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805+ (Hard)|   Geometry|      
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Bunuel
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Through a point on the hypotenuse of a right triangle, lines are drawn 18 Mar 2019, 01:52
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D
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95% (hard)

Question Stats:

32% (03:12) correct 68% (02:54) wrong based on 65 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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D
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Sanjeetgujrall
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Through a point on the hypotenuse of a right triangle, lines are drawn 09 Apr 2019, 03:00
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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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ajmekal
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Through a point on the hypotenuse of a right triangle, lines are drawn 09 Apr 2019, 06:12
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Sanjeetgujrall
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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Manas1212
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Through a point on the hypotenuse of a right triangle, lines are drawn 12 Apr 2019, 04:48
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Bunuel
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)
Show more

Please find the image attached.
\(\triangle ABD\) 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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Sanjeetgujrall
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Through a point on the hypotenuse of a right triangle, lines are drawn 12 Apr 2019, 05:59
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Manas1212
Bunuel
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)

Please find the image attached.
\(\triangle ABD\) 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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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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Through a point on the hypotenuse of a right triangle, lines are drawn 12 Apr 2019, 07:00
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Sanjeetgujrall
Manas1212
Bunuel
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)

Please find the image attached.
\(\triangle ABD\) 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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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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Through a point on the hypotenuse of a right triangle, lines are drawn 12 Apr 2019, 10:13
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Manas1212
Bunuel
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)

Please find the image attached.
\(\triangle ABD\) 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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How are you deriving 2m?
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Through a point on the hypotenuse of a right triangle, lines are drawn 12 Apr 2019, 11:56
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georgethomps
Manas1212
Bunuel
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)

Please find the image attached.
\(\triangle ABD\) 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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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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Through a point on the hypotenuse of a right triangle, lines are drawn 12 Apr 2019, 12:00
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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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Through a point on the hypotenuse of a right triangle, lines are drawn 12 Apr 2019, 18:58
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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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Let \(\angle BAD\)= y

So \(\angle ADB\)=90-y= \(\angle DFE\)

Hope It Helps Sanjeetgujrall
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Through a point on the hypotenuse of a right triangle, lines are drawn 02 Aug 2023, 07:04
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