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Licemer1 [7]
2 years ago
13

In the diagram, P1P2 and Q1Q2 are the perpendicular bisectors of line AB and BC , respectively. A1A2 and B1B2 are the angle bise

ctors of angle A and B , respectively. The center of the inscribed circle of ΔABC is point ____ , and the center of the circumscribed circle of ΔABC is point ____ .

Mathematics
2 answers:
Aliun [14]2 years ago
4 0
I hope I can show the image, but I hope this may help you, the answer is:

The center of the inscribed circle of ΔABC is <u>point S</u> , and the center of the circumscribed circle of ΔABC is <u>point P.</u>
Hatshy [7]2 years ago
3 0

Answer:  The complete statement is

The center of the inscribed circle of ΔABC is point S , and the center of the circumscribed circle of ΔABC is point P .

Step-by-step explanation:  Given that P_1P_2 and Q_1Q_2 are the perpendicular bisectors of the lines AB and BC respectively.

And, A_1A_2 and B_1B_2 are the angle bisectors of the angles A and B respectively.

We are to find the centre of the inscribed circle of ΔABC and the centre of the circumscribed circle of ΔABC.

<u>INCENTRE</u> : The point at which the angle bisectors of the three angles of a triangle meet is called the INCENTRE of the triangle. This point is the centre of the inscribed circle of the triangle.

The angle bisectors of angles A and B, that is A_1A_2 and B_1B_2  meet at the point S.

Also, since the three angle bisectors are concurrent, so the third angle bisector must also pass through S.

Thus, the point S is the incentre of ΔABC and so, the centre of the inscribed circle of ΔABC is point S.

<u>CIRCUMCENTRE</u> : The point at which the perpendicular bisectors of the three sides of a triangle meet is called the CIRCUMCENTRE of the triangle. This point is the centre of the circumscribed circle of the triangle.

The perpendicular bisectors of sides AB and BC, that is P_1P_2 and Q_1Q_2  meet at the point P.

Also, since the three perpendicular bisectors are concurrent, so the third perpendicular bisector must also pass through P.

Thus, the point P is the circumcentre of ΔABC and so, the centre of the circumscribed circle of ΔABC is point P.

Hence, the complete statement is

The center of the inscribed circle of ΔABC is point S , and the center of the circumscribed circle of ΔABC is point P .

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Answer:  \bold{9)\ \sin \theta=\dfrac{1}{3}\qquad 10)\ \sin \theta = \dfrac{4}{5}\qquad 11)\ \cos \theta = \dfrac{\sqrt{11}}{6}\qquad 12)\ \tan \theta = \dfrac{17\sqrt2}{26}}

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Pythagorean Theorem is: a² + b² = c²   , <em>where "c" is the hypotenuse</em>

<em />

9)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{4}{12}\quad \rightarrow \large\boxed{\dfrac{1}{3}}

Note: 4² + (8√2)² = hypotenuse²   →   hypotenuse = 12

10)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{16}{20}\quad \rightarrow \large\boxed{\dfrac{4}{5}}

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11)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{\sqrt{11}}{6}\quad =\large\boxed{\dfrac{\sqrt{11}}{6}}

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12)\ \tan \theta=\dfrac{\text{side opposite of}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{17}{13\sqrt2}\quad =\large\boxed{\dfrac{17\sqrt2}{26}}

Note: adjacent² + 7² = (13√2)²   →   adjacent = 17

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