If DE = 4cm, CE = 5cm and BC = 9cm. Find CD, AB and AC from the given figure.

Answer
324.9k+ views
Hint: First consider \[\vartriangle CDE\], find the length of the CD using the Pythagoras theorem. By using the similarity of both the triangles, the ratio of length of their sides will be proportional. Find length of AB. Then by using Pythagoras theorem in \[\vartriangle ABC\], find the length of AC.
Complete step-by-step answer:
Given to us a right angled triangle ABC and values of the sides of \[\vartriangle ABC\].
i.e. DE = 4cm, CE = 5cm and BC = 9cm.
Here, the length of BC = 9cm.
BC = CE + EB
We know the length of BC and CE.
Let’s find the length of EB.
\[\begin{align}
& 9=5+EB \\
&\Rightarrow EB=9-5=4cm \\
\end{align}\]
Now let us consider \[\vartriangle ABC\]and the smaller right angled triangle DEC.
The 2 triangles are similar i.e. the ratio of the length of sides of both triangles are proportional.
Here AB and ED are both altitudes in \[\vartriangle ABC\]and \[\vartriangle CDE\] and BC and DC are both bases of the triangle ABC and CDR.
Therefore as the triangles are similar, we can write that,
\[\dfrac{AB}{ED}=\dfrac{BC}{DC}-(1)\]
Now let us go back to \[\vartriangle CDE\], by using the Pythagoras theorem, we know that,
\[\begin{align}
& C{{E}^{2}}=C{{D}^{2}}+D{{E}^{2}} \\
& \Rightarrow C{{D}^{2}}=C{{E}^{2}}-D{{E}^{2}} \\
& CD=\sqrt{C{{E}^{2}}-D{{E}^{2}}} \\
\end{align}\]
From the figure we know that CE = 5cm and DE = 4cm.
\[\therefore \] Length of \[CD=\sqrt{{{5}^{2}}-{{4}^{2}}}=\sqrt{25-16}=\sqrt{9}=3cm\].
\[\therefore \] Length of CD = 3cm.
Now, let us consider equation (1), we know the lengths of ED, BC and DC.
Substitute and find the length of AB.
\[\begin{align}
& \dfrac{AB}{ED}=\dfrac{BC}{DC} \\
& \Rightarrow \dfrac{AB}{4}=\dfrac{9}{3} \\
& \therefore AB=\dfrac{4\times 9}{3}=4\times 3=12cm \\
\end{align}\]
\[\therefore \] Length of AB = 12cm.
Now we need to find the length of AC.
Consider \[\vartriangle ABC\], by Pythagoras theorem,
\[\begin{align}
& A{{C}^{2}}=A{{B}^{2}}+C{{D}^{2}} \\
& AC=\sqrt{A{{B}^{2}}+C{{D}^{2}}} \\
\end{align}\]
We know AB = 12cm and CD = 9cm. Substitute and find the value of AC.
\[\therefore AC=\sqrt{{{12}^{2}}+{{9}^{2}}}=\sqrt{144+81}=\sqrt{225}=15cm\]
\[\therefore \] Length of AC = 15cm.
Hence, we got all the required lengths as,
CD = 3cm, AB = 12cm and AC = 15cm.
Note: In our two right angles triangles, the corresponding sides are in the same ratio. Hence the matching angles are also the same. Thus we have considered the ratio of their sides \[\dfrac{AB}{ED}\] and \[\dfrac{BC}{DC}\], which are equal. The two triangles can be also proved by SSS congruence i.e. all the sides of the two triangles are similar.
Complete step-by-step answer:
Given to us a right angled triangle ABC and values of the sides of \[\vartriangle ABC\].
i.e. DE = 4cm, CE = 5cm and BC = 9cm.
Here, the length of BC = 9cm.
BC = CE + EB
We know the length of BC and CE.
Let’s find the length of EB.

\[\begin{align}
& 9=5+EB \\
&\Rightarrow EB=9-5=4cm \\
\end{align}\]
Now let us consider \[\vartriangle ABC\]and the smaller right angled triangle DEC.
The 2 triangles are similar i.e. the ratio of the length of sides of both triangles are proportional.
Here AB and ED are both altitudes in \[\vartriangle ABC\]and \[\vartriangle CDE\] and BC and DC are both bases of the triangle ABC and CDR.
Therefore as the triangles are similar, we can write that,
\[\dfrac{AB}{ED}=\dfrac{BC}{DC}-(1)\]
Now let us go back to \[\vartriangle CDE\], by using the Pythagoras theorem, we know that,
\[\begin{align}
& C{{E}^{2}}=C{{D}^{2}}+D{{E}^{2}} \\
& \Rightarrow C{{D}^{2}}=C{{E}^{2}}-D{{E}^{2}} \\
& CD=\sqrt{C{{E}^{2}}-D{{E}^{2}}} \\
\end{align}\]
From the figure we know that CE = 5cm and DE = 4cm.
\[\therefore \] Length of \[CD=\sqrt{{{5}^{2}}-{{4}^{2}}}=\sqrt{25-16}=\sqrt{9}=3cm\].
\[\therefore \] Length of CD = 3cm.
Now, let us consider equation (1), we know the lengths of ED, BC and DC.
Substitute and find the length of AB.
\[\begin{align}
& \dfrac{AB}{ED}=\dfrac{BC}{DC} \\
& \Rightarrow \dfrac{AB}{4}=\dfrac{9}{3} \\
& \therefore AB=\dfrac{4\times 9}{3}=4\times 3=12cm \\
\end{align}\]
\[\therefore \] Length of AB = 12cm.
Now we need to find the length of AC.
Consider \[\vartriangle ABC\], by Pythagoras theorem,
\[\begin{align}
& A{{C}^{2}}=A{{B}^{2}}+C{{D}^{2}} \\
& AC=\sqrt{A{{B}^{2}}+C{{D}^{2}}} \\
\end{align}\]
We know AB = 12cm and CD = 9cm. Substitute and find the value of AC.
\[\therefore AC=\sqrt{{{12}^{2}}+{{9}^{2}}}=\sqrt{144+81}=\sqrt{225}=15cm\]
\[\therefore \] Length of AC = 15cm.
Hence, we got all the required lengths as,
CD = 3cm, AB = 12cm and AC = 15cm.
Note: In our two right angles triangles, the corresponding sides are in the same ratio. Hence the matching angles are also the same. Thus we have considered the ratio of their sides \[\dfrac{AB}{ED}\] and \[\dfrac{BC}{DC}\], which are equal. The two triangles can be also proved by SSS congruence i.e. all the sides of the two triangles are similar.
Last updated date: 26th May 2023
•
Total views: 324.9k
•
Views today: 4.83k
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
