A rectangle is 6 cm by 9 cm. Find its diagonal to 3 significant figures.

1. The diagonal is the hypotenuse of a right-angled triangle.

2. Use Pythagoras.

   $$d^2=6^2+9^2=117$$

3. Take the square root and round.

   $$d=\sqrt{117}=10.816\dots \approx 10.8cm$$

A cuboid has length 9 cm, depth 5 cm and height 6 cm. Find the space diagonal $AG$ to 3 significant figures.

1. First find the base diagonal $AC$.

   $$A{C}^2=9^2+5^2=106$$

2. Now use $AC$ with the height 6 cm to find $AG$.

   $$A{G}^2=A{C}^2+6^2=106+36=142$$

3. Take the square root.

   $$AG=\sqrt{142}=11.916\dots \approx 11.9cm$$

A cuboid has width 5 cm, height 6 cm and space diagonal 12 cm. Find the missing depth $x$ to 3 significant figures.

1. Use the cuboid diagonal relationship.

   $$12^2=5^2+x^2+6^2$$

2. Rearrange by subtracting the known squares.

   $$x^2=12^2-5^2-6^2=144-25-36=83$$

3. Square root the result.

   $$x=\sqrt{83}=9.110\dots \approx 9.11cm$$

A cuboid has base dimensions 8 cm by 6 cm and height 4 cm. Find the angle between the sloping diagonal $EC$ and the base diagonal $AC$.

1. Find the base diagonal $AC$.

   $$AC=\sqrt{8^2+6^2}=10cm$$

2. In triangle $AEC$, the height $AE=4$ is opposite the angle at C, and $AC=10$ is adjacent.

3. Use tangent because you know opposite and adjacent.

   $$\tan \theta =\frac{4}{10},\theta ={\tan }^{-1}\left(\frac{4}{10}\right)=21.801{\dots }^{\circ }$$

4. The angle is 21.8° to 3 significant figures.

In a triangular prism, triangle $ADC$ is right-angled at C. The lengths are $AD=15$ cm and $CD=8$ cm. In another right-angled triangle, $\angle FDC=35^{\circ }$. Find $\angle AFC$ to 1 decimal place.

1. Find $AC$ from triangle $ADC$.

   $$AC=\sqrt{15^2-8^2}=\sqrt{161}$$

2. Find $FC$ from triangle $FDC$.

   $$FC=8\tan 35^{\circ }=5.601\dots$$

3. In triangle $AFC$, the angle at F has opposite side $AC$ and adjacent side $FC$.

   $$\theta ={\tan }^{-1}\left(\frac{\sqrt{161}}{5.601\dots }\right)=66.19{\dots }^{\circ }$$

4. So $\angle AFC=66.2^{\circ }$ to 1 decimal place.

In a prism, point F is vertically above C. The projection of $AF$ onto the base plane is $AC$. In the base, triangle $ADC$ is right-angled at C, with $AD=34$ cm and $CD=16$ cm. Also, in triangle $FCD$, $\angle FDC=35^{\circ }$. Find the angle between $AF$ and the base plane.

1. The required angle is $\angle FAC$, because $AC$ is the projection of $AF$ on the plane.

2. Find the projection length $AC$.

   $$AC=\sqrt{34^2-16^2}=30cm$$

3. Find the vertical height $FC$.

   $$FC=16\tan 35^{\circ }=11.203\dots$$

4. Use triangle $AFC$ to find the angle.

   $$\theta ={\tan }^{-1}\left(\frac{11.203\dots }{30}\right)=20.49{\dots }^{\circ }$$

5. The angle is 20.5° to 3 significant figures.

A square-based pyramid has side length 8 cm. The apex E is 12 cm vertically above the centre O of the base. Find $\angle EAC$.

1. Find the diagonal of the square base.

   $$AC=8\sqrt{2}$$

2. Since O is the centre of the square, $AO$ is half the diagonal.

   $$AO=4\sqrt{2}$$

3. In right-angled triangle $AOE$, $\angle EAC$ is the same as $\angle EAO$.

   $$\tan \theta =\frac{12}{4\sqrt{2}}$$

4. Use inverse tangent.

   $$\theta ={\tan }^{-1}\left(\frac{12}{4\sqrt{2}}\right)=64.760{\dots }^{\circ }$$

5. So $\angle EAC=64.8^{\circ }$ to 3 significant figures.

A square-based pyramid has base side length 15 cm. The apex P is vertically above the centre O, and $\angle PAC=65^{\circ }$. Find the volume of the pyramid.

1. Find $AO$, half the diagonal of the square.

   $$AO=\frac{15\sqrt{2}}{2}=10.606\dots$$

2. Use tangent to find the vertical height $PO$.

   $$PO=10.606\dots \tan 65^{\circ }=22.744\dots$$

3. Find the base area.

   $$base\; area=15^2=225$$

4. Use the pyramid volume formula.

   $$V=\frac{1}{3}\times 225\times 22.744\dots =1705.8\dots$$

5. The volume is about 1710 cm³ to 3 significant figures.