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Chapter 9: Q14. (page 342)

a. Given:WZ¯is a diameter of ⊙o;OX¯∥ZY¯

Prove: WXÁåœâ‰…XYÁåœ.

(Hint: draw OYÁåœ.)

Short Answer

Expert verified

(a)

The final answer is

Arc length WX≅arc length XY.

Step by step solution

01

Step 1. Given information:

WZ¯ is a diameter of circle and OX¯∥ZY¯

02

Step 2. Concept used:

We use basic geometric and angle concept

03

Step 3. Applying the concept:

First let us draw OY¯in the given figure.

Then OY¯and OZ¯becomes radii of the circles ⇒OY¯=OZ¯

Let ∠WOX=x°, since OX¯ ||ZY¯ then ∠OZY  =x°and also Δ OZYis isosceles triangles as OY¯=OZ¯then ∠OYZ  =x°

From Δ OZY, we have  ∠OZY+∠ZYO+∠YOZ=1800

⇒x∘+x∘+∠ZOY=180∘⇒∠ZOY=180∘−2x∘

Also from straight line WZ¯, we have  ∠WOX+∠XOY+∠ZOY=1800

⇒x∘+∠XOY+180∘−2x∘=180∘⇒∠XOY−x∘=180−180∴∠XOY=x∘

Thus, ∠WOX=x°=∠XOY

⇒Arcs made by same angle in a circle are congruent.

Hence, Arc length WX  ≅arc lengthXY.

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Most popular questions from this chapter

⊙Q and ⊙R are congruent circles the intersect at C andD. CD¯ is called the common chord of the circles.

  1. What kind of quadrilateral is QDRC? Why?
  2. CD¯Must be the perpendicular bisector of ⊙Q. Why?
  3. If QC=17 and QR=30, find CD.

Find the radius of the circle inscribed in the triangle.

Plane Z passes through the center of sphere Q.

  1. Explain why QR=QS=QT.
  2. Explain why the intersection of the plane and the sphere is a circle. (The intersection of a sphere with any plane passing through the center of the sphere is called a great circle of the sphere).

The diagram below shows tangent lines and circles. Find PD.

JK¯ is tangent to ⊙P and ⊙Q.

JK=?
See all solutions

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