1. Problem (18): The perimeter of a triangular field is 144 m and its sides are in the ratio 3 : 4 : 5. Find the area of the field. Step 1: Let the sides be $3x$, $4x$, $5x$. Since their sum is the perimeter, $$3x+4x+5x=144 \implies 12x=144 \implies x=12$$ Step 2: So sides are $36, 48,$ and $60$ meters. Step 3: This is a right-angled triangle since $36^2+48^2=1296+2304=3600 = 60^2$. Step 4: Area of right triangle = $\frac{1}{2} \times 36 \times 48 = \frac{1}{2} \times 1728 = 864$ m$^2$. 2. Problem (20): Find the area of the shaded region (triangle with right triangle where $AB=20$ cm). Given: - Lengths $AP=12$ cm, $PB=16$ cm - Right triangle $APB$ with $AB=20$ cm. Step 1: The triangle's base $AB=20$ cm. Step 2: Height from point perpendicular is not explicitly given. But since $AP^2 + PB^2 = AB^2$ using Pythagoras: $$12^2 + 16^2 = 144 + 256 = 400 = 20^2$$ Hence, the triangle is right angled at $P$. Step 3: Area = $\frac{1}{2} \times AP \times PB = \frac{1}{2} \times 12 \times 16 = 96$ cm$^2$. 3. Problem (24): The perimeter of a square and an equilateral triangle are same. The length of diagonal of a square is $12\sqrt{2}$ cm. Find the area of the triangle. Step 1: Diagonal of square $d = 12\sqrt{2}$ cm. Step 2: Side of square $s = \frac{d}{\sqrt{2}}=\frac{12\sqrt{2}}{\sqrt{2}}=12$ cm. Step 3: Perimeter of square = $4 \times 12 = 48$ cm. Step 4: Equilateral triangle side $= \frac{48}{3} = 16$ cm. Step 5: Area of equilateral triangle = $\frac{\sqrt{3}}{4} \times 16^2 = \frac{\sqrt{3}}{4} \times 256 = 64\sqrt{3}$ cm$^2$. 4. Problem (25): The length of median in an equilateral triangle is $x$ cm. Find its area. Step 1: In an equilateral triangle with side length $a$, median length $m = \frac{\sqrt{3}}{2}a$. Step 2: Given median length $m = x$, so $$x = \frac{\sqrt{3}}{2}a \implies a = \frac{2x}{\sqrt{3}}$$ Step 3: Area of equilateral triangle = $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \times \left(\frac{2x}{\sqrt{3}}\right)^2 = \frac{\sqrt{3}}{4} \times \frac{4x^2}{3} = \frac{\sqrt{3}}{4} \times \frac{4x^2}{3} = \frac{\sqrt{3} \cdot 4x^2}{12} = \frac{\sqrt{3}x^2}{3}$ Final answers: - Problem 18 area: $864$ m$^2$ - Problem 20 area: $96$ cm$^2$ - Problem 24 area: $64\sqrt{3}$ cm$^2$ - Problem 25 area: $\frac{\sqrt{3}x^2}{3}$ cm$^2$