1. A donkey is tied with a rope to a post which is 15 m from a fence. If the length of the rope between the donkey and the post is 17m, calculate the length...
Question 1 Report
1. A donkey is tied with a rope to a post which is 15 m from a fence. If the length of the rope between the donkey and the post is 17m, calculate the length of the fence within the reach of the donkey.
2. The base of a right pyramid with vertex, V, is a square, PQRS, of side 15 cm. If the slant height is 32 cm long:
3. represent the information in a diagram;
4. calculate its:
5. height, correct to one decimal place;
6. volume, correct to the nearest \(cm^3\)
Answer Details
1. Length of fence within reach
The donkey can reach every point 17 m from the post, so its path forms a circle of radius 17 m. The fence is 15 m from the post. This makes a right-angled triangle from the post to the nearest point on the fence and to one end of the reachable section.
Using Pythagoras’ theorem on one half of the reachable fence length:
\[ x^2=17^2-15^2=289-225=64 \] \[ x=8\text{ m} \]The total length of fence is twice this value:
\[ 2x=2(8)=\boxed{16\text{ m}} \]2. Right square pyramid
The reference answer treats the given 32 cm as the sloping edge from the vertex \(V\) to a corner of the square base, for example \(VP\). In a square, the centre \(O\) is halfway along a diagonal, so \(OP\) is half the diagonal. The vertical height is \(VO\).
Height
First find the diagonal \(PR\) of the square base:
\[ PR=\sqrt{15^2+15^2}=15\sqrt2\text{ cm} \]Since \(O\) is the centre of the square,
\[ OP=\frac{PR}{2}=\frac{15\sqrt2}{2}\approx10.6066\text{ cm} \]Triangle \(VOP\) is right-angled at \(O\), so:
\[ 32^2=h^2+10.6066^2 \] \[ h^2=1024-112.5=911.5 \] \[ h=\sqrt{911.5}=30.191\ldots \] \[ \boxed{h=30.2\text{ cm}} \]Volume
\[ \text{Volume}=\frac13\times\text{base area}\times\text{perpendicular height} \] \[ =\frac13\times15^2\times30.191\ldots \] \[ =2264.33\ldots\text{ cm}^3 \] \[ \boxed{2264\text{ cm}^3} \]The unit for volume must be \(\text{cm}^3\), not \(\text{cm}^2\). Also, distinguish between half a base edge, \(7.5\) cm, and the distance from the centre to a corner, \(OP=\frac{15\sqrt2}{2}\) cm. The latter is required here because the 32 cm line is \(VP\), ending at a corner of the base.
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