Question

show that the hypotenuse of a right angle triangle can be a diameter of a circle passing through the vertex containing its perpendicular



please I will Mark the brainliest

Answer 1

Let the perpendicular sides be AB=7cm and BC=24cm

According to the Pythagoras theorem

AB²+BC²=AC²

72+242=49+576=625

AC=25

A circle passes through three vertices of a triangle is called circumcircle. Centre of this circle will be intersection of perpendicular bisectors of any of two sides of the triangle. Perpendicular bisector of side BCwill be parallel to side AB. In a triangle if we draw a parallel line to one side of the triangle through the mid point of another side that line will bisect the the third side. Perpendicular bisector to side BCwill bisect the side AC at O. Perpendicular bisector of AB also meet AC at O. Ois also intersecting point of perpendicular bisectors of ABand BC. So Ois the circle of the centre. It is the midpoint of AC. So AOis the radius of the circle and half of the ACthat is 12.5 cm.

Radius of the circle is 12.5cm

Answer 2

HII....

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HERE IS YOUR ANSWER:--------

LET THE PERPENDICULAR SIDES BE AB=7CM AND BC=24CM

ACCORDING TO THE PYTHAGOREAN THEOREM WE GET,

AB^2+BC^2=AC^2

72+242=49+576

THEREFORE AC=25CM

NOW,

DIVIDING IT BY 2 WE GET,
12.5 CM AS THE RADIUS.


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