Wait, nevermind I worked it out, thanks!
– Zelda_01
yesterday
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Let the side of the square is $x$. Then, using similarity of triangles, $fracb-xx=fracxa-x$ or $(b-x)(a-x)=x^2$; $ab-ax-bx+x^2=x^2$; $ab=x(a+b)$ or $x=fracaba+b$. Thus, the shaded area is $frac(ab)^2(a+b)^2$.
The unshaded area is $$0.5x(b-x)+0.5x(a-x)=0.5left(fracaba+b(b-fracaba+b)+fracaba+b(a-fracaba+bright)=fracab(b^2+a^2)2(a+b)^2=fracabc^22(a+b)^2$$ so the final ratio is $$frac2abc^2$$
answered yesterday
Vasya
2,4601413
Let the side of the square is $x$. Then, using similarity of triangles, $fracb-xx=fracxa-x$ or $(b-x)(a-x)=x^2$; $ab-ax-bx+x^2=x^2$; $ab=x(a+b)$ or $x=fracaba+b$. Thus, the shaded area is $frac(ab)^2(a+b)^2$.
The unshaded area is $$0.5x(b-x)+0.5x(a-x)=0.5left(fracaba+b(b-fracaba+b)+fracaba+b(a-fracaba+bright)=fracab(b^2+a^2)2(a+b)^2=fracabc^22(a+b)^2$$ so the final ratio is $$frac2abc^2$$
answered yesterday
Vasya
2,4601413
answered yesterday
Vasya
2,4601413
answered yesterday
Vasya
2,4601413
answered yesterday
Vasya
2,4601413
2,4601413
Thanks for this, was just wondering how you went from working out both areas to then finding the final ratio? I haven't done ratios in a while sorry!
– Zelda_01
yesterday
Wait, nevermind I worked it out, thanks!
– Zelda_01
yesterday
add a comment |Â
Thanks for this, was just wondering how you went from working out both areas to then finding the final ratio? I haven't done ratios in a while sorry!
– Zelda_01
yesterday
Wait, nevermind I worked it out, thanks!
– Zelda_01
yesterday
Thanks for this, was just wondering how you went from working out both areas to then finding the final ratio? I haven't done ratios in a while sorry!
– Zelda_01
yesterday
Thanks for this, was just wondering how you went from working out both areas to then finding the final ratio? I haven't done ratios in a while sorry!
– Zelda_01
yesterday
Wait, nevermind I worked it out, thanks!
– Zelda_01
yesterday
Wait, nevermind I worked it out, thanks!
– Zelda_01
yesterday
add a comment |Â
up vote
0
down vote
Let $x$ be the side length of the square.
By similarity,
- The hypotenuse of the lower small right triangle is $bigl(largefraccbbigr)x$.$\[4pt]$
- The hypotenuse of the upper small right triangle is $bigl(largefraccabigr)x$.
Hence we get
$$left(fraccbright)x+left(fraccaright)x=c$$
which yields
$$x=fracaba+b$$
If $S,U$ are the respective areas of the shaded and unshaded regions, then
- $S=x^2$$\[4pt]$
- $U=bigl(largefrac12bigr)ab-x^2$
hence, the required ratio can be expressed as
beginalign*
fracUS
&=fracleft(frac12right)ab-x^2x^2\[4pt]
&=fracleft(frac12right)abx^2-1\[4pt]
&=left(
left(smallfrac12right)ab
right)
left(smallfrac1xright)^2
-1
\[4pt]
&=left(
left(smallfrac12right)ab
right)
left(smallfraca+babright)^2
-1
\[4pt]
&=frac(a+b)^22ab-1\[4pt]
&=frac(a+b)^2-2ab2ab\[4pt]
&=fraca^2+b^22ab\[4pt]
&=fracc^22ab\[4pt]
endalign*
answered yesterday
quasi
32.9k22258
add a comment |Â
up vote
0
down vote
Let $x$ be the side length of the square.
By similarity,
- The hypotenuse of the lower small right triangle is $bigl(largefraccbbigr)x$.$\[4pt]$
- The hypotenuse of the upper small right triangle is $bigl(largefraccabigr)x$.
Hence we get
$$left(fraccbright)x+left(fraccaright)x=c$$
which yields
$$x=fracaba+b$$
If $S,U$ are the respective areas of the shaded and unshaded regions, then
- $S=x^2$$\[4pt]$
- $U=bigl(largefrac12bigr)ab-x^2$
hence, the required ratio can be expressed as
beginalign*
fracUS
&=fracleft(frac12right)ab-x^2x^2\[4pt]
&=fracleft(frac12right)abx^2-1\[4pt]
&=left(
left(smallfrac12right)ab
right)
left(smallfrac1xright)^2
-1
\[4pt]
&=left(
left(smallfrac12right)ab
right)
left(smallfraca+babright)^2
-1
\[4pt]
&=frac(a+b)^22ab-1\[4pt]
&=frac(a+b)^2-2ab2ab\[4pt]
&=fraca^2+b^22ab\[4pt]
&=fracc^22ab\[4pt]
endalign*
answered yesterday
quasi
32.9k22258
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $x$ be the side length of the square.
By similarity,
- The hypotenuse of the lower small right triangle is $bigl(largefraccbbigr)x$.$\[4pt]$
- The hypotenuse of the upper small right triangle is $bigl(largefraccabigr)x$.
Hence we get
$$left(fraccbright)x+left(fraccaright)x=c$$
which yields
$$x=fracaba+b$$
If $S,U$ are the respective areas of the shaded and unshaded regions, then
- $S=x^2$$\[4pt]$
- $U=bigl(largefrac12bigr)ab-x^2$
hence, the required ratio can be expressed as
beginalign*
fracUS
&=fracleft(frac12right)ab-x^2x^2\[4pt]
&=fracleft(frac12right)abx^2-1\[4pt]
&=left(
left(smallfrac12right)ab
right)
left(smallfrac1xright)^2
-1
\[4pt]
&=left(
left(smallfrac12right)ab
right)
left(smallfraca+babright)^2
-1
\[4pt]
&=frac(a+b)^22ab-1\[4pt]
&=frac(a+b)^2-2ab2ab\[4pt]
&=fraca^2+b^22ab\[4pt]
&=fracc^22ab\[4pt]
endalign*
answered yesterday
quasi
32.9k22258
Let $x$ be the side length of the square.
By similarity,
- The hypotenuse of the lower small right triangle is $bigl(largefraccbbigr)x$.$\[4pt]$
- The hypotenuse of the upper small right triangle is $bigl(largefraccabigr)x$.
Hence we get
$$left(fraccbright)x+left(fraccaright)x=c$$
which yields
$$x=fracaba+b$$
If $S,U$ are the respective areas of the shaded and unshaded regions, then
- $S=x^2$$\[4pt]$
- $U=bigl(largefrac12bigr)ab-x^2$
hence, the required ratio can be expressed as
beginalign*
fracUS
&=fracleft(frac12right)ab-x^2x^2\[4pt]
&=fracleft(frac12right)abx^2-1\[4pt]
&=left(
left(smallfrac12right)ab
right)
left(smallfrac1xright)^2
-1
\[4pt]
&=left(
left(smallfrac12right)ab
right)
left(smallfraca+babright)^2
-1
\[4pt]
&=frac(a+b)^22ab-1\[4pt]
&=frac(a+b)^2-2ab2ab\[4pt]
&=fraca^2+b^22ab\[4pt]
&=fracc^22ab\[4pt]
endalign*
answered yesterday
quasi
32.9k22258
answered yesterday
quasi
32.9k22258
answered yesterday
quasi
32.9k22258
answered yesterday
quasi
32.9k22258
32.9k22258
add a comment |Â
add a comment |Â
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Do you see any similar triangles in the figure, and can you see how to use them to determine the length of the side of the square?
– Blue
yesterday
Hi, yeah all three triangles in the diagram are similar, but I don't know how to use them to determine the length of the side of the square. I ended up splitting a and b and calling them a1/a2 and b1/b2, but that made it confusing.
– Zelda_01
yesterday
Suppose the side of the square is $s$. What are the sides of the two smaller triangles?
– Blue
yesterday
(a-s) and (b-s)? Or am I missing something easier?
– Zelda_01
yesterday
Good! Specifically, $a-s$ and $b-s$ are two of the legs. (I meant "legs", not "sides".) The two other legs are ... what? And what proportion links all four of these values?
– Blue
yesterday