1. Let's analyze each house diagram carefully. 2. For the first house (top-left): - The triangle has sides 75s, 85s, and base 20s. - The rectangle height is labeled 3 min. 3. For the second house (top-right): - Triangle sides: 84cm, 32cm, base 84cm. - Rectangle height: 2 m. 4. For the third house (bottom-left): - Triangle sides: 134cm, 56cm, base unlabeled. - Rectangle height: 3 m. 5. For the fourth house (bottom-right): - Triangle sides: 200cm, 50cm, base 100cm. - Rectangle height: m (unknown). 6. To solve, we assume the triangles are right triangles and use the Pythagorean theorem to verify or find missing lengths. 7. Check first triangle: $$75s^2 + 20s^2 = 5625s^2 + 400s^2 = 6025s^2$$ $$85s^2 = 7225s^2$$ Since $$6025s^2 \neq 7225s^2$$, the triangle is not right angled as is. 8. Check second triangle: $$32cm^2 + 84cm^2 = 1024 + 7056 = 8080 cm^2$$ $$84cm^2 = 7056 cm^2$$ Not equal; triangle is not right angled. 9. Third triangle: base unknown, sides 134cm, 56cm. Let's assume 134cm is hypotenuse: $$134^2 = 17956$$ Find base $$b$$: $$b^2 + 56^2 = 17956$$ $$b^2 + 3136 = 17956$$ $$b^2 = 17956 - 3136 = 14820$$ $$b = \sqrt{14820} \approx 121.74 cm$$ 10. Fourth triangle: Hypotenuse maybe 200cm; one side 50cm; base 100cm given. Check Pythagorean theorem: $$50^2 + 100^2 = 2500 + 10000 = 12500$$ $$200^2 = 40000$$ Not equal; triangle is not right angled. 11. For rectangle heights: - Conversion needed to unify units or to find unknown $m$. 12. Given rectangle height in bottom-right is $m$: Compare with known heights in other diagrams. If 200cm (2m) corresponds to 3m height in third house, we estimate $m$ as 3m to maintain proportion. Final answers: - Missing base in third house is approximately 121.74 cm. - Rectangle height $m$ in fourth house estimated as 3 m. These calculations combine dimensional conversions and Pythagorean theorem to find unknown lengths and heights.