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Let ABCD be a rectangle with AB = 72 cm and BC = 30 cm. A circle passing through points A and C cuts the side AB at P such that AP = 56 cm. The radius, in cm, of the circle is

Solution

Correct Option: 1

Place the rectangle with A=(0,0)A = (0, 0), B=(72,0)B = (72, 0), C=(72,30)C = (72, 30). Since AP=56AP = 56, point P=(56,0)P = (56, 0) and PB=16PB = 16.

The circle passes through AA, PP, CC, so it is the circumcircle of triangle APCAPC.


Compute the relevant sides:

PC=PB2+BC2=162+302=1156=34PC = \sqrt{PB^2 + BC^2} = \sqrt{16^2 + 30^2} = \sqrt{1156} = 34

AC=AB2+BC2=722+302=6084=78AC = \sqrt{AB^2 + BC^2} = \sqrt{72^2 + 30^2} = \sqrt{6084} = 78


Apply the extended law of sines on triangle APCAPC:

2R=ACsinAPC2R = \dfrac{AC}{\sin \angle APC}

In right triangle PBCPBC (right-angled at BB):

sinBPC=BCPC=3034=1517\sin \angle BPC = \dfrac{BC}{PC} = \dfrac{30}{34} = \dfrac{15}{17}

Since AA, PP, BB are collinear, APC\angle APC and BPC\angle BPC are supplementary, so:

sinAPC=1517\sin \angle APC = \dfrac{15}{17}


2R=7815/17=78×1715=44252R = \dfrac{78}{15/17} = \dfrac{78 \times 17}{15} = \dfrac{442}{5}

R=2215R = \dfrac{221}{5}

Answer =2215= \dfrac{221}{5}

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