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805+ (Hard)|   Geometry|      
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In rectangle ABCD (below left), AB=3 and BC=5. The corner at C is fold 05 Sep 2019, 19:11
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95% (hard)

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30% (01:58) correct 70% (02:40) wrong based on 37 sessions

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In rectangle ABCD (below left), AB=3 and BC=5. The corner at C is folded up (below right) and lands on AB so that AC=2 and CB=1.
The area of quadrilateral CPNM can be written as \(\frac{m}{n}\), where m and n are positive, co-prime integers. Find m+n.

A. 5
B. 28
C. 65
D. 106
E. 138
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D

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chetan2u
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In rectangle ABCD (below left), AB=3 and BC=5. The corner at C is fold 06 Sep 2019, 06:26
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nick1816
In rectangle ABCD (below left), AB=3 and BC=5. The corner at C is folded up (below right) and lands on AB so that AC=2 and CB=1.
The area of quadrilateral CPNM can be written as \(\frac{m}{n}\), where m and n are positive, co-prime integers. Find m+n.

A. 5
B. 28
C. 65
D. 106
E. 138
Show more


PLease see attached figure..

Triangle BMC..
BC=5, take BM=x, then MC=5-x
Since it is a right angled triangle, we can find x from \(1^2+x^2=(5-x)^2\), so x=2.4 =BM and MC=5-2.4=2.6

Take triangle APC and BMC.. angle BMC=angle ACP, angle BMC=angle ACP, angle APC=angle BCM and third angle is 90 degree, so similar triangle...
Now we can find sides of ACP since we know AC=2...Rest sides will be as 2, 5/6 and 13/6..

Now, triangle APC and DPN, Both are same..

Area of these 3 triangles = \((\frac{1}{2})*2*(\frac{5}{6})*2+(\frac{1}{2})*1*2.4=\frac{5}{3}+\frac{6}{5}=\frac{43}{15}\)..
The area of ABCD = 3*5=15, so remaining area = \(15-\frac{43}{15}=\frac{182}{15}\)

But area of CMNP is HALF of it as it is superimposed, so \(\frac{182}{15}*\frac{1}{2}=\frac{91}{15}=\frac{m}{n}\).
Thus m+n=91+15=106

D
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Mayukhca2018
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In rectangle ABCD (below left), AB=3 and BC=5. The corner at C is fold 06 Sep 2019, 10:38
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chetan2u
nick1816
In rectangle ABCD (below left), AB=3 and BC=5. The corner at C is folded up (below right) and lands on AB so that AC=2 and CB=1.
The area of quadrilateral CPNM can be written as \(\frac{m}{n}\), where m and n are positive, co-prime integers. Find m+n.

A. 5
B. 28
C. 65
D. 106
E. 138


PLease see attached figure..

Triangle BMC..
BC=5, take BM=x, then MC=5-x
Since it is a right angled triangle, we can find x from \(1^2+x^2=(5-x)^2\), so x=2.4 =BM and MC=5-2.4=2.6

Take triangle APC and BMC.. angle BMC=angle ACP, angle BMC=angle ACP, angle APC=angle BCM and third angle is 90 degree, so similar triangle...
Now we can find sides of ACP since we know AC=2...Rest sides will be as 2, 5/6 and 13/6..

Now, triangle APC and DPN, Both are same..

Area of these 3 triangles = \((\frac{1}{2})*2*(\frac{5}{6})*2+(\frac{1}{2})*1*2.4=\frac{5}{3}+\frac{6}{5}=\frac{43}{15}\)..
The area of ABCD = 3*5=15, so remaining area = \(15-\frac{43}{15}=\frac{182}{15}\)

But area of CMNP is HALF of it as it is superimposed, so \(\frac{182}{15}*\frac{1}{2}=\frac{91}{15}=\frac{m}{n}\).
Thus m+n=91+15=106

D
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Why angle BMC = angle ACP? Just because it looks similar?

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In rectangle ABCD (below left), AB=3 and BC=5. The corner at C is fold 06 Sep 2019, 18:55
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Assume ACP= x
ACP+ PCM+ BCM=180
x+90+BCM=180
BCM=90-x

Triangle BCM is a right angled triangle at B; hence, angle BMC =90-(90-x)=x





Mayukhca2018
chetan2u
nick1816
In rectangle ABCD (below left), AB=3 and BC=5. The corner at C is folded up (below right) and lands on AB so that AC=2 and CB=1.
The area of quadrilateral CPNM can be written as \(\frac{m}{n}\), where m and n are positive, co-prime integers. Find m+n.

A. 5
B. 28
C. 65
D. 106
E. 138


PLease see attached figure..

Triangle BMC..
BC=5, take BM=x, then MC=5-x
Since it is a right angled triangle, we can find x from \(1^2+x^2=(5-x)^2\), so x=2.4 =BM and MC=5-2.4=2.6

Take triangle APC and BMC.. angle BMC=angle ACP, angle BMC=angle ACP, angle APC=angle BCM and third angle is 90 degree, so similar triangle...
Now we can find sides of ACP since we know AC=2...Rest sides will be as 2, 5/6 and 13/6..

Now, triangle APC and DPN, Both are same..

Area of these 3 triangles = \((\frac{1}{2})*2*(\frac{5}{6})*2+(\frac{1}{2})*1*2.4=\frac{5}{3}+\frac{6}{5}=\frac{43}{15}\)..
The area of ABCD = 3*5=15, so remaining area = \(15-\frac{43}{15}=\frac{182}{15}\)

But area of CMNP is HALF of it as it is superimposed, so \(\frac{182}{15}*\frac{1}{2}=\frac{91}{15}=\frac{m}{n}\).
Thus m+n=91+15=106

D

Why angle BMC = angle ACP? Just because it looks similar?

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In rectangle ABCD (below left), AB=3 and BC=5. The corner at C is fold 06 Sep 2019, 19:27
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Mayukhca2018
chetan2u
nick1816
In rectangle ABCD (below left), AB=3 and BC=5. The corner at C is folded up (below right) and lands on AB so that AC=2 and CB=1.
The area of quadrilateral CPNM can be written as \(\frac{m}{n}\), where m and n are positive, co-prime integers. Find m+n.

A. 5
B. 28
C. 65
D. 106
E. 138


PLease see attached figure..

Triangle BMC..
BC=5, take BM=x, then MC=5-x
Since it is a right angled triangle, we can find x from \(1^2+x^2=(5-x)^2\), so x=2.4 =BM and MC=5-2.4=2.6

Take triangle APC and BMC.. angle BMC=angle ACP, angle BMC=angle ACP, angle APC=angle BCM and third angle is 90 degree, so similar triangle...
Now we can find sides of ACP since we know AC=2...Rest sides will be as 2, 5/6 and 13/6..

Now, triangle APC and DPN, Both are same..

Area of these 3 triangles = \((\frac{1}{2})*2*(\frac{5}{6})*2+(\frac{1}{2})*1*2.4=\frac{5}{3}+\frac{6}{5}=\frac{43}{15}\)..
The area of ABCD = 3*5=15, so remaining area = \(15-\frac{43}{15}=\frac{182}{15}\)

But area of CMNP is HALF of it as it is superimposed, so \(\frac{182}{15}*\frac{1}{2}=\frac{91}{15}=\frac{m}{n}\).
Thus m+n=91+15=106

D

Why angle BMC = angle ACP? Just because it looks similar?

Posted from my mobile device
Show more

There is nothing known as ' looks similar' in maths, it is 'similar'.I have added angles too on the figure, will help you.
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In rectangle ABCD (below left), AB=3 and BC=5. The corner at C is fold 06 Sep 2019, 23:16
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[quote="chetan2u"][quote="Mayukhca2018"][quote="chetan2u"][quote="nick1816"]In rectangle ABCD (below left), AB=3 and BC=5. The corner at C is folded up (below right) and lands on AB so that AC=2 and CB=1.
The area of quadrilateral CPNM can be written as [m][fraction]m/n[/fraction][/m], where m and n are positive, co-prime integers. Find m+n.

A. 5
B. 28
C. 65
D. 106
E. 138[/quote]


PLease see attached figure..

Triangle BMC..
BC=5, take BM=x, then MC=5-x
Since it is a right angled triangle, we can find x from [m]1^2+x^2=(5-x)^2[/m], so x=2.4 =BM and MC=5-2.4=2.6

Take triangle APC and BMC.. angle BMC=angle ACP, angle BMC=angle ACP, angle APC=angle BCM and third angle is 90 degree, so similar triangle...
Now we can find sides of ACP since we know AC=2...Rest sides will be as 2, 5/6 and 13/6..

Now, triangle APC and DPN, Both are same..

Area of these 3 triangles = [m]([fraction]1/2[/fraction])*2*([fraction]5/6[/fraction])*2+([fraction]1/2[/fraction])*1*2.4=[fraction]5/3[/fraction]+[fraction]6/5[/fraction]=[fraction]43/15[/fraction][/m]..
The area of ABCD = 3*5=15, so remaining area = [m]15-[fraction]43/15[/fraction]=[fraction]182/15[/fraction][/m]

But area of CMNP is HALF of it as it is superimposed, so [m][fraction]182/15[/fraction]*[fraction]1/2[/fraction]=[fraction]91/15[/fraction]=[fraction]m/n[/fraction][/m].
Thus m+n=91+15=106

D[/quote]

Why angle BMC = angle ACP? Just because it looks similar?

[size=80][b][i]Posted from my mobile device[/i][/b][/size][/quote]

There is nothing known as ' looks similar' in maths, it is 'similar'.I have added angles too on the figure, will help you.[/quote]

Thank you Sir. Much clear now.
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In rectangle ABCD (below left), AB=3 and BC=5. The corner at C is fold 22 Oct 2019, 09:20
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chetan2u
nick1816
In rectangle ABCD (below left), AB=3 and BC=5. The corner at C is folded up (below right) and lands on AB so that AC=2 and CB=1.
The area of quadrilateral CPNM can be written as \(\frac{m}{n}\), where m and n are positive, co-prime integers. Find m+n.

A. 5
B. 28
C. 65
D. 106
E. 138


PLease see attached figure..

Triangle BMC..
BC=5, take BM=x, then MC=5-x
Since it is a right angled triangle, we can find x from \(1^2+x^2=(5-x)^2\), so x=2.4 =BM and MC=5-2.4=2.6

Take triangle APC and BMC.. angle BMC=angle ACP, angle BMC=angle ACP, angle APC=angle BCM and third angle is 90 degree, so similar triangle...
Now we can find sides of ACP since we know AC=2...Rest sides will be as 2, 5/6 and 13/6..

Now, triangle APC and DPN, Both are same..

Area of these 3 triangles = \((\frac{1}{2})*2*(\frac{5}{6})*2+(\frac{1}{2})*1*2.4=\frac{5}{3}+\frac{6}{5}=\frac{43}{15}\)..
The area of ABCD = 3*5=15, so remaining area = \(15-\frac{43}{15}=\frac{182}{15}\)

But area of CMNP is HALF of it as it is superimposed, so \(\frac{182}{15}*\frac{1}{2}=\frac{91}{15}=\frac{m}{n}\).
Thus m+n=91+15=106

D
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Can you please explain how the area of CMNP is HALF of it as it is superimposed?

Thanks.
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In rectangle ABCD (below left), AB=3 and BC=5. The corner at C is fold 02 Jul 2022, 14:26
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