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Circle Geometry, SMB-019

The diagram shows a large semicircle with diameter  `AB`  and two smaller semicircles with diameters  `AC`  and  `BC`, respectively, where  `C`  is a point on the diameter  `AB`. The point  `M`  is the centre of the semicircle with diameter  `AC`.

The line perpendicular to  `AB`  through  `C`  meets the largest semicircle at the point  `D`. The points  `S`  and  `T`  are the intersections of the lines  `AD`  and  `BD`  with the smaller semicircles. The point  `X`  is the intersection of the lines  `CD`  and  `ST`.
 

Explain why `CTDS` is a rectangle.   (3 marks)

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`text(Proof)\ \ text{(See Worked Solutions)}`

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`/_SDT = 90°\ text{(angle in semi-circle)}`

`/_ ASC = 90°\ \ text{(angle in semi-circle)}`

`=> /_ CSD = 180-90=90°\ \ text{(} /_ ASD\ text{is a straight line)}`

`text(Similarly,)`

`/_CTB = /_CTD=90°`

`/_SCT = 90°\ \ text{(angle sum of quadrilateral}\ CTDS text{)}`

`text(S)text(ince all angles are right angles,)`

`CTDS\ text(is a rectangle)` 

Circle Geometry, SMB-017

In the diagram, the points `A`, `B`, `C` and `D` are on the circumference of a circle, whose centre `O` lies on `BD`. The chord `AC` intersects the diameter `BD` at `Y`. The tangent at `D` passes through the point `X`.

It is given that  `∠CYB = 100^@`  and  `∠DCY = 30^@`.

 

 

  1. What is the size of  `∠ACB`?   (1 mark)

  2. What is the size of  `∠CBD`?   (2 marks)

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  1. `60^@`
  2. `20^@`
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i.    `∠DCB` `= 90^@\ \ text{(angle in semi-circle)}`
`∠ACB` `= 90-30`
  `= 60^@`

 
ii.    `∠CYD = 180-100=80^@\ \ text{(180° in straight line)}`

`∠CDY = 180-(80+30)=70^@\ \ text{(180° in Δ)}`

`∠CBD = 180-(90+70)=20^@\ \ text{(180° in Δ)}`

Circle Geometry, SMB-011

In the diagram, \(PR\) is a diameter of the circle centred \(O\) and \(\angle QPR = 15^{\circ} \).
 

Find \(\theta\).  (2 marks)   

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\(\theta = 75^{\circ}\)

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\(\angle PQR = 90^{\circ}\ \  \text{(angle in semicircle)} \)

\(\theta\) \(=180-(90+15)\ \ (180^{\circ}\ \text{in}\ \Delta) \)  
  \(=75^{\circ} \)