Answer
Verified
422.1k+ views
Hint: Use the concept of similar triangles i.e. “If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.” And write the relation between area and corresponding sides of the similar triangles and put the given values in it you will get the answer.
Complete step-by-step answer:
To solve the above problem we should draw the diagram with given notations as follows,
Now we will write the given data,
$\Delta ABC\sim \Delta DEF$
Area of $\Delta ABC$ = $A\left( \Delta ABC \right)$ = $9c{{m}^{2}}$
Area of $\Delta DEF$ = $A\left( \Delta DEF \right)$ = $16c{{m}^{2}}$
BC = 2.1 cm.
As the two triangles are similar therefore the relation between their corresponding sides is given by the concept given below,
Concept:
If two triangles are similar then their corresponding sides are in proportion.
By using the above concept we can write,
$\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DE}$ …………………………………………………. (1)
As we have given the value of one of the side BC and there’s no other side given therefore we have to use the area given to find the value of EF and for that we should know the concept given below,
Concept:
If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.
By using the above concept and squaring the equation (1) we will get,
$\dfrac{A\left( \Delta ABC \right)}{A\left( \Delta DEF \right)}=\dfrac{A{{B}^{2}}}{D{{E}^{2}}}=\dfrac{B{{C}^{2}}}{E{{F}^{2}}}=\dfrac{A{{C}^{2}}}{D{{E}^{2}}}$
Above equation can also be written as,
$\therefore \dfrac{A\left( \Delta ABC \right)}{A\left( \Delta DEF \right)}=\dfrac{B{{C}^{2}}}{E{{F}^{2}}}$
If we put the given values in above equation we will get,
$\therefore \dfrac{9}{16}=\dfrac{{{\left( 2.1 \right)}^{2}}}{E{{F}^{2}}}$
As we know that the square of 2.1 is 4.41 therefore we will get,
$\therefore \dfrac{9}{16}=\dfrac{4.41}{E{{F}^{2}}}$
If we shift the $E{{F}^{2}}$ on the left hand side of the equation we will get,
$\therefore E{{F}^{2}}\times \dfrac{9}{16}=4.41$
If we shift $\dfrac{9}{16}$ on the right hand side of the equation we will get,
$\therefore E{{F}^{2}}=4.41\times \dfrac{16}{9}$
If we do the further simplifications in the above equation we will get,
$\therefore E{{F}^{2}}=7.84$
Now if we take the square roots on both sides of the equations we will get,
$\therefore \sqrt{E{{F}^{2}}}=\sqrt{7.84}$
Therefore, EF = 2.8 cm
Therefore the value of EF is 2.8 cm.
Note: There are chances that you write the relation between area and sides of the similar triangles as, \[\dfrac{A\left( \Delta ABC \right)}{A\left( \Delta DEF \right)}=\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DE}\] but you should remember that it is $\dfrac{A\left( \Delta ABC \right)}{A\left( \Delta DEF \right)}=\dfrac{A{{B}^{2}}}{D{{E}^{2}}}=\dfrac{B{{C}^{2}}}{E{{F}^{2}}}=\dfrac{A{{C}^{2}}}{D{{E}^{2}}}$.
Complete step-by-step answer:
To solve the above problem we should draw the diagram with given notations as follows,
Now we will write the given data,
$\Delta ABC\sim \Delta DEF$
Area of $\Delta ABC$ = $A\left( \Delta ABC \right)$ = $9c{{m}^{2}}$
Area of $\Delta DEF$ = $A\left( \Delta DEF \right)$ = $16c{{m}^{2}}$
BC = 2.1 cm.
As the two triangles are similar therefore the relation between their corresponding sides is given by the concept given below,
Concept:
If two triangles are similar then their corresponding sides are in proportion.
By using the above concept we can write,
$\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DE}$ …………………………………………………. (1)
As we have given the value of one of the side BC and there’s no other side given therefore we have to use the area given to find the value of EF and for that we should know the concept given below,
Concept:
If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.
By using the above concept and squaring the equation (1) we will get,
$\dfrac{A\left( \Delta ABC \right)}{A\left( \Delta DEF \right)}=\dfrac{A{{B}^{2}}}{D{{E}^{2}}}=\dfrac{B{{C}^{2}}}{E{{F}^{2}}}=\dfrac{A{{C}^{2}}}{D{{E}^{2}}}$
Above equation can also be written as,
$\therefore \dfrac{A\left( \Delta ABC \right)}{A\left( \Delta DEF \right)}=\dfrac{B{{C}^{2}}}{E{{F}^{2}}}$
If we put the given values in above equation we will get,
$\therefore \dfrac{9}{16}=\dfrac{{{\left( 2.1 \right)}^{2}}}{E{{F}^{2}}}$
As we know that the square of 2.1 is 4.41 therefore we will get,
$\therefore \dfrac{9}{16}=\dfrac{4.41}{E{{F}^{2}}}$
If we shift the $E{{F}^{2}}$ on the left hand side of the equation we will get,
$\therefore E{{F}^{2}}\times \dfrac{9}{16}=4.41$
If we shift $\dfrac{9}{16}$ on the right hand side of the equation we will get,
$\therefore E{{F}^{2}}=4.41\times \dfrac{16}{9}$
If we do the further simplifications in the above equation we will get,
$\therefore E{{F}^{2}}=7.84$
Now if we take the square roots on both sides of the equations we will get,
$\therefore \sqrt{E{{F}^{2}}}=\sqrt{7.84}$
Therefore, EF = 2.8 cm
Therefore the value of EF is 2.8 cm.
Note: There are chances that you write the relation between area and sides of the similar triangles as, \[\dfrac{A\left( \Delta ABC \right)}{A\left( \Delta DEF \right)}=\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DE}\] but you should remember that it is $\dfrac{A\left( \Delta ABC \right)}{A\left( \Delta DEF \right)}=\dfrac{A{{B}^{2}}}{D{{E}^{2}}}=\dfrac{B{{C}^{2}}}{E{{F}^{2}}}=\dfrac{A{{C}^{2}}}{D{{E}^{2}}}$.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Select the word that is correctly spelled a Twelveth class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE