FPM(2023) January Paper 01R Q9 and solution

The right pyramid $OABCD$ has a square base $ABCD$ with side length 12 cm. The edges $OA = OB = OC = OD = x$ cm, and the angles $\angle OAC = \angle ODB = \angle OCA = \angle OBD = 30^\circ$.

(a) Find the exact length of $AC$

The diagonal $AC$ of the square base can be found using the Pythagorean theorem, as $AB = 12$ cm. Since $AC$ is the diagonal of the square:

$ AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 12^2} = \sqrt{144 + 144} = \sqrt{288} = 12\sqrt{2} \text{ cm}. $

(b) Show that $x = 4\sqrt{6}$

From the information given, $\angle OAC = 30^\circ$. We can use trigonometry to relate $x$ and $AC$. In the triangle $OAC$, using the cosine rule:

$ \cos(30^\circ) = \frac{AC}{OA}. $

Substitute the known values:

$ \cos(30^\circ) = \frac{12\sqrt{2}}{x}. $

Since $\cos(30^\circ) = \frac{\sqrt{3}}{2}$, we have:

$ \frac{\sqrt{3}}{2} = \frac{12\sqrt{2}}{x}. $

Now, solve for $x$:

$ x = \frac{12\sqrt{2} \times 2}{\sqrt{3}} = \frac{24\sqrt{2}}{\sqrt{3}} = 24\sqrt{\frac{2}{3}} = 4\sqrt{6}. $

Thus, $x = 4\sqrt{6}$ cm.

(c) Find the total surface area, to the nearest cm², of the pyramid

The total surface area of the pyramid consists of the area of the square base and the areas of the four triangular faces.

- The area of the square base:

$ \text{Area of base} = AB^2 = 12^2 = 144 \text{ cm}^2. $

- To find the area of one triangular face, we need to find the slant height $l$ of the triangle. The slant height is the perpendicular distance from the apex $O$ to the midpoint of the side of the square. In the right triangle $OAC$, the slant height $h$ can be found using:

$ h=\sqrt{(4\sqrt{6})^2- (12/2)^2}= 2\sqrt{15}= 7.7459 $

Now, use this to find the area of one triangle. The area of one triangular face is:

$ \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{slant height} = \frac{1}{2} \times 12 \times 2\sqrt{15}= 46.475. $

$ \text{Total surface Area of the pyramid } = 4\times 46.475+144=329.9=330. $

(d) Find the size of the obtuse angle, to the nearest degree, between the plane $OAB$ and the plane $OBC$

In triangle $\triangle OAB$, we are given:

$ OA = OB = 4\sqrt{6}, \quad AB = 12. $

Let $Y$ be the foot of the perpendicular from $A$ to $OB$. We aim to find $AY$ using the area method.

Step 1: Area of $\triangle OAB$ using $AB$ as the base

The area of $\triangle OAB$ can be expressed as:

$ \text{Area} = \frac{1}{2} \times AB \times h, $

where $h = 2\sqrt{15}$ is the perpendicular distance from $O$ to $AB$. Substituting:

$ \text{Area} = \frac{1}{2} \times 12 \times 2\sqrt{15} = 12\sqrt{15}. $

Step 2: Area of $\triangle OAB$ using $OB$ as the base

The area can also be expressed as:

$ \text{Area} = \frac{1}{2} \times OB \times AY, $

where $AY$ is the perpendicular from $A$ to $OB$. Substituting $OB = 4\sqrt{6}$:

$ \text{Area} = \frac{1}{2} \times 4\sqrt{6} \times AY = 2\sqrt{6} \cdot AY. $

Step 3: Equating the two expressions for the area

Equating the two expressions:

$ 12\sqrt{15} = 2\sqrt{6} \cdot AY. $

Solve for $AY$:

$ AY = \frac{12\sqrt{15}}{2\sqrt{6}}. $

Step 4: Simplify $AY$

Simplify the fraction:

$ AY = 6 \cdot \frac{\sqrt{15}}{\sqrt{6}} = 6 \cdot \sqrt{\frac{15}{6}} = 6 \cdot \sqrt{\frac{5}{2}} = 3\sqrt{10}. $

In triangle $\triangle AYC$, we are given:

$ AC = 12\sqrt{2}, \quad AY = YC = 3\sqrt{10}. $

We aim to find $\angle AYC$.

Step 5: Apply the Law of Cosines

Using the Law of Cosines in $\triangle AYC$:

$ AC^2 = AY^2 + YC^2 - 2 \cdot AY \cdot YC \cdot \cos\angle AYC. $

Substitute the given values:

$ (12\sqrt{2})^2 = (3\sqrt{10})^2 + (3\sqrt{10})^2 - 2 \cdot (3\sqrt{10}) \cdot (3\sqrt{10}) \cdot \cos\angle AYC. $

Step 6: Simplify the equation

First, compute the squares:

$ (12\sqrt{2})^2 = 288, \quad (3\sqrt{10})^2 = 90. $

Substitute these into the equation:

$ 288 = 90 + 90 - 2 \cdot 3\sqrt{10} \cdot 3\sqrt{10} \cdot \cos\angle AYC. $

Simplify further:

$ 288 = 180 - 2 \cdot 90 \cdot \cos\angle AYC. $

$ 288 = 180 - 180 \cdot \cos\angle AYC. $

Step 7: Solve for $\cos\angle AYC$

Rearrange to isolate $\cos\angle AYC$:

$ 288 - 180 = -180 \cdot \cos\angle AYC. $

$ 108 = -180 \cdot \cos\angle AYC. $

$ \cos\angle AYC = -\frac{108}{180} = -\frac{3}{5}. $

Step 8: Find $\angle AYC$

Since $\cos\angle AYC = -\frac{3}{5}$, the angle is in the second quadrant:

$ \angle AYC = \cos^{-1}\left(-\frac{3}{5}\right). $

Final Answer

The size of the obtuse angle, to the nearest degree, between the plane $OAB$ and the plane $OBC$ is

$ \boxed{\angle AYC = \cos^{-1}\left(-\frac{3}{5}\right)=126.87=127} $