1. The radius of a circle with centre O is $$\sqrt{50}$$cm. A and C are two points on the circle, and B is a point inside the circle. The length of AB is 6 cm, and the length of BC is 2 cm. The angle ABC is a right angle. Find the square of the distance OB.
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By: anil on 05 May 2019 03.05 pm
We know that ABC is a right angled triangle.
=> $$AC=sqrt{6^2+2^2}$$
$$AC=sqrt{40}$$
$$AC=2*sqrt{10}$$
Let the coordinates of A be (0,0)
We know that the radius of the circle, OA = $$sqrt{50}$$cm
Let OD be the height of the triangle AOC.
By applying Pythagoras theorem, we get,
=> $$AC/2 = sqrt{50-10}$$
$$AC/2=sqrt{40}$$cm
=> Coordinates of point O = $$(sqrt{10},sqrt{40})$$
Area of triangle ABC = $$0.5*6*2$$ = $$6$$ square units.
Let the height of triangle ABC be h.
0.5*h*AC=6
h*AC = 12
h*$$2*sqrt{10}$$ = $$12$$
h = $$frac{6}{sqrt{10}}$$
X-coordinate of point B =$$sqrt{6^2-frac{36}{10}}$$
= $$sqrt{32.4}$$ cm
Distance between points O and B = $$sqrt{(sqrt{10}-sqrt{32.4})^2+(sqrt{40}-frac{6}{sqrt{10})^2}$$
Expanding, we get,
Distance between points O and B = $$26$$ cm.
Therefore, option A is the right answer.
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=> $$AC=sqrt{6^2+2^2}$$
$$AC=sqrt{40}$$
$$AC=2*sqrt{10}$$
Let the coordinates of A be (0,0)
We know that the radius of the circle, OA = $$sqrt{50}$$cm
Let OD be the height of the triangle AOC.
By applying Pythagoras theorem, we get,
=> $$AC/2 = sqrt{50-10}$$
$$AC/2=sqrt{40}$$cm
=> Coordinates of point O = $$(sqrt{10},sqrt{40})$$
Area of triangle ABC = $$0.5*6*2$$ = $$6$$ square units.
Let the height of triangle ABC be h.
0.5*h*AC=6
h*AC = 12
h*$$2*sqrt{10}$$ = $$12$$
h = $$frac{6}{sqrt{10}}$$
X-coordinate of point B =$$sqrt{6^2-frac{36}{10}}$$
= $$sqrt{32.4}$$ cm
Distance between points O and B = $$sqrt{(sqrt{10}-sqrt{32.4})^2+(sqrt{40}-frac{6}{sqrt{10})^2}$$
Expanding, we get,
Distance between points O and B = $$26$$ cm.
Therefore, option A is the right answer.