Parallelogram Construction 59328C

1. **State the problem:** Construct parallelogram $ABCD$ with $AB=10$ cm, $AD=6.5$ cm, $\angle BAD=110^\circ$, and $\angle ABC=70^\circ$. Then measure diagonal $AC$ and construct loci of points inside the parallelogram based on given distance conditions. 2. **Construct parallelogram $ABCD$:** - Draw segment $AB=10$ cm. - At point $A$, construct angle $110^\circ$ and draw ray $AD$. - On ray $AD$, mark point $D$ such that $AD=6.5$ cm. - Since $ABCD$ is a parallelogram, $BC$ is parallel and equal to $AD$, so draw $BC$ parallel to $AD$ with length $6.5$ cm. - Connect points $C$ and $D$ to complete parallelogram. 3. **Measure diagonal $AC$:** - Use the law of cosines in triangle $ABC$ or measure directly. - Using law of cosines: $$AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\angle ABC)$$ $$= 10^2 + 6.5^2 - 2 \times 10 \times 6.5 \times \cos 70^\circ$$ $$= 100 + 42.25 - 130 \times 0.3420 = 142.25 - 44.46 = 97.79$$ $$AC = \sqrt{97.79} \approx 9.89 \text{ cm}$$ 4. **Construct locus of points:** - (a) Points $x$ cm from $AB$: This is a line parallel to $AB$ at distance $x$ inside the parallelogram. - (b) Points $y$ cm from $C$: This is a circle centered at $C$ with radius $y$. - (c) Points equidistant from $BC$ and $CD$: This is the angle bisector of angle $C$ inside the parallelogram. 5. **Point $Z$ conditions:** - Distance from $AB \leq x$ cm. - Distance from $C \leq y$ cm. - Closer to $CD$ than $BC$. 6. **Shade the region:** - Shade the area inside the parallelogram satisfying all above conditions. - Label sides $BC$ and $CD$ clearly. **Final answer:** - Length $AC \approx 9.89$ cm. - Loci constructed as described. - Region for point $Z$ shaded accordingly.