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# In right angle triangle ABC, AB = 6, BC = 8 and AC = 10. A square PQRS

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Senior Manager
Joined: 22 Feb 2018
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In right angle triangle ABC, AB = 6, BC = 8 and AC = 10. A square PQRS  [#permalink]

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06 Jul 2018, 05:25
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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:

Capture.PNG [ 4.97 KiB | Viewed 2189 times ]

A) $$3$$
B) $$\frac{120}{37}$$
C) $$4$$
D) $$\frac{40}{9}$$
E) $$\frac{120}{23}$$
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Re: In right angle triangle ABC, AB = 6, BC = 8 and AC = 10. A square PQRS  [#permalink]

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06 Jul 2018, 06:02
This question can be solved by using similarity of triangles, see the sketch attached.
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WhatsApp Image 2018-07-06 at 19.30.14.jpeg [ 67.04 KiB | Viewed 2156 times ]

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In right angle triangle ABC, AB = 6, BC = 8 and AC = 10. A square PQRS  [#permalink]

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07 Jul 2018, 00:48
Princ wrote:
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) $$\frac{120}{23}$$

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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Re: In right angle triangle ABC, AB = 6, BC = 8 and AC = 10. A square PQRS  [#permalink]

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07 Jul 2018, 01:45
1
OA: B
Given AB=6 ,BC = 8 , AC =10
Attachment:

Capture Ans.PNG [ 14.06 KiB | Viewed 1986 times ]

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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Re: In right angle triangle ABC, AB = 6, BC = 8 and AC = 10. A square PQRS  [#permalink]

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19 Jan 2020, 15:49
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Re: In right angle triangle ABC, AB = 6, BC = 8 and AC = 10. A square PQRS   [#permalink] 19 Jan 2020, 15:49
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