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605-655 (Medium)|   Geometry|      
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sairam595
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There is a right-angled triangle yard ABC. A and B are where sprinkler 24 Apr 2016, 08:08
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74% (02:57) correct 26% (03:02) wrong based on 115 sessions

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There is a right-angled triangle yard ABC. A and B are where sprinklers are located, and the radius of the range of sprinklers is 20 as shown in the above figure. What is the area of parts of the yard where water from the sprinklers does not reach?
A. 384−100π
B. 384−10π
C. 284−100π
D. 384−80π
E. 284−80π
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A

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There is a right-angled triangle yard ABC. A and B are where sprinkler 24 Apr 2016, 09:28
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Area that couldn't get the water = Area of triangular yard - Area of the portion that received water

Area of triangular yard = \(\frac{1}{2}* 32 * 24\)
\(=> 384\)

Area of the portion that received water = Sum of the areas of two arcs with radii 20 and angles A and B
=> \(\frac{angleA}{360} * \pi*20*20 + \frac{angleB}{360} * \pi*20*20\)
=> \(\frac{\pi*20*20}{360} (angleA + angleB)\)
=> \(\frac{\pi*20*20}{360} * 90\)
=> \(100 \pi\)

Required area: \(384 - 100\pi\)

Option A
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There is a right-angled triangle yard ABC. A and B are where sprinkler 24 Apr 2016, 09:18
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area of circle = pie r2
1 - (45/360)*20(sq2) = 50pie
2- (45/360)*20(sq2) = 50pie
area of ABC = 384
so 384 - 100pie
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There is a right-angled triangle yard ABC. A and B are where sprinkler 24 Apr 2016, 09:31
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area of circle = pie r2
1 - (45/360)*20(sq2) = 50pie
2- (45/360)*20(sq2) = 50pie
area of ABC = 384
so 384 - 100pie
Show more

Hey brother,

The two angles cannot be 45 each as its not an isosceles triangle :)
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There is a right-angled triangle yard ABC. A and B are where sprinkler 24 Apr 2016, 09:36
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arhumsid
paidlukkha
area of circle = pie r2
1 - (45/360)*20(sq2) = 50pie
2- (45/360)*20(sq2) = 50pie
area of ABC = 384
so 384 - 100pie

Hey brother,

The two angles cannot be 45 each as its not an isosceles triangle :)
Show more

Thanks got save on this question will keep an eye in the future :-D :-D :-D :-D
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There is a right-angled triangle yard ABC. A and B are where sprinkler 24 Apr 2016, 18:51
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arhumsid
Area that couldn't get the water = Area of triangular yard - Area of the portion that received water

Area of triangular yard = \(\frac{1}{2}* 32 * 24\)
\(=> 384\)

Area of the portion that received water = Sum of the areas of two arcs with radii 20 and angles A and B
=> \(\frac{angleA}{360} * \pi*20*20 + \frac{angleB}{360} * \pi*20*20\)
=> \(\frac{\pi*20*20}{360} (angleA + angleB)\)
=> \(\frac{\pi*20*20}{360} * 90\)
=> \(100 \pi\)

Required area: \(384 - 100\pi\)

Option A
Show more

Hi friend,

can you please explain how you derive below step:

\(\frac{angleA}{360} * \pi*20*20 + \frac{angleB}{360} * \pi*20*20\)
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There is a right-angled triangle yard ABC. A and B are where sprinkler 24 Apr 2016, 19:07
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smartguy595
arhumsid
Area that couldn't get the water = Area of triangular yard - Area of the portion that received water

Area of triangular yard = \(\frac{1}{2}* 32 * 24\)
\(=> 384\)

Area of the portion that received water = Sum of the areas of two arcs with radii 20 and angles A and B
=> \(\frac{angleA}{360} * \pi*20*20 + \frac{angleB}{360} * \pi*20*20\)
=> \(\frac{\pi*20*20}{360} (angleA + angleB)\)
=> \(\frac{\pi*20*20}{360} * 90\)
=> \(100 \pi\)

Required area: \(384 - 100\pi\)

Option A

Hi friend,

can you please explain how you derive below step:

\(\frac{angleA}{360} * \pi*20*20 + \frac{angleB}{360} * \pi*20*20\)
Show more

Sure! That's how we find the area of a sector of a circle: (\(\frac{angle.formed.at.the.center}{360}\) * \(\pi r^2\))
We can see the area covered by the sprinklers is nothing but-
Area of the sector of a circle with center A and radius 20 + Area of the sector of a circle with center B and radius 20

Hope it's clear. Lemme know if not :)
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There is a right-angled triangle yard ABC. A and B are where sprinkler 24 Apr 2016, 19:11
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arhumsid
smartguy595
arhumsid
Area that couldn't get the water = Area of triangular yard - Area of the portion that received water

Area of triangular yard = \(\frac{1}{2}* 32 * 24\)
\(=> 384\)

Area of the portion that received water = Sum of the areas of two arcs with radii 20 and angles A and B
=> \(\frac{angleA}{360} * \pi*20*20 + \frac{angleB}{360} * \pi*20*20\)
=> \(\frac{\pi*20*20}{360} (angleA + angleB)\)
=> \(\frac{\pi*20*20}{360} * 90\)
=> \(100 \pi\)

Required area: \(384 - 100\pi\)

Option A

Hi friend,

can you please explain how you derive below step:

\(\frac{angleA}{360} * \pi*20*20 + \frac{angleB}{360} * \pi*20*20\)

Sure! That's how we find the area of a sector of a circle: (\(\frac{angle.formed.at.the.center}{360}\) * \(\pi r^2\))
We can see the area covered by the sprinklers is nothing but-
Area of the sector of a circle with center A and radius 20 + Area of the sector of a circle with center B and radius 20

Hope it's clear. Lemme know if not :)
Show more

Hi friend,

thank you!
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There is a right-angled triangle yard ABC. A and B are where sprinkler 24 Apr 2016, 19:32
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arhumsid
smartguy595
arhumsid
Area that couldn't get the water = Area of triangular yard - Area of the portion that received water

Area of triangular yard = \(\frac{1}{2}* 32 * 24\)
\(=> 384\)

Area of the portion that received water = Sum of the areas of two arcs with radii 20 and angles A and B
=> \(\frac{angleA}{360} * \pi*20*20 + \frac{angleB}{360} * \pi*20*20\)
=> \(\frac{\pi*20*20}{360} (angleA + angleB)\)
=> \(\frac{\pi*20*20}{360} * 90\)
=> \(100 \pi\)

Required area: \(384 - 100\pi\)

Option A

Hi friend,

can you please explain how you derive below step:

\(\frac{angleA}{360} * \pi*20*20 + \frac{angleB}{360} * \pi*20*20\)

Sure! That's how we find the area of a sector of a circle: (\(\frac{angle.formed.at.the.center}{360}\) * \(\pi r^2\))
We can see the area covered by the sprinklers is nothing but-
Area of the sector of a circle with center A and radius 20 + Area of the sector of a circle with center B and radius 20

Hope it's clear. Lemme know if not :)
Show more

Hi,

you do not require to find and work out for the two angles..
this is a right angle triangle, so sum of the two angles being talked about will always be 90..
so the area covered by sprinkler = \(\frac{90}{360}*pi*20^2 = 100*p\)i..

now lets see the triangle
Hyp = 40. one side = 32, so other side =\(\sqrt{40^2-32^2} = \sqrt{8^2*(5^2-4^2)}=\sqrt{8^2*9} = 24\)
another way \(H=40= 8*5... S_1=8*4\).. so th eratio of sides = 5:4:3.. second side = 3*8 = 24

area of triangle = 1/2 * 32*24 = 384..
our answer = 384-100pi
A
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There is a right-angled triangle yard ABC. A and B are where sprinkler 24 Apr 2016, 19:43
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Hi chetan2u

other side '24' is already given in figure:

Please explain why to do below steps again!
"now lets see the triangle
Hyp = 40. one side = 32, so other side =402−322−−−−−−−−√=82∗(52−42)−−−−−−−−−−−√=82∗9−−−−−√=24402−322=82∗(52−42)=82∗9=24
another way H=40=8∗5...S1=8∗4H=40=8∗5...S1=8∗4.. so th eratio of sides = 5:4:3.. second side = 3*8 = 24"
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There is a right-angled triangle yard ABC. A and B are where sprinkler 24 Apr 2016, 19:45
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smartguy595
Hi chetan2u

other side '24' is already given in figure:

Please explain why to do below steps again!
"now lets see the triangle
Hyp = 40. one side = 32, so other side =402−322−−−−−−−−√=82∗(52−42)−−−−−−−−−−−√=82∗9−−−−−√=24402−322=82∗(52−42)=82∗9=24
another way H=40=8∗5...S1=8∗4H=40=8∗5...S1=8∗4.. so th eratio of sides = 5:4:3.. second side = 3*8 = 24"
Show more

hi,
missed out on 4 written below, its become alll the more easy now..
But now we know even if we didn't know the third side, we could have taken that out :wink: :wink:
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There is a right-angled triangle yard ABC. A and B are where sprinkler 24 Apr 2016, 19:49
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chetan2u
smartguy595
Hi chetan2u

other side '24' is already given in figure:

Please explain why to do below steps again!
"now lets see the triangle
Hyp = 40. one side = 32, so other side =402−322−−−−−−−−√=82∗(52−42)−−−−−−−−−−−√=82∗9−−−−−√=24402−322=82∗(52−42)=82∗9=24
another way H=40=8∗5...S1=8∗4H=40=8∗5...S1=8∗4.. so th eratio of sides = 5:4:3.. second side = 3*8 = 24"

hi,
missed out on 4 written below, its become alll the more easy now..
But now we know even if we didn't know the third side, we could have taken that out :wink: :wink:
Show more

chetan2u,

Noted. thank you
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There is a right-angled triangle yard ABC. A and B are where sprinkler 06 Jul 2017, 22:34
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If sides of a triangle are in the ratio- 3:4:5, then its not necessary the angles will be 30-60-90 ?
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There is a right-angled triangle yard ABC. A and B are where sprinkler 06 Jul 2017, 22:45
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If sides of a triangle are in the ratio- 3:4:5, then its not necessary the angles will be 30-60-90 ?
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You are mixing two different triangles.

If the ratio of the sides of a triangle is 3:4:5, then the angels in the triangle are 90°, ~36.87° and ~53.13.

If a triangle's angles are 30, 60 and 90 degrees, then the ratio of the sides is \(1:\sqrt{3}:2\)
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There is a right-angled triangle yard ABC. A and B are where sprinkler 07 Apr 2025, 04:19
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