Geometry - Construction and Loci

A gardener wants to plant a tree that is: (1) More than 5m from a fixed post P, and (2) Closer to fence AB than to fence AC. Using a scale of 1cm : 1m, shade the region where the tree can be planted.

1. Draw a circle with center P and radius 5cm. The area outside this circle satisfies condition 1. 2. Construct the angle bisector of the angle at vertex A (where fences AB and AC meet). The region on the side of the bisector closer to AB satisfies condition 2. 3. Shade the intersection of the area outside the circle and the side of the bisector closer to AB.

Condition 1 defines a locus based on a fixed point, which is a circle. 'More than' means the exterior region. Condition 2 defines a locus equidistant from two intersecting lines, which is an angle bisector. The 'closer to' requirement indicates one side of that bisector.

Construct a triangle ABC where AB = 8cm, AC = 6cm, and BC = 7cm. Then, construct the locus of points inside the triangle that are exactly 3cm from vertex A.

1. Draw line AB = 8cm. 2. Set compass to 6cm, draw an arc from A. 3. Set compass to 7cm, draw an arc from B. The intersection is point C. Join AC and BC. 4. To find the locus 3cm from A, set the compass to 3cm and draw an arc inside the triangle with the needle at point A.

This problem combines basic triangle construction using SSS (Side-Side-Side) criteria with the definition of a locus from a fixed point. The locus is a circular arc centered at A with a radius of 3cm.

The scale of a map is 1:25,000. If the distance between two towns on the map is 4cm, calculate the actual distance in kilometers.

1. Actual distance in cm = $4 \times 25,000 = 100,000$ cm. 2. Convert to meters: $100,000 \div 100 = 1,000$ m. 3. Convert to km: $1,000 \div 1,000 = 1$ km.

To convert map distance to real distance, multiply by the scale factor $n$. Then, use the conversions $100\text{cm} = 1\text{m}$ and $1000\text{m} = 1\text{km}$ to reach the required units.

1. Draw the rectangle $ABCD$ with dimensions $8\text{ cm} \times 5\text{ cm}$.
2. The locus of points equidistant from $AD$ and $BC$ is the perpendicular bisector of the sides $AB$ and $CD$. This is a vertical line $x = 4$ if $A$ is at $(0,5)$ and $B$ is at $(8,5)$. Shading to the left of this line satisfies 'closer to $AD$'.
3. The locus of points exactly $6\text{ cm}$ from $A$ is an arc of a circle with center $A$ and radius $6\text{ cm}$.
4. The final region is the intersection of the left half of the rectangle and the interior of the arc centered at $A$.

To solve this, we combine the 'perpendicular bisector' logic (applied to parallel lines) with the 'circle' locus centered at a vertex. Only the overlapping area within the rectangle is shaded.

1. Mark points $T1$ and $T2$ at a distance of $10\text{ cm}$.
2. Construct the perpendicular bisector of the line segment $T1T2$.
3. Draw a circle around $T1$ with a radius of $4\text{ cm}$ ($40\text{ km}$).
4. Shade the region that is to the left of the bisector (closer to $T1$) but outside the circle (at least $40\text{ km}$ from $T1$).

The boundary for 'closer to' is the perpendicular bisector. The boundary for 'at least' is the circle itself, including all points on the circumference and everything outside of it.