Tag Archives: geometry

November 16, 2025 · 5:54 am

Geometry Circle Question

In the diagram below, $A, B, C$ and $D$ lie on the circle with centre $O$ . If $\angle{DBC} = 41^{\circ}$ and $\angle{ACD} = 53^{\circ}$ , determine with reasoning $\angle{BAC}$ and $\angle{AOB}$

We know $OA=OB=OD$ – radii of the circle.

Which means, $\Delta{AOB}$ is isosceles and $\angle{OAB}=\angle{OBA}$ – equal angles isosceles triangle.

$\angle{AOD}=2\angle{ACD}$ – angle at the centre twice the angle at the circumference.

$\angle{AOB}=106^{\circ}$

This means $\angle{AOB}=74^{\circ}$ – angles on a straight line are supplementary

$\angle{OAD}=\angle{ODA}=37^{\circ}$ – equal angles isosceles triangle and the angle sum of a triangle.

$\angle{DBA}=\angle{DCA}=53^{\circ}$ – angle at the circumference subtended by the same arc are congruent.

$\angle{CAD}=\angle{CBD}=41^{\circ}$ – angles at the circumference subtended by the same arc are congruent.

$\angle{OAB}=53^{\circ}$ – equal angle isosceles triangle

Hence $\angle{BAC}=12^{\circ}$

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 = $0.0733L^2+0.0035L^2=0.108L^2$

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).