Question

Please Solve This And Get Thanks​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

∠EAB = 90°, ∠DCB = 90°, ∠EBD = 90°

In right angle triangle ABE, AB = 6 cm, AE = 9 cm

In right angle triangle DBC, BC = 6 cm, DC = 4 cm

Now, Consider right-angled triangle ABE

By using Pythagoras Theorem, we have

$\rm \: {BE}^{2} = {AB}^{2} + {AE}^{2}$

So, on substituting the values of AB and AE, we get

$\rm \: {BE}^{2} = {6}^{2} + {9}^{2}$

$\rm \: {BE}^{2} =36 + 81$

$\rm \: {BE}^{2} = 117 \:$

$\rm\implies \:\boxed{\sf{ \: \: \rm \: {EB}^{2} \: = \: 117 \: cm \: \: \: }}$

Now, Consider right-angled triangle DCB

By using Pythagoras Theorem, we have

$\rm \: {DB}^{2} = {CB}^{2} + {DC}^{2}$

On substituting the values of CB and DC, we get

$\rm \: {DB}^{2} = {6}^{2} + {4}^{2}$

$\rm \: {DB}^{2} = 36 + 16$

$\rm \: {DB}^{2} = 52 \:$

$\rm\implies \: \: \: \boxed{\sf{ \: \: \: \rm \: {BD}^{2} = 52 \: cm \: \: \: }}$

Now, Consider right-angled triangle EBD

By using Pythagoras Theorem, we have

$\rm \: {ED}^{2} = {EB}^{2} + {BD}^{2}$

So, on substituting the values, we get

$\rm \: {ED}^{2} = 117 + 52$

$\rm \: {ED}^{2} = 169$

$\rm \: {ED}^{2} = {13}^{2}$

$\rm\implies \:\boxed{\sf{ \: \: \rm \: \: ED \: = \: 13 \: cm \: \: \: }}$

Hence,

$\begin{gathered}\begin{gathered}\bf\: \rm\implies \:\begin{cases} &\sf{ {EB}^{2} = 117 \: cm } \\ \\ &\sf{ {BD}^{2} = 52 \: cm } \\ \\ &\sf{ED \: = \: 13 \: cm} \end{cases}\end{gathered}\end{gathered}$

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Additional Information :-

1. Pythagoras Theorem :-

This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

2. Converse of Pythagoras Theorem :-

This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

3. Area Ratio Theorem :-

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

4. Basic Proportionality Theorem :-

This theorem states that : If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.

Answer 2

Question :-

From the figure , Calculate :-

$\rm \: i) {EB}^{2} \: \: ii) {BD}^{2} \: \: iii) {ED}^{2}$

$\rm \huge{ \underline{ \underline{Solution :-}}}$

Given :-

• ∠BAE = 90°
• ∠DCB = 90°
• ∠DBE = 90°

Now , can you notice two right - angled triangle ABE and DBC .

1. In the right - angle triangle ABE , Given :-

• AB = 6 cm
• AE = 9 cm

    2. In the right - angle triangle DCB , Given :-

• BC = 6 cm
• CD = 4 cm

Now , According to Pythagoras Theorem ,

$\boxed{ \tt \: AB^{2} = AC ^{2} + BC ^{2} }$

But , here it

In right angle triangle ABE ,

$\boxed{ \tt \: EB ^{2} = AB ^{2} + AE ^{2} }$

$\tt \implies \: EB^{2} = {6}^{2} + {9}^{2}$

$\tt \implies \: EB^{2} = 36 + 81$

$\tt \implies \:EB^{2} = 117 \: cm$

$\therefore \tt \: EB ^{2} = 117 \: cm$

Now finding ii that is BD whole Square.

Now , According to Pythagoras Theorem,

In right angle triangle DCB ,

$\boxed{ \tt \: DB ^{2} = CB^{2} + DC ^{2} }$

$\implies\tt \: DB ^{2} = 6^{2} + 4 ^{2}$

$\implies\tt \: DB ^{2} = 3 6 + 16$

$\implies\tt \: DB ^{2} = 52 \: cm$

$\therefore \tt \:DB ^{2} = 52\: cm$

Now , According to Pythagoras Theorem ,

In right angle triangle EBD ,

$\boxed{ \tt \: ED ^{2} = EB ^{2} + BD ^{2} }$

$\implies\tt \: ED ^{2} = 117+ 52$

$\implies\tt \: ED ^{2} = 169$

$\implies\tt \: ED ^{2} = 13 \times 13$

$\implies\tt \: ED ^{2} = 13 ^{2}$

$\therefore \tt \:ED ^{2} = 13\: cm$

Therefore,

$\longmapsto \: \begin{cases} \tt EB^{2} = 117 \: cm \: \\ \tt \: BD ^{2} = 52 \: cm \\ \tt ED = 13 \: cm\end{cases}$

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More :-

$\star\boxed{\tt\bf \: Hypotenuse ^2 = Base ^2+Perpendicular ^2}$

• Always remember Hypotenuse is the biggest side