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Through a point on the hypotenuse of a right triangle, lines are drawn

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Through a point on the hypotenuse of a right triangle, lines are drawn  [#permalink]

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18 Mar 2019, 01:52
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95% (hard)

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25% (03:08) correct 75% (02:49) wrong based on 44 sessions

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Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is m times the area of the square. The ratio of the area of the other small right triangle to the area of the square is

(A) 1/(2m + 1)

(B) m

(C) 1 - m

(D) 1/(4m)

(E) 1/(8m^2)

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Re: Through a point on the hypotenuse of a right triangle, lines are drawn  [#permalink]

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09 Apr 2019, 03:00
Can someone explain why its D?

For it to construct a square doesnt the point on the hypotenuse have to be the mid point of it ?

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Re: Through a point on the hypotenuse of a right triangle, lines are drawn  [#permalink]

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09 Apr 2019, 06:12
Sanjeetgujrall wrote:
Can someone explain why its D?

For it to construct a square doesnt the point on the hypotenuse have to be the mid point of it ?

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This condition may or may not be true.
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Through a point on the hypotenuse of a right triangle, lines are drawn  [#permalink]

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12 Apr 2019, 04:48
2
Bunuel wrote:
Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is m times the area of the square. The ratio of the area of the other small right triangle to the area of the square is

(A) 1/(2m + 1)

(B) m

(C) 1 - m

(D) 1/(4m)

(E) 1/(8m^2)

$$\triangle ABD$$ and $$\triangle DEF$$ are similar

So AB/DE = BD/EF

So $$\frac{x}{1}$$=$$\frac{1}{2m}$$
x= $$\frac{1}{2m}$$

Area of $$\triangle ABD$$ = $$\frac{x}{2}$$ =$$\frac{1}{4m}$$

So the ratio of $$\triangle ABD$$ to area of Square BCDE to is = $$\frac{1}{4m}$$

ajmekal Sanjeetgujrall Hope this helps
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Re: Through a point on the hypotenuse of a right triangle, lines are drawn  [#permalink]

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12 Apr 2019, 05:59
Manas1212 wrote:
Bunuel wrote:
Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is m times the area of the square. The ratio of the area of the other small right triangle to the area of the square is

(A) 1/(2m + 1)

(B) m

(C) 1 - m

(D) 1/(4m)

(E) 1/(8m^2)

$$\triangle ABD$$ and $$\triangle DEF$$ are similar

So AB/DE = BD/EF

So $$\frac{x}{1}$$=$$\frac{1}{2m}$$
x= $$\frac{1}{2m}$$

Area of $$\triangle ABD$$ = $$\frac{x}{2}$$ =$$\frac{1}{4m}$$

So the ratio of $$\triangle ABD$$ to area of Square BCDE to is = $$\frac{1}{4m}$$

ajmekal Sanjeetgujrall Hope this helps

Hi mamas, how are they similar ?
Both have 1 side equal because of square and angle ABC = angle DEF....what's the third thing ?

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Re: Through a point on the hypotenuse of a right triangle, lines are drawn  [#permalink]

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12 Apr 2019, 07:00
1
Sanjeetgujrall wrote:
Manas1212 wrote:
Bunuel wrote:
Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is m times the area of the square. The ratio of the area of the other small right triangle to the area of the square is

(A) 1/(2m + 1)

(B) m

(C) 1 - m

(D) 1/(4m)

(E) 1/(8m^2)

$$\triangle ABD$$ and $$\triangle DEF$$ are similar

So AB/DE = BD/EF

So $$\frac{x}{1}$$=$$\frac{1}{2m}$$
x= $$\frac{1}{2m}$$

Area of $$\triangle ABD$$ = $$\frac{x}{2}$$ =$$\frac{1}{4m}$$

So the ratio of $$\triangle ABD$$ to area of Square BCDE to is = $$\frac{1}{4m}$$

ajmekal Sanjeetgujrall Hope this helps

Hi mamas, how are they similar ?
Both have 1 side equal because of square and angle ABC = angle DEF....what's the third thing ?

Posted from my mobile device

Take any angle as y , other angle of the triangle would be 90-y. So it would be angle-angle-angle Similar

Hope it helps

P.S It's Manas
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Re: Through a point on the hypotenuse of a right triangle, lines are drawn  [#permalink]

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12 Apr 2019, 10:13
Manas1212 wrote:
Bunuel wrote:
Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is m times the area of the square. The ratio of the area of the other small right triangle to the area of the square is

(A) 1/(2m + 1)

(B) m

(C) 1 - m

(D) 1/(4m)

(E) 1/(8m^2)

$$\triangle ABD$$ and $$\triangle DEF$$ are similar

So AB/DE = BD/EF

So $$\frac{x}{1}$$=$$\frac{1}{2m}$$
x= $$\frac{1}{2m}$$

Area of $$\triangle ABD$$ = $$\frac{x}{2}$$ =$$\frac{1}{4m}$$

So the ratio of $$\triangle ABD$$ to area of Square BCDE to is = $$\frac{1}{4m}$$

ajmekal Sanjeetgujrall Hope this helps
How are you deriving 2m?
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Re: Through a point on the hypotenuse of a right triangle, lines are drawn  [#permalink]

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12 Apr 2019, 11:56
georgethomps wrote:
Manas1212 wrote:
Bunuel wrote:
Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is m times the area of the square. The ratio of the area of the other small right triangle to the area of the square is

(A) 1/(2m + 1)

(B) m

(C) 1 - m

(D) 1/(4m)

(E) 1/(8m^2)

$$\triangle ABD$$ and $$\triangle DEF$$ are similar

So AB/DE = BD/EF

So $$\frac{x}{1}$$=$$\frac{1}{2m}$$
x= $$\frac{1}{2m}$$

Area of $$\triangle ABD$$ = $$\frac{x}{2}$$ =$$\frac{1}{4m}$$

So the ratio of $$\triangle ABD$$ to area of Square BCDE to is = $$\frac{1}{4m}$$

ajmekal Sanjeetgujrall Hope this helps
How are you deriving 2m?

Its given the area of the square to one of the triangles is m. lets assume thats $$\triangle DEF$$

$$\frac{Area of triangle DEF}{Area of Square}$$ = $$\frac{m}{1}$$

$$\frac{1/2*DE*AB}{1}$$= $$\frac{m}{1}$$ where DE = 1

Solving this AB = 2m

georgethomps Hope that helps
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The three great essentials to achieve anything worthwhile are, first, hard work; second, stick-to-itiveness; third, common sense."
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Re: Through a point on the hypotenuse of a right triangle, lines are drawn  [#permalink]

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12 Apr 2019, 12:00
Manas1212

Actually that will not work. If we take any angle as y , other angle of the triangle would be 90-y.......agree

But it does not have to be the same for the other triangle. Other triangle could have an angle z and 90-z

But y does not have to be z

How is it AAA?
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Re: Through a point on the hypotenuse of a right triangle, lines are drawn  [#permalink]

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12 Apr 2019, 18:58
1
Sanjeetgujrall wrote:
Manas1212

Actually that will not work. If we take any angle as y , other angle of the triangle would be 90-y.......agree

But it does not have to be the same for the other triangle. Other triangle could have an angle z and 90-z

But y does not have to be z

How is it AAA?

Let $$\angle BAD$$= y

So $$\angle ADB$$=90-y= $$\angle DFE$$

Hope It Helps Sanjeetgujrall
Re: Through a point on the hypotenuse of a right triangle, lines are drawn   [#permalink] 12 Apr 2019, 18:58
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