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Right Triangle
it is a right triangle because a rhombus' diagonals are perpendicular
let the two circles with centre O and P are congruent circles, therefore their radius will be equal. given: AB and CD are the chords of the circles with centres O and P respectively. ∠AOB=∠CPD TPT: AB=CD proof: in the ΔAOB and ΔCPD AO=CP=r and OB=PD=r ∠AOB=∠CPD therefore by SAS congruency, ΔAOB and ΔCPD are congruent triangle. therefore AB=CD
Draw the circle O, and the chord AB. From the center, draw the radius OC which passes though the midpoint, D, of AB. Since the radius OC bisects the chord AB, it is perpendicular to AB. So that CD is the required height, whose length equals to the difference of the length of the radius OC and the length of its part OD. Draw the radius OA and OB. So that OD is the median and the height of the isosceles triangle AOB, whose length equals to √(r2 - AB2/4) (by the Pythagorean theorem). Thus, the length of CD equals to r - √(r2 - AB2/4).