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kaplan problem
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akshatgupta87
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Anurag@Gurome
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Solution:
Consider triangles BDC and BEC.
Angle BCD = angle BEC = 90.
Angle DBC = angle EBC.
So, obviously, angle BDC = angle BCE.
This makes triangle BDC similar to triangle BEC.
So, (BD^2)/(BC^2) = (Area of triangle BDC)/(Area of triangle BEC).
Now, BD = sqrt(20^2 +15^2) = 25.
So, (25^2)/(20^2) = 25/16 = (½ * 20 * 15)/ (Area of triangle BEC).
So, area of triangle BEC = 96.
Consider triangles BDC and BEC.
Angle BCD = angle BEC = 90.
Angle DBC = angle EBC.
So, obviously, angle BDC = angle BCE.
This makes triangle BDC similar to triangle BEC.
So, (BD^2)/(BC^2) = (Area of triangle BDC)/(Area of triangle BEC).
Now, BD = sqrt(20^2 +15^2) = 25.
So, (25^2)/(20^2) = 25/16 = (½ * 20 * 15)/ (Area of triangle BEC).
So, area of triangle BEC = 96.
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manpsingh87
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BD=sqrt(12^2+20^2);akshatgupta87 wrote:someone explain this.
BD=25;
let EC=x; and ED=y;
therefore, area of triangle BDC= Area of triangle BEC+ area of traingle CED;
1/2*15*20= 1/2*x*y+1/2*x*(25-y);
300=xy+x(25-y);
300=xy+25x-xy;
x=12;
also in triangle BEC we have; x^2+y^2=15^2;
y^2=225-144;
y^2=81;
y=9;
thusBE=25-9=16;
hence area of shaded region= 1/2*12*16=96 C
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