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

Question

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)

(A) 1/(2m + 1)

(B) m

(C) 1 - m

(D) 1/(4m)

(E) 1/(8m^2)

Sanjeetgujrall · Answer

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 ?

Posted from my mobile device

ajmekal · Answer

This condition may or may not be true.

Manas1212 · Answer

Please find the image attached.
 	riangle ABD  and  	riangle DEF  are similar

So AB/DE = BD/EF

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

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

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

ajmekal   Sanjeetgujrall  Hope this helps

Sanjeetgujrall · Answer

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

Manas1212 · Answer

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

georgethomps · Answer

How are you deriving 2m?

Sanjeetgujrall · Answer

Its given the area of the square to one of the triangles is m. lets assume thats  	riangle 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

Sanjeetgujrall · Answer

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?

Manas1212 · Answer

Let  \angle BAD = y

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

Hope It Helps  Sanjeetgujrall

bumpbot · Answer

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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

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