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805+ (Hard)|   Geometry|      
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In right angle triangle ABC, AB = 6, BC = 8 and AC = 10. A square PQRS 06 Jul 2018, 06:25
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95% (hard)

Question Stats:

41% (03:03) correct 59% (03:22) wrong based on 44 sessions

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In right angle triangle ABC, AB = 6, BC = 8 and AC = 10. A square PQRS is inscribed in the triangle as shown in the figure. Find the side of the square?
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A) \(3\)
B) \(\frac{120}{37}\)
C) \(4\)
D) \(\frac{40}{9}\)
E) \(\frac{120}{23}\)
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B
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In right angle triangle ABC, AB = 6, BC = 8 and AC = 10. A square PQRS 06 Jul 2018, 07:02
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This question can be solved by using similarity of triangles, see the sketch attached.
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In right angle triangle ABC, AB = 6, BC = 8 and AC = 10. A square PQRS 07 Jul 2018, 01:48
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In right angle triangle ABC, AB = 6, BC = 8 and AC = 10. A square PQRS is inscribed in the triangle as shown in the figure. Find the side of the square?
Attachment:
The attachment Capture.PNG is no longer available

A) \(3\)
B) \(\frac{120}{37}\)
C) \(4\)
D) \(\frac{40}{9}\)
E) \(\frac{120}{23}\)
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If we look at sides as of \(\triangle{ABC}\), 6, 8, 10, it tells us that it is 3:4:5 triangle..

Now let side of square be x and PQ is || to AC (opposite sides of square)
in \(\triangle{BPQ}\), the sides too will be in 3:4:5 = BQ : BP : x
so BP = \(\frac{4x}{5}\)

now \(\triangle{CPS}\) and \(\triangle{ABC}\) are also similar and hence 3:4:5 \(\triangle\)
now 3:4:5=PS:CS:CP=x:CS:CP
so 3 = x, so 5 = CP = \(\frac{5x}{3}\)

Now BP+CP=AB...........\(\frac{4x}{5}+\frac{5x}{3}=8.................\frac{12x+25x}{15}=8..............\frac{37x}{15}=8\).
\(x=\frac{8*15}{37}=\frac{120}{37}\)

B
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In right angle triangle ABC, AB = 6, BC = 8 and AC = 10. A square PQRS 07 Jul 2018, 02:45
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OA: B
Given AB=6 ,BC = 8 , AC =10
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As can be seen from figure above, all four \(\triangle\) i.e. \(\triangle\)ABC ,\(\triangle\)ARQ ,\(\triangle\)QBP ,\(\triangle\) PSC are similar.
Because of AA similarity as their corresponding angles are equal in measure.

Considering \(\triangle\) ABC and \(\triangle\) QBP ,we get
\(\frac{BP}{BC}=\frac{BQ}{AB}=\frac{PQ}{AC} = k\) (a constant)

We get \(QP=10k\) and \(BQ=6k\)

As all sides of a square are equal,QR will be also equal to QP i.e \(QR =10k\).

Considering \(\triangle\) ABC and \(\triangle\) QBP ,we get
\(\frac{QR}{BC}=\frac{AR}{AB}=\frac{AQ}{AC} = p\) (a constant)
\(\frac{QR}{BC}=\frac{AQ}{AC}\)
\(\frac{{10k}}{8}=\frac{AQ}{10}\)
\(AQ =\frac{{100k}}{8}=\frac{{25k}}{2}\)

\(AB =BQ+AQ\)
\(6= 6k + \frac{{25k}}{2}\)
\(k = \frac{12}{37}\)

Side of Square\(= 10k = 10* \frac{12}{37} =\frac{120}{37}\)
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In right angle triangle ABC, AB = 6, BC = 8 and AC = 10. A square PQRS 19 Jan 2020, 16:49
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