Category Archives: Geometry

Intersecting Secant Theorem

CD is a tangent to the circle.

Prove c^2=a(a+b)

I am going to add two chords to the circle

Chord AD and BD are added

\angle{BDC}=\angle{CAD} (angles in alternate segments are congruent)

\angle{BCD}=\angle{DCA} (shared angle)

\therefore \Delta BDC\cong \Delta{DAC} (AA)

Hence

\frac{DC}{AC}=\frac{BC}{DC} (Corresponding sides in similar triangles)

\frac{c}{a+b}=\frac{a}{c}

\therefore c^2=a(a+b)

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Filed under Circle Theorems, Geometry, Year 11 Specialist Mathematics

Circle Geometry Question 2

One of my Year 11 Specialist students had this question

Triangle ABC touches the given circle at Points P, Q, R and S only. The secant BW touches the circle at V and W.

Diagram not drawn to scale

(a) Determine the lengths of the line segments marked x, y and z, leaving your answers as exact values.

(b) If the length of the line segment QW is 4 units, determine the exact radius of the circle.

(a) We are going to use the Intersecting Secant Theorem – the tangent version

c^2=a\times(a+b)

Hence, we have

    \begin{equation*}30^2=25(25+x+6)\end{equation}

    \begin{equation*}900=25(31+x)\end{equation}

    \begin{equation*}x=5\end{equation}

Then we can use the intersecting chord theorem to find y.

    \begin{equation*}10\times y=6 \times x\end{equation}

    \begin{equation*}10y=30\end{equation}

    \begin{equation*}y=3\end{equation}

Back to the Intersecting Secant Theorem to find z

    \begin{equation*}z^2=4\times 17\end{equation}

    \begin{equation*}z=2\sqrt{17}\end{equation}

(b)


QW is part of a 3-4-5 triangle, therefore \angle{Q}=90^\circ

This is definitely the case of the diagram not being drawn to scale. If \angle{Q}=90^\circ, then the purple line must be the diameter.

We can use pythagoras to find the length of the diameter

    \begin{equation*}(2r)^2=13^2+4^2\end{equation}

    \begin{equation*}4r^2=185\end{equation}

    \begin{equation*}r=\frac{\sqrt{285}}{2}\end{equation}

The radius of the circle is \frac{\sqrt{285}}{2}

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Filed under Circle Theorems, Geometry, Pythagoras, Year 11 Specialist Mathematics

Circle Geometry Question

In the above diagram O is the centre of the larger circle. A, B,D and E are points on the circumference of the larger circle. A, C, E and 0 are points on the circumference of the smaller circle. Show that \angle{CAB}=\angle{ABC}. AB, AC and BC are straight lines.

AO=OB (radii of the larger circle)

At a line from O to E (it is also a radius of the larger circle)

Let \angle{CAB}=\alpha.

ACEO is a cyclic quadrilateral.

Hence, \angle{CED}=180-\alpha (AECO is a cyclic quadrilateral)

As CB is a straight line \angle{OEB}=180-(180-\alpha)=\alpha.

\Delta OEB is an isosceles triangle.

Therefore, \angle{ABC}=\alpha

Therefore \angle{ABC}=\angle{CAB}

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Filed under Circle Theorems, Finding an angle, Geometry, Year 11 Specialist Mathematics

Tricky Area Ratio Question

A circle of radius r, is inscribed within an isosceles triangle ABC. CA=CB=5r. Given that \angle{ACB} is acute, find the ratio of the area of the circle to that of triangle ABC.

Mathematics Specialist 3AB Question15 page 55

I came across this question while searching for an area of sectors and segments question.

Here’s a diagram

We know AC, AB and BC are tangents to the circle. Because the triangle is isosceles, the distance from A to the circle is the same as the distance from B to the circle.

CP is perpendicular to AB (because the triangle is isosceles).

Because it is proportional, i.e. r and 5r, we can let r=1

Let h=CP

(1)   \begin{equation*}h=\sqrt{25-a^2}\end{equation*}

but h also equals

(2)   \begin{equation*}h=1+\sqrt{(5-a)^2+1}\end{equation*}

Set equation 1 equal to equation 2

    \begin{equation*}\sqrt{25-a^2}={1+\sqrt{(5-a)^2+1}\end{equation}

Square both sides

    \begin{equation*}25-a^2=1+(5-a)^2+1+2\sqrt{(5-a)^2+1}\end{equation}

Expand and simplify

    \begin{equation*}25-a^2=2+25-10a+a^2+2\sqrt{(5-a)^2+1}\end{equation}

    \begin{equation*}0=2-10a+2a^2+2\sqrt{(5-a)^2+1}\end{equation}

Divide by 2

    \begin{equation*}0=1-5a+a^2+\sqrt{(5-a)^2+1}\end{equation}

    \begin{equation*}-\sqrt{(5-a)^2+1}=a^2-5a+1\end{equation}

Square both sides

    \begin{equation*}(5-a)^2+1=a^4-5a^3+a^2-5a^3+25a^2-5a+a^2-5a+1\end{equation}

    \begin{equation*}25-10a+a^2+1=a^4-10a^3+27a^2-10a+1\end{equation}

    \begin{equation*}0=a^4-10a^3+26a^2-25 \end{equation}

I solved this using a graphics calculator

a=-0.8434, a=1.3068, a=4.5367, a=5

We can reject a=-0.8434, a=4.5367, and a=5. If a=5, there isn’t a triangle, and if a=4.5367 \angle{ACB} is not acute.

Hence the area of triangle ABC=a\times h

    \begin{equation*}A_{\Delta}=1.3068\times\sqrt{25-1.3068^2}\end{equation}

(3)   \begin{equation*}A_{\Delta}=6.3069\end{equation*}

(4)   \begin{equation*}A_{\circ}=\pi\end{equation*}

Hence, equation 4 divided by equation 3 is

    \begin{equation*}\frac{\pi}{6.3069}\approx 0.5\end{equation}

Perhaps I approached this question in the wrong way. Is there an easier process?

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Filed under Area, Geometry, Pythagoras

Intersecting Circles

Two circles of radius L and 2L intersect as shown. What is the area of the shaded region?

From Professor Povey’s Perplexing Problems

My plan is to find the sum of the area of the two segments (see below).

Construct triangles

The diagonal of the square (the pink line above) has length

=\sqrt{(2L)^2+(2L)^2}=\sqrt{8L^2}=2\sqrt{2}L

From the pink triangle in the above diagram, I am going to find the angles using the cosine rule.

cos\theta=\frac{L^2+(2\sqrt{2}L)^2-(2L)^2}{2(L)(2\sqrt{2}L)}

cos\theta=\frac{5L^2}{4\sqrt{2}L^2}

cos\theta=\frac{5}{4\sqrt{2}}

\theta=0.487

cos\alpha=\frac{(2L)^2+(2\sqrt{2}L)^2-L^2}{(2\sqrt{2}L)(2L)}

cos\alpha=\frac{11L^2}{8\sqrt{2}L^2}

cos\alpha=\frac{11}{8\sqrt{2}}

\alpha=0.236

The green quadrilateral is a kite, which means the diagonals are perpendicular.

This means the segment angles are 2\theta and 2\alpha (because the triangles are isosceles and the diagonal is perpendicular to the base of the triangles).

Area of green segment

A=\frac{1}{2}r^2(\theta-sin\theta)

A=\frac{1}{2}L^2(0.9734-sin(0.9734))=0.0733L^2

Area of yellow segment

A=\frac{1}{2}4L^2(0.4721-sin(0.4721))=0.0035

Total Area =0.0733L^2+0.0035L^2=0.108L^2

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Filed under Finding an area, Geometry, Professor Povey

Geometry Problem

This problem is from Geometry Snacks by Ed Southall and Vincent Pantaloni – it’s a great book.

Two squares are constructed such that three vertices are collinear as shown. Find the value of the marked angle.

I started by marking in the right angles. And I added the diagonal of the larger square (pink line).

Because there are right angles at O and P, we know there is a circle, which has the diagonal of the square as its diameter (see second image below).

\angle{RSP} is 45^{\circ} (Angle between the diagonal and side of a square)

PORS is a cyclic quadrilateral.

In cyclic quadrilaterals opposite angles are supplementary.

Hence, \angle{ROP}=180^{\circ}-45^{\circ}=135^{\circ}

As \angle{ROS}=90^{\circ}, \angle{SOP} must be 45^{\circ}

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Filed under Finding an angle, Geometry