Answer

Verified

448.5k+ views

Hint: Use the property of similar triangles which is related to the area of triangles and sides of the similar triangles which is given as “the ratio of area of similar triangles is equal to the ratio of the square of sides.”

Complete step-by-step answer:

Here, we have two triangles $\Delta ABC$ and $\Delta DEF$ which are similar to each other. And side BC of $\Delta ABC$ is given as 3cm and side EF of $\Delta DEF$ is given as 4cm. And we need to determine the area of $\Delta DEF$ if the area of $\Delta ABC$ be $54c{{m}^{2}}$.

Now, we need to use the property of similar triangles with respect to the area of them. Relation between sides and area of similar triangles can be given as

Area of triangle 1/Area of triangle 2=\[{{\left( \dfrac{\text{Side of triangle 1}}{\text{side of triangle 2}} \right)}^{2}}\ldots \ldots (1)\]

So, by applying the above property with the similar triangles ABC and DEF, we get

\[\dfrac{area(\Delta ABC)}{are(\Delta DEF)}={{\left( \dfrac{AB}{DE} \right)}^{2}}={{\left( \dfrac{BC}{EF} \right)}^{2}}={{\left( \dfrac{AC}{DF} \right)}^{2}}\ldots \ldots (2)\]

Now we have values of sides BC and EF and area of $\Delta ABC$, so, by substituting values of BC, EF and area($\Delta ABC$),

We have,

$area(\Delta ABC)=54c{{m}^{2}}$

BC=3cm

EF=4cm

Hence,

\[\begin{align}

& \dfrac{54}{area\left( \Delta DEF \right)}={{\left( \dfrac{3}{4} \right)}^{2}} \\

& \dfrac{54}{area\left( \Delta DEF \right)}=\dfrac{9}{16} \\

\end{align}\]

On cross-multiplying, we get

\[\begin{align}

& area\left( \Delta DEF \right)=\dfrac{16\times 5}{9} \\

& area\left( \Delta DEF \right)=16\times 6=96c{{m}^{2}}. \\

\end{align}\]

Hence, the area of $\Delta DEF$can be given as $96c{{m}^{2}}$.

Therefore, option D is the correct answer.

Note: One can apply property of similar triangles as, \[\dfrac{area(\Delta ABC)}{are(\Delta DEF)}=\dfrac{BC}{EF}\] which is wrong.

So, we need to take care with the relations of areas of similar triangles and their sides.

One can prove the given property by writing area of two similar triangles as

Area =$\dfrac{1}{2}\times base\times height$.

And use the property of similar triangles that the ratio of sides of similar triangles is equal.

Complete step-by-step answer:

Here, we have two triangles $\Delta ABC$ and $\Delta DEF$ which are similar to each other. And side BC of $\Delta ABC$ is given as 3cm and side EF of $\Delta DEF$ is given as 4cm. And we need to determine the area of $\Delta DEF$ if the area of $\Delta ABC$ be $54c{{m}^{2}}$.

Now, we need to use the property of similar triangles with respect to the area of them. Relation between sides and area of similar triangles can be given as

Area of triangle 1/Area of triangle 2=\[{{\left( \dfrac{\text{Side of triangle 1}}{\text{side of triangle 2}} \right)}^{2}}\ldots \ldots (1)\]

So, by applying the above property with the similar triangles ABC and DEF, we get

\[\dfrac{area(\Delta ABC)}{are(\Delta DEF)}={{\left( \dfrac{AB}{DE} \right)}^{2}}={{\left( \dfrac{BC}{EF} \right)}^{2}}={{\left( \dfrac{AC}{DF} \right)}^{2}}\ldots \ldots (2)\]

Now we have values of sides BC and EF and area of $\Delta ABC$, so, by substituting values of BC, EF and area($\Delta ABC$),

We have,

$area(\Delta ABC)=54c{{m}^{2}}$

BC=3cm

EF=4cm

Hence,

\[\begin{align}

& \dfrac{54}{area\left( \Delta DEF \right)}={{\left( \dfrac{3}{4} \right)}^{2}} \\

& \dfrac{54}{area\left( \Delta DEF \right)}=\dfrac{9}{16} \\

\end{align}\]

On cross-multiplying, we get

\[\begin{align}

& area\left( \Delta DEF \right)=\dfrac{16\times 5}{9} \\

& area\left( \Delta DEF \right)=16\times 6=96c{{m}^{2}}. \\

\end{align}\]

Hence, the area of $\Delta DEF$can be given as $96c{{m}^{2}}$.

Therefore, option D is the correct answer.

Note: One can apply property of similar triangles as, \[\dfrac{area(\Delta ABC)}{are(\Delta DEF)}=\dfrac{BC}{EF}\] which is wrong.

So, we need to take care with the relations of areas of similar triangles and their sides.

One can prove the given property by writing area of two similar triangles as

Area =$\dfrac{1}{2}\times base\times height$.

And use the property of similar triangles that the ratio of sides of similar triangles is equal.

Recently Updated Pages

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Let x1x2xn be in an AP of x1 + x4 + x9 + x11 + x20-class-11-maths-CBSE

Let x1x2x3 and x4 be four nonzero real numbers satisfying class 11 maths CBSE

Trending doubts

How many crores make 10 million class 7 maths CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

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

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

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

Give 10 examples of Material nouns Abstract nouns Common class 10 english CBSE

What are the public facilities provided by the government? Also explain each facility

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