Answer to Question #241565 in Geometry for Mohammad Khursheed

Question

ABCD is a right trapezoid (\angle A = 90

Answer 1

complete question :ABCD is a right trapezoid (\angle A = 90°), its diagonals intersect at point G. The circle with diameter AB is tangent to lateral side CD of the trapezoid at point H. Find the length of base BC provided that GH = 5, AD = 7

solution

in a trapezoid AB || CD.

Radius of the circle=7 and diameter=14

i.e. AB=14 and AD=7

We find BC

We will take $\Delta$ ADC such that

Angles D and H are  90° 

area of triangle ADC=area of triangle CHG and area of trapezium ADHG

$\frac{1}{2}\times AD\times CD=\frac{1}{2}\times CH \times GH+\frac{1}{2}\times (AD+GH)\times HD\\ \frac{1}{2}\times 7\times CD=\frac{1}{2}\times (CD-DH) \times 5+\frac{1}{2}\times (7+5)\times 7\\7CD=5CD-5DH+84\\7CD-5CD=-5\times 7+84\\2CD=49\\CD=\frac{49}{2}=24.5$

now CD=24.5, BE=AD=7

DE=AB=14

EC=CD-DE

$24-1.4=10.5$

Now in $\Delta BEC$

$BC^2=BE^2+CE^2\\ BC^2=7^2+10.5^2\\BC^2=159.5\\BC=12.619$