PS: Geometry (triangle)

Question

If ABCD is a rectangle, what is the area of the shaded region?
64
82
96
120
150

The figure is attached. Sorry, but I do nod know how to put the figure in the forum

superman · Answer

Area of the rectangle is 20*15 = 300

Area of the triangle BDC = 1/2 of rectangle = 300/2

Area of unshaded region of triangle = 1/2bh = 1/2*12*9 = 54 (( note it is a right angled triangle and if the hypotenuse is 15 then the other two sides are 12 and 9)

Area of the shaded region = Area of the triangle BDC - Area of unshaded region of triangle = 150-54 = 96

samrus98 · Answer

SOL:
The diagonal = Sqrt(20^2 + 15^2) = 25 units
Now Tr(CBD) is a right tr with sides 15-20-25 i.e 3-4-5 right tr but only scaled up. The perpendicular from the base to the hypotenuse of a 3-4-5 tr is 2.4 units (this can be proven by equating the area of tr CBD by considering DC as the base and DB as the base.)

But since this is a scaled up tr, the length of the perpendicular will also get scaled up by factor of 5 => 2.4 * 5 = 12

Assume that the perpendicular intersects with BD at E. Thus Tr CED is also a 3-4-5 right tr scaled up and DE = Sqrt(15^2 - 12^2) = 9

Now A(CEB) = 1/2 * EB * CE
= 1/2 * 16 * 12
= 96

ANS: C

samrus98 · Answer

You could try attaching a snapshot ie jpg of the question instead of a doc or a pdf.....it would be directly visible in the browser if you do tht....

samrus98 · Answer

Hey gmanjesh
Clean solution! Could you tell me, how you were able to deduce that unshaded triangle would surely be a 3-4-5 triangle?

superman · Answer

The line from C to BD is at right angle to BD. Therefor the unshaded region is a right-angle tirangle with the hypotenuse being 15. Does that explain, or am I wrong in my assumption

samrus98 · Answer

Well in this case it seemed like a calculated guess coz it worked. But unless its stated that the sides of the tr are integers, I guess its better to refrain from assuming such details.

superman · Answer

got it, my bad. I wish such luck favors me in the exams :D

samrus98 · Answer

Well...I would say if you don't make an 'already-made' mistake...u might not need luck in the test!! Amen to that eh!

Economist · Answer

First of all thanks for the solution. You mentioned that triangle CED is also a 3-4-5 triangle. Is this generic? i.e. If we draw a perpendicular to the hypotenuse of a 3-4-5 triangle from the right angle then it divides the triangle into two 3-4-5 triangles?

samrus98 · Answer

Yes, a perpendicular to the hypotenuse of 3-4-5 right triangle will always divide the it into two triangles with similar properties, in this case two 3-4-5 right triangles.

This property holds for any right triangle, be it a 1:1: \sqrt{2}  tr, a 5-12-13 tr, a 30-60-90 tr, a 45-45-90 tr, etc. that is a perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle!!

Economist · Answer

Thanks for the clarification buddy. I found this property by googling. i.e. the perpendicular to hypotenuse will create 3 similar triangles. Also the following properties hold true in such scenarios for all right triangles:

1) CE^2 = BE*ED
2) BC^2 = BE*BD
3) CD^2 = BD*DE , we can use them as well to come up with our figures:)

1) CE^2 = BE*ED
2) BC^2 = BE*BD
3) CD^2 = BD*DE , we can use them as well to come up with our figures:)

samrus98 · Answer

Yeah, well there are a lot of relations in geometry man! There is seriously no end....you google a little more, you will find a lot more worth remembering! But I feel, such stuff isnt really required for GMAT......

gmac25 · Answer

length of diagonal = 25 acts as base

area of triangle =150 i.e. half of rectangle so 1/2*25*h = 150 => h=12

so base of small triangle = 9 => base of large triangle =25-9 =16

so area of shaded region = 1/2*12*16 = 96

crejoc · Answer

that is pretty simple and cool...

PS: Geometry (triangle)

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