Triangular Prism 5Bdff2

1. **Problem Statement:** Calculate the following for the triangular prism with given lengths: (a) $PR$ (b) $PT$ (c) $PV$ (d) $\angle TPR$ (e) $\angle VPS$ (f) $\angle PVQ$ Given: $PQ=12$ cm, $QR=10$ cm, $RT=7.5$ cm, $UV=\frac{1}{4}UT$ 2. **Step (a): Calculate $PR$** - $P$, $Q$, $R$ lie on the base plane. - Use triangle $PQR$ with sides $PQ=12$, $QR=10$. - $PR$ can be found using the triangle inequality or coordinate geometry. Assuming $PQRS$ is a rectangle or parallelogram, but since no angle is given, use the triangle inequality or Pythagoras if right angle. Since no angle is given, assume $PQR$ is a right triangle with right angle at $Q$ (common in such problems). Then, $$PR=\sqrt{PQ^2 + QR^2} = \sqrt{12^2 + 10^2} = \sqrt{144 + 100} = \sqrt{244} = 2\sqrt{61} \approx 15.62$$ 3. **Step (b): Calculate $PT$** - $T$ is vertically above $R$ by $RT=7.5$ cm. - $PT$ is the distance from $P$ to $T$. - Since $T$ is above $R$, $PT$ is the hypotenuse of right triangle with base $PR$ and height $RT$. $$PT = \sqrt{PR^2 + RT^2} = \sqrt{(2\sqrt{61})^2 + 7.5^2} = \sqrt{244 + 56.25} = \sqrt{300.25} \approx 17.33$$ 4. **Step (c): Calculate $PV$** - $V$ lies on $UT$ such that $UV=\frac{1}{4}UT$. - $U$ is vertically above $S$, $T$ above $R$. - $UT$ is vertical segment from $U$ to $T$. - Length $UT = RS$ (height of prism), but not given directly. Since $V$ divides $UT$ in ratio $1:3$ from $U$ to $T$, coordinate of $V$ is: $$V = U + \frac{1}{4} (T - U)$$ Distance $PV$ can be found using 3D distance formula. Coordinates (assuming $S$ at origin): - $S=(0,0,0)$ - $R=(x,0,0)$ with $QR=10$ along x-axis, $PQ=12$ along y-axis - $P=(0,12,0)$ - $Q=(0,0,0)$ - $U=(0,0,h)$ above $S$ - $T=(x,0,h)$ above $R$ $V=(x,0,h - \frac{3}{4}h) = (x,0,\frac{h}{4})$ Calculate $PV$: $$PV = \sqrt{(x-0)^2 + (0-12)^2 + \left(\frac{h}{4} - 0\right)^2} = \sqrt{x^2 + 144 + \frac{h^2}{16}}$$ But $x=QR=10$, $h=RT=7.5$: $$PV = \sqrt{10^2 + 144 + \frac{7.5^2}{16}} = \sqrt{100 + 144 + \frac{56.25}{16}} = \sqrt{244 + 3.515625} = \sqrt{247.515625} \approx 15.73$$ 5. **Step (d): Calculate $\angle TPR$** - Triangle $TPR$ with points $T$, $P$, $R$. - Use vectors $\overrightarrow{PT}$ and $\overrightarrow{PR}$. Vectors: $$\overrightarrow{PT} = T - P = (10 - 0, 0 - 12, 7.5 - 0) = (10, -12, 7.5)$$ $$\overrightarrow{PR} = R - P = (10 - 0, 0 - 12, 0 - 0) = (10, -12, 0)$$ Dot product: $$\overrightarrow{PT} \cdot \overrightarrow{PR} = 10 \times 10 + (-12) \times (-12) + 7.5 \times 0 = 100 + 144 + 0 = 244$$ Magnitudes: $$|\overrightarrow{PT}| = \sqrt{10^2 + (-12)^2 + 7.5^2} = \sqrt{100 + 144 + 56.25} = \sqrt{300.25}$$ $$|\overrightarrow{PR}| = \sqrt{10^2 + (-12)^2 + 0^2} = \sqrt{100 + 144} = \sqrt{244}$$ Angle: $$\cos \theta = \frac{244}{\sqrt{300.25} \times \sqrt{244}} = \frac{244}{\sqrt{300.25 \times 244}}$$ Calculate denominator: $$\sqrt{300.25 \times 244} = \sqrt{73261} \approx 270.68$$ So, $$\cos \theta = \frac{244}{270.68} \approx 0.9017$$ $$\theta = \cos^{-1}(0.9017) \approx 25.0^\circ$$ 6. **Step (e): Calculate $\angle VPS$** - Points $V$, $P$, $S$. - Vectors: $$\overrightarrow{PV} = V - P = (10 - 0, 0 - 12, 1.875 - 0) = (10, -12, 1.875)$$ $$\overrightarrow{PS} = S - P = (0 - 0, 0 - 12, 0 - 0) = (0, -12, 0)$$ Dot product: $$10 \times 0 + (-12) \times (-12) + 1.875 \times 0 = 144$$ Magnitudes: $$|\overrightarrow{PV}| = \sqrt{10^2 + (-12)^2 + 1.875^2} = \sqrt{100 + 144 + 3.515625} = \sqrt{247.515625} \approx 15.73$$ $$|\overrightarrow{PS}| = \sqrt{0^2 + (-12)^2 + 0^2} = 12$$ Angle: $$\cos \theta = \frac{144}{15.73 \times 12} = \frac{144}{188.76} \approx 0.7629$$ $$\theta = \cos^{-1}(0.7629) \approx 40.1^\circ$$ 7. **Step (f): Calculate $\angle PVQ$** - Points $P$, $V$, $Q$. - Vectors: $$\overrightarrow{VP} = P - V = (0 - 10, 12 - 0, 0 - 1.875) = (-10, 12, -1.875)$$ $$\overrightarrow{VQ} = Q - V = (0 - 10, 0 - 0, 0 - 1.875) = (-10, 0, -1.875)$$ Dot product: $$(-10)(-10) + 12(0) + (-1.875)(-1.875) = 100 + 0 + 3.515625 = 103.515625$$ Magnitudes: $$|\overrightarrow{VP}| = \sqrt{(-10)^2 + 12^2 + (-1.875)^2} = \sqrt{100 + 144 + 3.515625} = \sqrt{247.515625} \approx 15.73$$ $$|\overrightarrow{VQ}| = \sqrt{(-10)^2 + 0^2 + (-1.875)^2} = \sqrt{100 + 0 + 3.515625} = \sqrt{103.515625} \approx 10.18$$ Angle: $$\cos \theta = \frac{103.515625}{15.73 \times 10.18} = \frac{103.515625}{160.1} \approx 0.6465$$ $$\theta = \cos^{-1}(0.6465) \approx 49.7^\circ$$ **Final answers:** - (a) $PR \approx 15.62$ cm - (b) $PT \approx 17.33$ cm - (c) $PV \approx 15.73$ cm - (d) $\angle TPR \approx 25.0^\circ$ - (e) $\angle VPS \approx 40.1^\circ$ - (f) $\angle PVQ \approx 49.7^\circ$