Question

In a right triangle, prove that the square of the hypotenuse is equal to the sum of squares of the other two sides.

Using the above, solve the following:

From figure, find the length of $CAifCD\perp DBandAB\perp DB$

Answer 1

Step 1: Note the given data and draw a diagram

Given a right angle triangle.

Let $△ABC$ be a right angled triangle at $B$.

Construction: Draw a perpendicular from $B$ on $AC$.

Step 2: Check similarity for $△ABC$ and $△ABD$; $△ABC$ and $△BCD$

AA similarity: If two angles of a triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

$\angle ABC=\angle ADB=90°$

$\angle BAC=\angle BAD$ (common angle)

According to AA similarity

$△ABC~$$△ABD$

$$\begin{gathered} \Rightarrow \frac{AD}{AB}=\frac{AB}{AC} \\ \Rightarrow A{B}^2=AD\times AC...\left(i\right) \end{gathered}$$

In $△ABC$ and $△BCD$

$\angle ABC=\angle CDB=90°$

$\angle BCA=\angle BCD$ (common angle)

According to AA similarity

$△ABC~$$△BCD$

$$\begin{gathered} \Rightarrow \frac{BC}{DC}=\frac{AC}{BC} \\ \Rightarrow B{C}^2=DC\times AC..\left(ii\right) \end{gathered}$$

Step 3: Adding equations (i) and (ii)

$$\begin{gathered} A{B}^2+B{C}^2=AD\times AC+DC\times AC \\ \Rightarrow A{B}^2+B{C}^2=AC\left(AD+DC\right) \\ \Rightarrow A{B}^2+B{C}^2=AC\times AC \\ \Rightarrow A{B}^2+B{C}^2=A{C}^2 \end{gathered}$$

Hence the square of the hypotenuse is equal to the sum of squares of the other two sides.(proved)

Step 4: Find the length of $CE$ and $AE$

Construct: Draw $CE\parallel DB$

Construct: Draw $CE\parallel DB$

$CDBE$is a rectangle since $CD\perp DB,AB\perp DB$ and $CE\parallel DB$.

$\Rightarrow CE=12cm$

$$\begin{gathered} AE=AB-EB \\ \Rightarrow AE=AB-CD(CDBE\mathrm{is}\mathrm{a}\mathrm{rectangle},CD=EB) \\ \Rightarrow AE=11-6 \\ \Rightarrow AE=5cm \end{gathered}$$

Step 5: Finding the length of $CA$ by applying Pythagoras' theorem in $△AEC$

In a right triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.

In $△AEC$, base$=CE$, height$=AE$ and hypotenuse$=CA$

$$\begin{gathered} C{A}^2=A{E}^2+C{E}^2 \\ \Rightarrow C{A}^2=5^2+{12}^2 \\ \Rightarrow C{A}^2=25+144 \\ \Rightarrow C{A}^2=169 \\ \Rightarrow CA=13cm \end{gathered}$$

Hence, the length of $CA=13cm$