Let O be the center of the rectangular base PQRS. Since the pyramid is a right pyramid, the vertex V is directly above O, meaning VO is the height of the pyramid and is perpendicular to the base PQRS.

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Let O be the center of the rectangular base PQRS. Since the pyramid is a right pyramid, the vertex V is directly above O, meaning VO is the height of the pyramid and is perpendicular to the base PQRS.

Asked on April 16, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-04-16T07:53:47.874Z

Here's the solution to question 11: Let O be the center of the rectangular base PQRS. Since the pyramid is a right pyramid, the vertex V is directly above O, meaning VO is the height of the pyramid and is perpendicular to the base PQRS. Given: $VP = VQ = VR = VS = 18$ cm $PQ = 16$ cm $QR = 12$ cm a) The length of the projection of line VP on the plane PQRS. Step 1: Identify the projection. The projection of line VP onto the plane PQRS is the line segment connecting P to the foot of the perpendicular from V to the plane, which is O. So, the projection is OP. Step 2: Calculate the length of the diagonal of the base. In the rectangular base PQRS, consider the right-angled triangle PQR. Using the Pythagorean theorem: $$PR^2 = PQ^2 + QR^2$$ $$PR^2 = 16^2 + 12^2$$ $$PR^2 = 256 + 144$$ $$PR^2 = 400$$ $$PR = \sqrt{400} = 20 \text{ cm}$$ Step 3: Calculate the length of OP. Since O is the center of the rectangle, it is the midpoint of the diagonal PR. $$OP = \frac{1}{2} PR$$ $$OP = \frac{1}{2} \times 20$$ $$OP = 10 \text{ cm}$$ The length of the projection of line VP on the plane PQRS is $\boxed{\text{10 cm}}$. b) The size of the angle between line VP and the plane PQRS. Step 1: Identify the angle. The angle between line VP and the plane PQRS is the angle between VP and its projection OP. This is $\angle VPO$. Step 2: Use trigonometry in the right-angled triangle VOP. Triangle VOP is a right-angled triangle with the right angle at O. We know $VP = 18$ cm (hypotenuse) and $OP = 10$ cm (adjacent to $\angle VPO$). $$\cos(\angle VPO) = \frac{OP}{VP}$$ $$\cos(\angle VPO) = \frac{10}{18}$$ $$\cos(\angle VPO) = \frac{5}{9}$$ $$\angle VPO = \arccos\left(\frac{5}{9}\right)$$ $$\angle VPO \approx 56.25^\circ$$ The size of the angle between line VP and the plane PQRS is $\boxed{56.3^\circ}$ (to one decimal place). c) The size of the angle between the planes VQR and PQRS. Step 1: Identify the line of intersection and perpendiculars. The line of intersection of plane VQR and plane PQRS is the line segment QR. Let M be the midpoint of QR. Since triangle VQR is an isosceles triangle ($VQ = VR = 18$ cm), the line segment VM from the vertex V to the midpoint M of the base QR is perpendicular to QR. So, $VM \perp QR$. In the rectangular base PQRS, the line segment OM (connecting the center O to the midpoint M of QR) is perpendicular to QR. So, $OM \perp QR$. Therefore, the angle between the planes VQR and PQRS is $\angle VMO$. Step 2: Calculate the length of OM. Since O is the center of the rectangle and M is the midpoint of QR, OM is parallel to PQ and its length is half of PQ. $$OM = \frac{1}{2} PQ$$ $$OM = \frac{1}{2} \times 16$$ $$OM = 8 \text{ cm}$$ Step 3: Calculate the height of the pyramid VO. In the right-angled triangle VOP (from part b), we can find VO using the Pythagorean theorem: $$VO^2 + OP^2 = VP^2$$ $$VO^2 + 10^2 = 18^2$$ $$VO^2 + 100 = 324$$ $$VO^2 = 324 - 100$$ $$VO^2 = 224$$ $$VO = \sqrt{224} = \sqrt{16 \times 14} = 4\sqrt{14} \text{ cm}$$ Step 4: Use trigonometry in the right-angled triangle VOM. Triangle VOM is a right-angled triangle with the right angle at O. We know $VO = 4\sqrt{14}$ cm (opposite to $\angle VMO$) and $OM = 8$ cm (adjacent to $\angle VMO$). $$\tan(\angle VMO) = \frac{VO}{OM}$$ $$\tan(\angle VMO) = \frac{4\sqrt{14}}{8}$$ $$\tan(\angle VMO) = \frac{\sqrt{14}}{2}$$ $$\angle VMO = \arctan\left(\frac{\sqrt{14}}{2}\right)$$ $$\angle VMO \approx 61.86^\circ$$ The size of the angle between the planes VQR and PQRS is $\boxed{61.9^\circ}$ (to one decimal place). What's next?

Accepted Answer · 2026-04-16T07:53:47.874Z

Here's the solution to question 11: Let O be the center of the rectangular base PQRS. Since the pyramid is a right pyramid, the vertex V is directly above O, meaning VO is the height of the pyramid and is perpendicular to the base PQRS. Given: $VP = VQ = VR = VS = 18$ cm $PQ = 16$ cm $QR = 12$ cm a) The length of the projection of line VP on the plane PQRS. Step 1: Identify the projection. The projection of line VP onto the plane PQRS is the line segment connecting P to the foot of the perpendicular from V to the plane, which is O. So, the projection is OP. Step 2: Calculate the length of the diagonal of the base. In the rectangular base PQRS, consider the right-angled triangle PQR. Using the Pythagorean theorem: $$PR^2 = PQ^2 + QR^2$$ $$PR^2 = 16^2 + 12^2$$ $$PR^2 = 256 + 144$$ $$PR^2 = 400$$ $$PR = \sqrt{400} = 20 \text{ cm}$$ Step 3: Calculate the length of OP. Since O is the center of the rectangle, it is the midpoint of the diagonal PR. $$OP = \frac{1}{2} PR$$ $$OP = \frac{1}{2} \times 20$$ $$OP = 10 \text{ cm}$$ The length of the projection of line VP on the plane PQRS is $\boxed{\text{10 cm}}$. b) The size of the angle between line VP and the plane PQRS. Step 1: Identify the angle. The angle between line VP and the plane PQRS is the angle between VP and its projection OP. This is $\angle VPO$. Step 2: Use trigonometry in the right-angled triangle VOP. Triangle VOP is a right-angled triangle with the right angle at O. We know $VP = 18$ cm (hypotenuse) and $OP = 10$ cm (adjacent to $\angle VPO$). $$\cos(\angle VPO) = \frac{OP}{VP}$$ $$\cos(\angle VPO) = \frac{10}{18}$$ $$\cos(\angle VPO) = \frac{5}{9}$$ $$\angle VPO = \arccos\left(\frac{5}{9}\right)$$ $$\angle VPO \approx 56.25^\circ$$ The size of the angle between line VP and the plane PQRS is $\boxed{56.3^\circ}$ (to one decimal place). c) The size of the angle between the planes VQR and PQRS. Step 1: Identify the line of intersection and perpendiculars. The line of intersection of plane VQR and plane PQRS is the line segment QR. Let M be the midpoint of QR. Since triangle VQR is an isosceles triangle ($VQ = VR = 18$ cm), the line segment VM from the vertex V to the midpoint M of the base QR is perpendicular to QR. So, $VM \perp QR$. In the rectangular base PQRS, the line segment OM (connecting the center O to the midpoint M of QR) is perpendicular to QR. So, $OM \perp QR$. Therefore, the angle between the planes VQR and PQRS is $\angle VMO$. Step 2: Calculate the length of OM. Since O is the center of the rectangle and M is the midpoint of QR, OM is parallel to PQ and its length is half of PQ. $$OM = \frac{1}{2} PQ$$ $$OM = \frac{1}{2} \times 16$$ $$OM = 8 \text{ cm}$$ Step 3: Calculate the height of the pyramid VO. In the right-angled triangle VOP (from part b), we can find VO using the Pythagorean theorem: $$VO^2 + OP^2 = VP^2$$ $$VO^2 + 10^2 = 18^2$$ $$VO^2 + 100 = 324$$ $$VO^2 = 324 - 100$$ $$VO^2 = 224$$ $$VO = \sqrt{224} = \sqrt{16 \times 14} = 4\sqrt{14} \text{ cm}$$ Step 4: Use trigonometry in the right-angled triangle VOM. Triangle VOM is a right-angled triangle with the right angle at O. We know $VO = 4\sqrt{14}$ cm (opposite to $\angle VMO$) and $OM = 8$ cm (adjacent to $\angle VMO$). $$\tan(\angle VMO) = \frac{VO}{OM}$$ $$\tan(\angle VMO) = \frac{4\sqrt{14}}{8}$$ $$\tan(\angle VMO) = \frac{\sqrt{14}}{2}$$ $$\angle VMO = \arctan\left(\frac{\sqrt{14}}{2}\right)$$ $$\angle VMO \approx 61.86^\circ$$ The size of the angle between the planes VQR and PQRS is $\boxed{61.9^\circ}$ (to one decimal place). What's next?

Let O be the center of the rectangular base PQRS. Since the pyramid is a right pyramid, the vertex V is directly above O, meaning VO is the height of the pyramid and is perpendicular to the base PQRS.

Let O be the center of the rectangular base PQRS. Since the pyramid is a right pyramid, the vertex V is directly above O, meaning VO is the height of the pyramid and is perpendicular to the base PQRS.

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Let O be the center of the rectangular base PQRS. Since the pyramid is a right pyramid, the vertex V is directly above O, meaning VO is the height of the pyramid and is perpendicular to the base PQRS.

Let O be the center of the rectangular base PQRS. Since the pyramid is a right pyramid, the vertex V is directly above O, meaning VO is the height of the pyramid and is perpendicular to the base PQRS.

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