ratio of area

• Jan 9th 2010, 06:07 AM
Vicky1997
ratio of area
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 with two smaller right triangles. The area of one of the two small right triangle 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) 1/8msquared

I got the answer d) but I am not sure how I got the same answer for the different triangles.

Could someone please take a look at my work?

Vicky.
• Jan 9th 2010, 06:46 AM
Soroban
Hello, Vicky!

Quote:

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 with two smaller right triangles.
The area of one of the two small right triangle 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

. . $\displaystyle a)\;\frac{1}{2m+1} \qquad b)\;m\qquad c)\;1-m\qquad d)\;\frac{1}{4m}\qquad e)\;\frac{1}{8m^2}$

Code:

    - *       |::*     b |:::::*       |:::A2:::*     - *-----------*       |    x    |::*       |          |:::::*     x |    S  x |::::::::*       |          |:::::A1::::*       |          | :::::::::::::*     - *-----------*-----------------*       :    x    :        a        :

The side of the square is $\displaystyle x.$
. . Its area is: .$\displaystyle S \:=\:x^2$

The area of $\displaystyle A_1$ is: .$\displaystyle A_1 \:=\:\tfrac{1}{2}ax$

We are told: .$\displaystyle A_1 \:=\:m\cdot S \quad\Rightarrow\quad \tfrac{1}{2}ax \:=\:mx^2 \quad\Rightarrow\quad a \:=\:2mx$ .[1]

The area of $\displaystyle A_2$ is: .$\displaystyle A_2 \:=\:\tfrac{1}{2}xb$ .[2]

From the similar right triangles: .$\displaystyle \frac{b}{x} \:=\:\frac{x}{a} \quad\Rightarrow\quad b \:=\:\frac{x^2}{a}$

Substitute into [2]: .$\displaystyle A_2 \:=\:\tfrac{1}{2}x\left(\frac{x^2}{a}\right) \:=\:\frac{x^3}{2a}$

The ratio of $\displaystyle A_2$ to $\displaystyle S$ is: .$\displaystyle \frac{A_2}{S} \:=\:\frac{\frac{x^3}{2a}}{x^2} \:=\:\frac{x}{2a}$

Substitute [1]: .$\displaystyle \frac{A_2}{S} \;=\;\frac{x}{2(2mx)} \;=\;\frac{1}{4m}$

• Jan 9th 2010, 06:51 AM
HallsofIvy
Quote:

Originally Posted by Vicky1997
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 with two smaller right triangles. The area of one of the two small right triangle 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) 1/8msquared

I got the answer d) but I am not sure how I got the same answer for the different triangles.

Could someone please take a look at my work?

Vicky.

Yes, (d) is correct.
• Jan 10th 2010, 07:11 AM
nikhil
aliter
ok so let ABC be a right angle triangle where B is 90 degrees.
take any point (T) on AC and drop perpendicular on AB and AC and name them D and E
(this is just the construction that you have done)
let AB=a and BC=b
now let area of triangle TEC =m times area of square or
(1/2) (b-x)x=m(x^2) or
b-x=2mx or
x=b/(2m+1)..........................................1
now
area of triangle ABC=ar triangle ADT +ar of square +ar of triangle TEC
(1/2)ab=(1/2)(b-x)x +x^2+(1/2)(a-x)x or
ab=x(a+b) or
x=ab/(a+b)........................................2
from 1 and 2 we have
ab/(a+b)=b/(2m+1)
or
a/b=1/2m
now we are suppose to calculate ratio of area of other triangle to the area of square
=(a-x)x/[2(x^2)]
=(a-x)/2x
now put x=b/(2m+1)
=(a-[b/(2m+1)])/2(b/(2m+1))
=[2ma+a-b]/2b
=m(a/b)+(a/2b)-(1/2)
now put a/b=1/2m
=m(1/2m)+(1/4m)-(1/2)
=1/4m