For every quadrilateral, usage a compass to see if friend can attract a circle that passes with all 4 the the quadrilateral’s vertices.

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For every one, highlight the arc from \(S\) to \(Q\) passing through \(P\). Then, discover the steps of:the arc friend highlightedthe various other arc from \(S\) to \(Q\)angle \(SPQ\)Here is one more quadrilateral through a circumscribed circle. What is the value of \(\alpha + \beta\)? describe or display your reasoning.

Brahmagupta’s formula claims that because that a square whose vertices all lie top top the very same circle, the area the the square is \(\sqrt(s-a)(s-b)(s-c)(s-d)\) where \(a,b,c,\) and also \(d\) are the lengths that the quadrilateral’s sides and also \(s\) is half its perbsci-ch.orgeter.

In the cyclic square in the bsci-ch.orgage, allude \(O\) is the center of the quadrilateral’s circumscribed circle. Validate Brahmagupta’s formula for this certain quadrilateral by very first finding the amount of the areas of the top and also bottom triangles. Then, calculation the area again using Brahmagupta’s formula.

**Description:**Quadrilateral through 2 opposite, best angles. Optimal side, 4 5th units, ideal side, 8 seventeenths units, bottom side, 15 seventeenths units, left next 3 fifths units. Allude O near the facility of figure.

Draw diagonal \(BD\). Exactly how will this diagonal relate come the circumscribed circle? explain your reasoning.Construct the center of the circumscribed circle for square \(ABCD\). Label this suggest \(O\). Explain why your an approach worked.Construct the circumscribed one for quadrilateral \(ABCD\).Could we follow this procedure to build a circumscribed circle because that

*any*cyclic quadrilaterals? define your reasoning.

A one is stated to be **circumscribed** about a polygon if every the vertices of the polygon lie on the circle. If it is possible to draw a circumscribed circle for a quadrilateral, the figure is referred to as a **cyclic quadrilateral**. No all quadrilaterals have this property.

We deserve to prove the the opposite angles of a cyclic quadrilateral space supplementary. Think about opposite angles \(BCD\) and also \(BAD\), labeling \(\alpha\) and also \(\beta\).

Angle \(BAD\) is inscriptions in the arc native \(B\) to \(D\) through \(C\). Angle \(BCD\) is inscriptions in the arc indigenous \(B\) come \(D\) through \(A\). Together, the 2 arcs trace out the whole circumference that the circle, so your measures add to 360 degrees. By the Inscribed edge Theorem, the sum of \(\alpha\) and also \(\beta\) should be half of 360 degrees, or 180 degrees. So angle \(BAD\) and \(BCD\) are supplementary. The same debate can be used to the various other pair of opposite angles.

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We to speak a polygon is circumscribed by a one if it fits inside the circle and also every peak of the polygon is on the circle.