1. **Problem Statement:** Given a rectangular land plot of dimensions 8 m by 7 m divided into four parts labeled (1), (2), (3), and (4) with the following sizes: - (1) has width $x+2$ and height 3 m - (2) has width $x+1$ and height 2 m - (3) has width $x$ and height 1 m - (4) has width $x$ and height 3 m The width 7 m satisfies $x + (x+1) + (x+2) = 7$. We are asked: (a) Find $x$ such that the area of (3) equals the sum of areas of (1), (2), and (4). (b) If the total land area is 100 m² larger than the sum of the four parts, find $x$. --- 2. **Step 1: Write area expressions** $$\text{Area}(1) = 3(x+2) = 3x + 6$$ $$\text{Area}(2) = 2(x+1) = 2x + 2$$ $$\text{Area}(3) = 1 \times x = x$$ $$\text{Area}(4) = 3 \times x = 3x$$ --- 3. **Step 2: Use width sum constraint** $$x + (x+1) + (x+2) = 7 \implies 3x + 3 = 7 \implies 3x = 4 \implies x = \frac{4}{3}$$ So the widths satisfy the original problem's dimension. --- 4. **Step 3: Condition that Area(3) equals sum of others: $$x = (3x + 6) + (2x + 2) + 3x = 3x+6 + 2x+2 + 3x = (3+2+3)x + 8 = 8x + 8$$ Rewrite: $$x = 8x + 8 \implies 0 = 7x + 8 \implies 7x = -8 \implies x = -\frac{8}{7}$$ This negative value is not possible for a length, so no valid $x$ satisfies this. --- 5. **Step 4: Check if $x=4/3$ satisfies the original land width** With $x=4/3$: $$\text{Total width} = x + (x+1)+ (x+2) = 7$$ (verified above) --- 6. **Step 5: Condition on total area difference** Total land area = $8 \times 7 = 56$ m² Sum of parts' areas: $$(3x + 6) + (2x + 2) + x + 3x = 9x + 8$$ According to the problem, land area is 100 m² greater than the sum of parts: $$56 = 9x + 8 + 100 \implies 9x + 8 = 56 - 100 = -44$$ Which is impossible since $x$ can't be negative. But the problem states "piece of land is 100 m² larger than the sum of parts," so: $$56 = ext{sum of parts} + 100 \to ext{sum of parts} = 56 - 100 = -44$$ which is invalid. If the problem means the land is 100 m² larger than the sum of the parts, then total land area must be: $$\text{Land area} = \text{sum of parts} + 100$$ But the total land area is fixed 56. So this condition can't hold. --- 7. **Step 6: Alternative interpretation: The land area is 100 m² larger than the sum of the four parts, so:** $$8 \times 7 = (3x + 6) + (2x + 2) + x + 3x + 100$$ $$56 = 9x + 8 + 100$$ $$56 = 9x + 108$$ $$9x = 56 - 108 = -52$$ $$x = -\frac{52}{9}$$ negative length again. Hence, no positive solution exists for this case. --- 8. **Step 7: Summary** - From the width constraint $3x+3=7$, $x=\frac{4}{3}$. - This $x$ gives areas: - Area(1) = $3(\frac{4}{3}+2) = 3(\frac{10}{3}) = 10$ - Area(2) = $2(\frac{4}{3}+1) = 2(\frac{7}{3}) = \frac{14}{3} \approx 4.67$ - Area(3) = $\frac{4}{3} \approx 1.33$ - Area(4) = $3 \times \frac{4}{3} = 4$ Sum parts areas: $10 + 4.67 + 1.33 + 4 = 20$ Land area is 56, difference is $56 - 20 = 36$ and not 100. - The condition that Area(3) equals the sum of (1)+(2)+(4) leads to no positive $x$. - The condition that the land area exceeds sum of parts by 100 leads to no positive $x$. --- **Final answer:** \begin{cases} x = \frac{4}{3} \text{ satisfies the widths matching the entire land width} \\ \text{No positive }x \text{ satisfies }\text{Area}(3)=\text{Area}(1)+\text{Area}(2)+\text{Area}(4) \\ \text{No positive }x \text{ satisfies } \text{Land area} = \text{sum parts} + 100 \end{cases}