Mixing
Find the area of the shaded region in the figure below:
Clash Royale CLAN TAG#URR8PPP
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Find the area of the shaded region in the figure below:
I am completely stuck on how to start off this question. Please help on some guidance on how to start it off.
geometry area
asked 5 hours ago
Meghan C
19227
add a comment |Â
up vote
3
down vote
favorite
Find the area of the shaded region in the figure below:
I am completely stuck on how to start off this question. Please help on some guidance on how to start it off.
geometry area
asked 5 hours ago
Meghan C
19227
Unless I am very mistaken, Mind Your Decisions made a video on this exact problem.
– Raptor
5 hours ago
@Raptor, I've checked his YouTube channel, you're right thanks. An alternative approach will also be appreciated.
– Meghan C
5 hours ago
Hint: the sum of areas of two quadrilaterals on upper-left and lower-right equal to the sum of areas of the two quadrilaterals on upper-right and lower-left.
– achille hui
5 hours ago
2
@achillehui, thanks, I will attempt that approach. May I ask why this is true?
– Meghan C
5 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Find the area of the shaded region in the figure below:
I am completely stuck on how to start off this question. Please help on some guidance on how to start it off.
geometry area
asked 5 hours ago
Meghan C
19227
Find the area of the shaded region in the figure below:
I am completely stuck on how to start off this question. Please help on some guidance on how to start it off.
geometry area
geometry area
asked 5 hours ago
Meghan C
19227
asked 5 hours ago
Meghan C
19227
asked 5 hours ago
Meghan C
19227
asked 5 hours ago
Meghan C
19227
asked 5 hours ago
Meghan C
19227
19227
Unless I am very mistaken, Mind Your Decisions made a video on this exact problem.
– Raptor
5 hours ago
@Raptor, I've checked his YouTube channel, you're right thanks. An alternative approach will also be appreciated.
– Meghan C
5 hours ago
Hint: the sum of areas of two quadrilaterals on upper-left and lower-right equal to the sum of areas of the two quadrilaterals on upper-right and lower-left.
– achille hui
5 hours ago
2
@achillehui, thanks, I will attempt that approach. May I ask why this is true?
– Meghan C
5 hours ago
add a comment |Â
Unless I am very mistaken, Mind Your Decisions made a video on this exact problem.
– Raptor
5 hours ago
@Raptor, I've checked his YouTube channel, you're right thanks. An alternative approach will also be appreciated.
– Meghan C
5 hours ago
Hint: the sum of areas of two quadrilaterals on upper-left and lower-right equal to the sum of areas of the two quadrilaterals on upper-right and lower-left.
– achille hui
5 hours ago
2
@achillehui, thanks, I will attempt that approach. May I ask why this is true?
– Meghan C
5 hours ago
Unless I am very mistaken, Mind Your Decisions made a video on this exact problem.
– Raptor
5 hours ago
Unless I am very mistaken, Mind Your Decisions made a video on this exact problem.
– Raptor
5 hours ago
@Raptor, I've checked his YouTube channel, you're right thanks. An alternative approach will also be appreciated.
– Meghan C
5 hours ago
@Raptor, I've checked his YouTube channel, you're right thanks. An alternative approach will also be appreciated.
– Meghan C
5 hours ago
Hint: the sum of areas of two quadrilaterals on upper-left and lower-right equal to the sum of areas of the two quadrilaterals on upper-right and lower-left.
– achille hui
5 hours ago
Hint: the sum of areas of two quadrilaterals on upper-left and lower-right equal to the sum of areas of the two quadrilaterals on upper-right and lower-left.
– achille hui
5 hours ago
2
2
@achillehui, thanks, I will attempt that approach. May I ask why this is true?
– Meghan C
5 hours ago
@achillehui, thanks, I will attempt that approach. May I ask why this is true?
– Meghan C
5 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Split the square into $8$ triangles, convince yourself you can group them into 4 pairs
and each pair has same area. Let the area of the triangles be $a, b, c, d$ as illustrated above.
You are given $c + d = 20$, $b + c = 32$ and $a + d = 16$. The area of the quadrilateral (in cyan) is
$$a + b = (a + d) + ( b + c) - ( c + d) = 16 + 32 -20 = 28$$
answered 4 hours ago
achille hui
91.6k5127248
+1 Nice Answer!!
– the_candyman
4 hours ago
@achillehui, I really appreciate your help. Thank you.
– Meghan C
4 hours ago
add a comment |Â
up vote
1
down vote
Alternatively, refer to the figure:
$hspace4cm$
Let $x$ be the half of the side of the large square. Then the side of the smaller oblique square is $xsqrt2$, how:
Pythagorean theorem.
The total green area is $x^2$, how:
$$frac12 cdot xsqrt2cdot h_1+frac12 cdot xsqrt2 cdot h_2=frac12cdot xsqrt2cdot (h_1+h_2)=frac12cdot xsqrt2cdot xsqrt2=x^2.$$
The total area of grey and green regions is $2x^2=16+32=48$, how:
Green area is $x^2$ and grey area is $2cdot fracx^22=x^2$.
Hence, the required area is $96-(16+20+32)=28$, how:
the area of the large square $(4x^2)$ minus the total area of grey, green and white regions $16+20+32$.
answered 3 hours ago
farruhota
16.8k2735
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Split the square into $8$ triangles, convince yourself you can group them into 4 pairs
and each pair has same area. Let the area of the triangles be $a, b, c, d$ as illustrated above.
You are given $c + d = 20$, $b + c = 32$ and $a + d = 16$. The area of the quadrilateral (in cyan) is
$$a + b = (a + d) + ( b + c) - ( c + d) = 16 + 32 -20 = 28$$
answered 4 hours ago
achille hui
91.6k5127248
+1 Nice Answer!!
– the_candyman
4 hours ago
@achillehui, I really appreciate your help. Thank you.
– Meghan C
4 hours ago
add a comment |Â
up vote
3
down vote
accepted
Split the square into $8$ triangles, convince yourself you can group them into 4 pairs
and each pair has same area. Let the area of the triangles be $a, b, c, d$ as illustrated above.
You are given $c + d = 20$, $b + c = 32$ and $a + d = 16$. The area of the quadrilateral (in cyan) is
$$a + b = (a + d) + ( b + c) - ( c + d) = 16 + 32 -20 = 28$$
answered 4 hours ago
achille hui
91.6k5127248
+1 Nice Answer!!
– the_candyman
4 hours ago
@achillehui, I really appreciate your help. Thank you.
– Meghan C
4 hours ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Split the square into $8$ triangles, convince yourself you can group them into 4 pairs
and each pair has same area. Let the area of the triangles be $a, b, c, d$ as illustrated above.
You are given $c + d = 20$, $b + c = 32$ and $a + d = 16$. The area of the quadrilateral (in cyan) is
$$a + b = (a + d) + ( b + c) - ( c + d) = 16 + 32 -20 = 28$$
answered 4 hours ago
achille hui
91.6k5127248
Split the square into $8$ triangles, convince yourself you can group them into 4 pairs
and each pair has same area. Let the area of the triangles be $a, b, c, d$ as illustrated above.
You are given $c + d = 20$, $b + c = 32$ and $a + d = 16$. The area of the quadrilateral (in cyan) is
$$a + b = (a + d) + ( b + c) - ( c + d) = 16 + 32 -20 = 28$$
answered 4 hours ago
achille hui
91.6k5127248
answered 4 hours ago
achille hui
91.6k5127248
answered 4 hours ago
achille hui
91.6k5127248
answered 4 hours ago
achille hui
91.6k5127248
91.6k5127248
+1 Nice Answer!!
– the_candyman
4 hours ago
@achillehui, I really appreciate your help. Thank you.
– Meghan C
4 hours ago
add a comment |Â
+1 Nice Answer!!
– the_candyman
4 hours ago
@achillehui, I really appreciate your help. Thank you.
– Meghan C
4 hours ago
+1 Nice Answer!!
– the_candyman
4 hours ago
+1 Nice Answer!!
– the_candyman
4 hours ago
@achillehui, I really appreciate your help. Thank you.
– Meghan C
4 hours ago
@achillehui, I really appreciate your help. Thank you.
– Meghan C
4 hours ago
add a comment |Â
up vote
1
down vote
Alternatively, refer to the figure:
$hspace4cm$
Let $x$ be the half of the side of the large square. Then the side of the smaller oblique square is $xsqrt2$, how:
Pythagorean theorem.
The total green area is $x^2$, how:
$$frac12 cdot xsqrt2cdot h_1+frac12 cdot xsqrt2 cdot h_2=frac12cdot xsqrt2cdot (h_1+h_2)=frac12cdot xsqrt2cdot xsqrt2=x^2.$$
The total area of grey and green regions is $2x^2=16+32=48$, how:
Green area is $x^2$ and grey area is $2cdot fracx^22=x^2$.
Hence, the required area is $96-(16+20+32)=28$, how:
the area of the large square $(4x^2)$ minus the total area of grey, green and white regions $16+20+32$.
answered 3 hours ago
farruhota
16.8k2735
add a comment |Â
up vote
1
down vote
Alternatively, refer to the figure:
$hspace4cm$
Let $x$ be the half of the side of the large square. Then the side of the smaller oblique square is $xsqrt2$, how:
Pythagorean theorem.
The total green area is $x^2$, how:
$$frac12 cdot xsqrt2cdot h_1+frac12 cdot xsqrt2 cdot h_2=frac12cdot xsqrt2cdot (h_1+h_2)=frac12cdot xsqrt2cdot xsqrt2=x^2.$$
The total area of grey and green regions is $2x^2=16+32=48$, how:
Green area is $x^2$ and grey area is $2cdot fracx^22=x^2$.
Hence, the required area is $96-(16+20+32)=28$, how:
the area of the large square $(4x^2)$ minus the total area of grey, green and white regions $16+20+32$.
answered 3 hours ago
farruhota
16.8k2735
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Alternatively, refer to the figure:
$hspace4cm$
Let $x$ be the half of the side of the large square. Then the side of the smaller oblique square is $xsqrt2$, how:
Pythagorean theorem.
The total green area is $x^2$, how:
$$frac12 cdot xsqrt2cdot h_1+frac12 cdot xsqrt2 cdot h_2=frac12cdot xsqrt2cdot (h_1+h_2)=frac12cdot xsqrt2cdot xsqrt2=x^2.$$
The total area of grey and green regions is $2x^2=16+32=48$, how:
Green area is $x^2$ and grey area is $2cdot fracx^22=x^2$.
Hence, the required area is $96-(16+20+32)=28$, how:
the area of the large square $(4x^2)$ minus the total area of grey, green and white regions $16+20+32$.
answered 3 hours ago
farruhota
16.8k2735
Alternatively, refer to the figure:
$hspace4cm$
Let $x$ be the half of the side of the large square. Then the side of the smaller oblique square is $xsqrt2$, how:
Pythagorean theorem.
The total green area is $x^2$, how:
$$frac12 cdot xsqrt2cdot h_1+frac12 cdot xsqrt2 cdot h_2=frac12cdot xsqrt2cdot (h_1+h_2)=frac12cdot xsqrt2cdot xsqrt2=x^2.$$
The total area of grey and green regions is $2x^2=16+32=48$, how:
Green area is $x^2$ and grey area is $2cdot fracx^22=x^2$.
Hence, the required area is $96-(16+20+32)=28$, how:
the area of the large square $(4x^2)$ minus the total area of grey, green and white regions $16+20+32$.
answered 3 hours ago
farruhota
16.8k2735
answered 3 hours ago
farruhota
16.8k2735
answered 3 hours ago
farruhota
16.8k2735
answered 3 hours ago
farruhota
16.8k2735
16.8k2735
add a comment |Â
add a comment |Â
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Unless I am very mistaken, Mind Your Decisions made a video on this exact problem.
– Raptor
5 hours ago
@Raptor, I've checked his YouTube channel, you're right thanks. An alternative approach will also be appreciated.
– Meghan C
5 hours ago
Hint: the sum of areas of two quadrilaterals on upper-left and lower-right equal to the sum of areas of the two quadrilaterals on upper-right and lower-left.
– achille hui
5 hours ago
2
@achillehui, thanks, I will attempt that approach. May I ask why this is true?
– Meghan C
5 hours ago