Pythagoras Theorem

Step 1: Determine Altitude $BD$
In right $\triangle ABC$ with $BD \perp AC$, by similarity of $\triangle ABD$ and $\triangle DBC$: $$\frac{AD}{BD} = \frac{BD}{CD} \implies BD^2 = AD \cdot CD$$ Substitute $AD = 4\text{ cm}$ and $CD = 9\text{ cm}$: $$BD^2 = 4 \times 9 = 36 \implies BD = \sqrt{36} = 6\text{ cm}$$ Step 2: Calculate Side $AB$
The hypotenuse $AC = AD + CD = 4 + 9 = 13\text{ cm}$.
By the leg-hypotenuse relation ($AB^2 = AD \cdot AC$): $$AB^2 = 4 \times 13 = 52 \implies AB = \sqrt{52} = 2\sqrt{13} \approx 7.21\text{ cm}$$ Step 3: Calculate Side $BC$
By the second leg relation ($BC^2 = CD \cdot AC$): $$BC^2 = 9 \times 13 = 117 \implies BC = \sqrt{117} = 3\sqrt{13} \approx 10.82\text{ cm}$$ Step 4: Verification via Pythagoras Theorem
$$AB^2 + BC^2 = 52 + 117 = 169 = 13^2 = AC^2$$ Matches perfectly. Final values: $BD = 6\text{ cm}$, $AB = 2\sqrt{13}\text{ cm}$, $BC = 3\sqrt{13}\text{ cm}$.

Step 1: Initial Distance of Foot from Building
Let length of ladder $L = 25\text{ m}$ (hypotenuse).
Height of window $y_1 = 20\text{ m}$.
Let initial distance of foot from wall be $x_1$.
By Pythagoras Theorem in the right-angled triangle formed by wall, ground, and ladder: $$x_1^2 + y_1^2 = L^2 \implies x_1^2 + 20^2 = 25^2$$ $$x_1^2 + 400 = 625 \implies x_1^2 = 225 \implies x_1 = \sqrt{225} = 15\text{ m}$$ Step 2: New Position after Top Slides Down
The top of the ladder slides down by $4\text{ m}$, so the new height is: $$y_2 = 20 - 4 = 16\text{ m}$$ The length of the ladder remains constant at $L = 25\text{ m}$.
Let the new distance of the foot from the building be $x_2$: $$x_2^2 + y_2^2 = L^2 \implies x_2^2 + 16^2 = 25^2$$ $$x_2^2 + 256 = 625 \implies x_2^2 = 369 \implies x_2 = \sqrt{369} \approx 19.21\text{ m}$$ Step 3: Distance the Foot Slides Outwards
$$\text{Outward slide } \Delta x = x_2 - x_1 = \sqrt{369} - 15 \approx 19.21 - 15 = 4.21\text{ m}$$ Conclusion: The initial distance of the foot was $15\text{ m}$, and it slides outward by approximately $4.21\text{ m}$.

Step 1: Formulate Right Triangles
Since $AD \perp BC$, $D$ is the foot of the altitude on $BC$. This creates two right-angled triangles: $\triangle ABD$ and $\triangle ACD$, both sharing the common altitude $AD$. Step 2: Apply Pythagoras Theorem to $\triangle ABD$
In right $\triangle ABD$, $\angle ADB = 90^\circ$: $$AB^2 = AD^2 + BD^2 \implies AD^2 = AB^2 - BD^2 \quad \text{--- (Equation 1)}$$ Step 3: Apply Pythagoras Theorem to $\triangle ACD$
In right $\triangle ACD$, $\angle ADC = 90^\circ$: $$AC^2 = AD^2 + CD^2 \implies AD^2 = AC^2 - CD^2 \quad \text{--- (Equation 2)}$$ Step 4: Equating Both Expressions for $AD^2$
From Equation (1) and Equation (2): $$AB^2 - BD^2 = AC^2 - CD^2$$ Rearranging terms by transposing $-BD^2$ and $-CD^2$: $$AB^2 + CD^2 = AC^2 + BD^2$$ Hence Proved (Q.E.D.).

Step 1: Identify the Right Triangles Involved
Since $\angle C = 90^\circ$, any triangle having $C$ as a vertex and two legs lying along $CA$ and $CB$ is a right-angled triangle. We identify four such triangles:
1. $\triangle ACE$ with hypotenuse $AE$.
2. $\triangle BCD$ with hypotenuse $BD$.
3. $\triangle ACB$ with hypotenuse $AB$.
4. $\triangle DCE$ with hypotenuse $DE$. Step 2: Express Hypotenuse Squares Using Pythagoras Theorem
In right $\triangle ACE$: $$AE^2 = AC^2 + CE^2 \quad \text{--- (1)}$$ In right $\triangle BCD$: $$BD^2 = CD^2 + BC^2 \quad \text{--- (2)}$$ Step 3: Sum the Left-Hand Side
Adding Equation (1) and Equation (2): $$AE^2 + BD^2 = (AC^2 + CE^2) + (CD^2 + BC^2)$$ Regrouping the terms on the right-hand side: $$AE^2 + BD^2 = (AC^2 + BC^2) + (CD^2 + CE^2) \quad \text{--- (3)}$$ Step 4: Substitute from the Remaining Right Triangles
In right $\triangle ACB$: $$AC^2 + BC^2 = AB^2$$ In right $\triangle DCE$: $$CD^2 + CE^2 = DE^2$$ Substituting these into Equation (3): $$AE^2 + BD^2 = AB^2 + DE^2$$ Hence Proved (Q.E.D.).

Step 1: Properties of Isosceles Triangle Altitude
In an isosceles triangle with $AB = AC$, the altitude $AD \perp BC$ bisects the base $BC$.
Therefore: $$BD = CD = \frac{BC}{2} = \frac{10}{2} = 5\text{ cm}$$ Step 2: Apply Pythagoras Theorem to $\triangle ABD$
In right-angled triangle $\triangle ABD$, $\angle ADB = 90^\circ$: $$AB^2 = AD^2 + BD^2$$ Substitute $AB = 13\text{ cm}$ and $BD = 5\text{ cm}$: $$13^2 = AD^2 + 5^2 \implies 169 = AD^2 + 25$$ $$AD^2 = 169 - 25 = 144 \implies AD = \sqrt{144} = 12\text{ cm}$$ Step 3: Calculate Area of $\triangle ABC$
$$\text{Area}(\triangle ABC) = \frac{1}{2} \times \text{base} \times \text{altitude} = \frac{1}{2} \times BC \times AD$$ $$\text{Area} = \frac{1}{2} \times 10 \times 12 = 60\text{ cm}^2$$ Conclusion: Length of altitude $AD = 12\text{ cm}$ and $\text{Area} = 60\text{ cm}^2$.