3D Trigonometry with the Pythagorean Theorem (B) Worksheet | PDF ... - Free Printable

Let's solve each problem step-by-step using 3D trigonometry and the Pythagorean Theorem, as required.

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Problem 1: Volume of Pyramid ABCDE

We are given:
- $ AE = DE = CE = BE = 10 \text{ cm} $
- $ AB = DC = 6 \text{ cm} $
- $ BC = AD = 4 \text{ cm} $
- Base $ABCD$ is a rectangle (since opposite sides equal and all angles are right angles).
- The apex $E$ is equidistant from all base vertices → this suggests $E$ is directly above the center of the base.

We need to find the volume of pyramid $ABCDE$, using:

$$
\text{Volume} = \frac{1}{3} \times \text{Base Area} \times \text{Perpendicular Height}
$$

#### Step 1: Area of the base
The base $ABCD$ is a rectangle with:
- Length $AB = 6$ cm
- Width $BC = 4$ cm

So,
$$
\text{Area of base} = 6 \times 4 = 24 \text{ cm}^2
$$

#### Step 2: Find the perpendicular height from $E$ to the base

Let’s denote $O$ as the center of the rectangle $ABCD$. Since $AE = BE = CE = DE = 10$ cm, point $E$ lies directly above $O$, so $EO$ is the perpendicular height.

We need to find $EO$.

First, find the distance from $O$ to any corner (e.g., $A$). In a rectangle, the diagonal half-length is:
$$
OA = \frac{1}{2} \times \sqrt{AB^2 + BC^2} = \frac{1}{2} \times \sqrt{6^2 + 4^2} = \frac{1}{2} \times \sqrt{36 + 16} = \frac{1}{2} \times \sqrt{52} = \frac{\sqrt{52}}{2}
$$

$$
\sqrt{52} = 2\sqrt{13}, \quad \Rightarrow OA = \frac{2\sqrt{13}}{2} = \sqrt{13} \approx 3.6056 \text{ cm}
$$

Now, in triangle $AOE$, we have:
- $AE = 10$ cm (hypotenuse)
- $OA = \sqrt{13}$ cm (base)
- $EO = h$ (height) — what we want

Using Pythagoras:
$$
AE^2 = AO^2 + EO^2 \\
10^2 = (\sqrt{13})^2 + h^2 \\
100 = 13 + h^2 \\
h^2 = 87 \\
h = \sqrt{87} \approx 9.327 \text{ cm}
$$

#### Step 3: Compute volume
$$
V = \frac{1}{3} \times 24 \times \sqrt{87} = 8 \times \sqrt{87}
$$

Calculate:
$$
\sqrt{87} \approx 9.327 \\
V \approx 8 \times 9.327 = 74.616 \text{ cm}^3
$$

Rounded to 2 decimal places:
$$
\boxed{74.62} \text{ cm}^3
$$

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✔ Answer for Problem 1:

$$
\boxed{74.62} \text{ cm}^3
$$

---

Problem 2: Tent Shape – Triangle Ends & Rectangle Base

Given:
- Base $ABCD$ is a rectangle: $9 \text{ m} \times 4 \text{ m}$
- Triangular ends are isosceles triangles: $DX = AX = 7 \text{ m}$, $CY = BY = 7 \text{ m}$
- $XY = 5 \text{ m}$
- $M$ is midpoint of $AD$

We are to find:
a. Length of $MX$
b. Angle between $MX$ and base $ABCD$
c. Angle $XDC$

Let’s analyze carefully.

#### First, sketch mentally:
- $ABCD$: rectangle with $AB = CD = 9$ m, $AD = BC = 4$ m
- $X$ and $Y$ are top points of triangular ends at $A$ and $C$
- $AX = DX = 7$ m → triangle $ADX$ is isosceles
- Similarly, $BY = CY = 7$ m
- $XY = 5$ m (top edge)
- $M$ is midpoint of $AD$, so since $AD = 4$ m, $AM = MD = 2$ m

---

#### a. Length of MX

We need $MX$. We know:
- $M$ is on $AD$, $2$ m from $A$ and $D$
- $X$ is such that $AX = DX = 7$ m

So triangle $ADX$ is isosceles with $AX = DX = 7$, base $AD = 4$ m.

Let’s find coordinates to make it easier.

##### Assign Coordinates:

Let’s place the tent in 3D space.

Set:
- $A = (0, 0, 0)$
- $D = (0, 4, 0)$ → since $AD = 4$ m along y-axis
- $B = (9, 0, 0)$, $C = (9, 4, 0)$
- Then $X$ is above the line $AD$, forming triangle $ADX$ with $AX = DX = 7$

Since $AX = DX$, $X$ lies in the plane perpendicular to $AD$ at its midpoint.

Midpoint of $AD$ is $M = (0, 2, 0)$

But wait: $M$ is midpoint of $AD$, which is $(0, 2, 0)$

But $X$ is not necessarily above $M$ unless symmetric — but $AX = DX$, so yes, $X$ lies vertically above the midpoint of $AD$? Wait — no, because $AD$ is vertical in our coordinate system?

Wait — let's redefine more clearly.

Better: Let’s define:

Let’s set:
- $A = (0, 0, 0)$
- $B = (9, 0, 0)$
- $C = (9, 4, 0)$
- $D = (0, 4, 0)$

Then $AD$ goes from $(0,0,0)$ to $(0,4,0)$ → along y-axis

So $AD = 4$ m, $AB = 9$ m

Now $X$ is such that:
- $AX = 7$ m
- $DX = 7$ m
- So $X$ lies in 3D space, and $AX = DX = 7$, $AD = 4$

Let $X = (x, y, z)$

Then:
$$
AX^2 = x^2 + y^2 + z^2 = 49 \\
DX^2 = (x - 0)^2 + (y - 4)^2 + z^2 = 49
$$

Subtract equations:
$$
[x^2 + y^2 + z^2] - [x^2 + (y - 4)^2 + z^2] = 0 \\
y^2 - (y^2 - 8y + 16) = 0 \\
8y - 16 = 0 \Rightarrow y = 2
$$

So $X$ has $y = 2$, i.e., lies in the plane $y = 2$

Now plug back into $AX^2 = 49$:
$$
x^2 + (2)^2 + z^2 = 49 \Rightarrow x^2 + z^2 = 45
$$

Now, we also know that $XY = 5$ m, and $Y$ is similarly defined on the other end.

Similarly, $Y$ satisfies $BY = CY = 7$, so same logic applies.

For $Y$: $B = (9,0,0), C = (9,4,0)$

So $Y = (p, q, r)$, then:
$$
BY^2 = (p - 9)^2 + q^2 + r^2 = 49 \\
CY^2 = (p - 9)^2 + (q - 4)^2 + r^2 = 49
$$

Subtract:
$$
q^2 - (q - 4)^2 = 0 \Rightarrow q = 2
$$

So $Y$ has $q = 2$

Then $BY^2 = (p - 9)^2 + 4 + r^2 = 49 \Rightarrow (p - 9)^2 + r^2 = 45$

Also, $XY = 5$ m

$X = (x, 2, z)$, $Y = (p, 2, r)$, so distance:
$$
(x - p)^2 + (2 - 2)^2 + (z - r)^2 = 25
\Rightarrow (x - p)^2 + (z - r)^2 = 25
$$

But we don’t know $x, p, z, r$. However, the tent is symmetric — likely $X$ and $Y$ are symmetric across the center.

Assume symmetry: the tent is symmetric about the midplane $x = 4.5$

So $X$ should be at $x = a$, $Y$ at $x = 9 - a$, and both have same $z$-coordinate?

But we can simplify.

From earlier: $x^2 + z^2 = 45$ for $X$

Similarly, for $Y$: $(p - 9)^2 + r^2 = 45$

But if symmetric, then $X = (a, 2, h)$, $Y = (9 - a, 2, h)$

Then $XY = |(9 - a) - a| = |9 - 2a|$ in x-direction, and $z$ difference = 0

So $XY = |9 - 2a| = 5$ ⇒ $9 - 2a = \pm 5$

Case 1: $9 - 2a = 5$ → $a = 2$

Case 2: $9 - 2a = -5$ → $a = 7$

Try $a = 2$: $X = (2, 2, h)$

Then $x^2 + z^2 = 4^2 + h^2 = 16 + h^2 = 45$ → $h^2 = 29$ → $h = \sqrt{29} \approx 5.385$

Similarly, $Y = (7, 2, \sqrt{29})$

Then $XY = \sqrt{(7 - 2)^2 + 0 + 0} = 5$ → correct!

Perfect.

So:
- $X = (2, 2, \sqrt{29})$
- $M$ is midpoint of $AD$: $A = (0,0,0)$, $D = (0,4,0)$ → $M = (0, 2, 0)$

Now compute $MX$

$$
MX = \sqrt{(2 - 0)^2 + (2 - 2)^2 + (\sqrt{29} - 0)^2} = \sqrt{4 + 0 + 29} = \sqrt{33} \approx 5.7446
$$

Rounded to 2 decimal places:
$$
\boxed{5.74} \text{ m}
$$

✔ Answer a: $ \boxed{5.74} $ m

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#### b. Angle between $MX$ and the base $ABCD$

The base $ABCD$ lies in the $xy$-plane ($z = 0$)

The angle between a line and a plane is the complement of the angle between the line and the normal to the plane.

But simpler: the angle $\theta$ between the line $MX$ and the base is the angle between $MX$ and its projection onto the base.

That is:
$$
\sin \theta = \frac{\text{vertical component}}{\text{length of } MX}
$$

Actually, better: the angle between the line and the plane is:
$$
\theta = \arctan\left( \frac{\text{height}}{\text{horizontal distance}} \right)
$$

In vector terms:
- Vector $ \vec{MX} = X - M = (2, 0, \sqrt{29}) $
- Horizontal component: $ \sqrt{2^2 + 0^2} = 2 $
- Vertical component: $ \sqrt{29} $

So:
$$
\tan \theta = \frac{\text{vertical}}{\text{horizontal}} = \frac{\sqrt{29}}{2} \approx \frac{5.385}{2} = 2.6925
$$

$$
\theta = \arctan(2.6925) \approx 69.7^\circ
$$

(Rounded to 1 decimal place — but let’s check exact)

Use calculator:
$$
\arctan(2.6925) \approx 69.7^\circ
$$

So:
$$
\boxed{69.7^\circ}
$$

✔ Answer b: $ \boxed{69.7^\circ} $

---

#### c. Size of angle $XDC$

Points:
- $X = (2, 2, \sqrt{29})$
- $D = (0, 4, 0)$
- $C = (9, 4, 0)$

We want angle at $D$: $ \angle XDC $

Vectors:
- $ \vec{DX} = X - D = (2 - 0, 2 - 4, \sqrt{29} - 0) = (2, -2, \sqrt{29}) $
- $ \vec{DC} = C - D = (9 - 0, 4 - 4, 0) = (9, 0, 0) $

Now use dot product:
$$
\vec{DX} \cdot \vec{DC} = (2)(9) + (-2)(0) + (\sqrt{29})(0) = 18
$$

Magnitude:
- $ |\vec{DX}| = \sqrt{2^2 + (-2)^2 + (\sqrt{29})^2} = \sqrt{4 + 4 + 29} = \sqrt{37} $
- $ |\vec{DC}| = \sqrt{81} = 9 $

So:
$$
\cos \theta = \frac{18}{\sqrt{37} \times 9} = \frac{2}{\sqrt{37}}
$$

$$
\theta = \arccos\left( \frac{2}{\sqrt{37}} \right)
$$

Compute:
- $ \sqrt{37} \approx 6.0828 $
- $ \frac{2}{6.0828} \approx 0.3286 $
- $ \arccos(0.3286) \approx 70.9^\circ $

So:
$$
\boxed{70.9^\circ}
$$

✔ Answer c: $ \boxed{70.9^\circ} $

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✔ Problem 2 Summary:

a. $ \boxed{5.74} $ m
b. $ \boxed{69.7^\circ} $
c. $ \boxed{70.9^\circ} $

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Problem 3: Hexagonal Pyramid

Given:
- Regular hexagon base, side length = 5 cm
- Vertical edges = 13 cm (from vertex to each base vertex)
- Vertex is directly above the center of the base
- So the pyramid is regular and symmetrical

Find:
a. Height of pyramid
b. Surface area

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#### a. Height of the pyramid

Let $O$ be the center of the base hexagon.

Let $V$ be the apex. Then $VO$ is the height $h$, and $VA = 13$ cm (given), where $A$ is any vertex of the base.

We need to find the distance from center $O$ to vertex $A$, i.e., the radius of the circumcircle of the regular hexagon.

For a regular hexagon with side length $s$, the distance from center to vertex is equal to the side length.

Why? Because a regular hexagon is made of 6 equilateral triangles.

So:
$$
OA = 5 \text{ cm}
$$

Now in right triangle $VOA$:
- $VA = 13$ cm (hypotenuse)
- $OA = 5$ cm
- $VO = h$ (height)

By Pythagoras:
$$
h^2 + 5^2 = 13^2 \\
h^2 + 25 = 169 \\
h^2 = 144 \\
h = \sqrt{144} = 12 \text{ cm}
$$

✔ Answer a: $ \boxed{12.00} $ cm

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#### b. Surface Area

Surface area = Base area + Lateral surface area

##### 1. Base area

Regular hexagon with side $s = 5$ cm

Formula:
$$
\text{Area} = \frac{3\sqrt{3}}{2} s^2 = \frac{3\sqrt{3}}{2} \times 25 = \frac{75\sqrt{3}}{2}
$$

Compute:
- $ \sqrt{3} \approx 1.732 $
- $ 75 \times 1.732 = 129.9 $
- Divide by 2: $ \approx 64.95 $

So base area ≈ $64.95$ cm²

But keep exact for now.

##### 2. Lateral surface area

There are 6 triangular faces, each with:
- Base = 5 cm
- Height = slant height from apex to base edge

We need the slant height of each triangular face.

Each lateral face is an isosceles triangle with two sides = 13 cm (edges from apex to base vertices), and base = 5 cm.

We can find the height of one such triangle.

Let’s call the slant height $l$.

In one triangular face (say $VAB$):
- $VA = VB = 13$ cm
- $AB = 5$ cm
- Let $M$ be midpoint of $AB$, then $VM$ is the height of the triangle

Then $AM = 2.5$ cm

In triangle $VMA$:
- $VA = 13$
- $AM = 2.5$
- $VM = l$ (unknown)

By Pythagoras:
$$
l^2 + (2.5)^2 = 13^2 \\
l^2 + 6.25 = 169 \\
l^2 = 162.75 \\
l = \sqrt{162.75} \approx 12.756
$$

Now area of one triangle:
$$
= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times \sqrt{162.75}
$$

But better: compute numerically.

$$
l \approx \sqrt{162.75} \approx 12.756 \\
\text{Area of one triangle} = \frac{1}{2} \times 5 \times 12.756 = 2.5 \times 12.756 \approx 31.89
$$

Times 6:
$$
6 \times 31.89 = 191.34 \text{ cm}^2
$$

Now add base area:
- Base ≈ $ \frac{3\sqrt{3}}{2} \times 25 = \frac{75\sqrt{3}}{2} \approx \frac{75 \times 1.732}{2} = \frac{129.9}{2} = 64.95 $

Total surface area:
$$
191.34 + 64.95 = 256.29 \text{ cm}^2
$$

Round to 2 decimal places:
$$
\boxed{256.29} \text{ cm}^2
$$

Alternatively, do it exactly:

- Lateral area: $6 \times \frac{1}{2} \times 5 \times \sqrt{13^2 - (2.5)^2} = 15 \times \sqrt{169 - 6.25} = 15 \times \sqrt{162.75} $

Note: $162.75 = \frac{651}{4}$, but better to compute:

$$
\sqrt{162.75} = \sqrt{651/4} = \frac{\sqrt{651}}{2}
$$

But numerically:
$$
\sqrt{162.75} \approx 12.756 \\
15 \times 12.756 = 191.34
$$

Same as before.

So total surface area:
$$
\boxed{256.29} \text{ cm}^2
$$

✔ Answer b: $ \boxed{256.29} $ cm²

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✔ Problem 3 Summary:

a. $ \boxed{12.00} $ cm
b. $ \boxed{256.29} $ cm²

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🔚 Final Answers:

#### 1) Volume of pyramid:
$$
\boxed{74.62} \text{ cm}^3
$$

#### 2)
a. $ \boxed{5.74} $ m
b. $ \boxed{69.7^\circ} $
c. $ \boxed{70.9^\circ} $

#### 3)
a. $ \boxed{12.00} $ cm
b. $ \boxed{256.29} $ cm²

All answers rounded to 2 decimal places as requested.