Question

Abc and def are similar triangles. If area of abc is 100 and def are 49 ..Altitude of abc is 10 then find altitude of def?

Answer 1

Answer:

$\large{\underline{\boxed{\sf Altitude \: of \: \triangle D EF = 7}}}$

Step-by-step explanation:

Given that -

ΔABC and ΔDEF are similar triangles.

⇒ ΔABC ~ΔDEF

We know that -

• The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Also, it is given that -

• Altitude of ΔABC = 10

Let the altitude of ΔDEF be x.

$\sf \frac{ \triangle ABC}{\triangle DE F } = \frac{(Altitude \: of \: ABC) {}^{2} }{(Altitude \: of \: DE F) {}^{2} } \\ \\ \sf \implies \frac{100}{49} = \frac{(10)^2 }{x^2} \\ \\ \bf On \: cross \: multiplying : \\ \\ \implies \sf 100x^2 = 49 * 100 \\ \\ \implies \sf x^2 = \frac{4900}{100} \\ \\ \implies \sf x^2 = 49\\\\ \sf \implies x = \sqrt{49} \\\\ \sf \implies x = 7$

Hence, the altitude of DEF is 7 unit.

Answer 2

Question:

ABC and DEF are similar triangles. If area of ABC is 100 and DEF is 49 and altitude of ABC is 10. Then find altitude of DEF?

Solution:

Assume a ∆ABC and ∆DEF.

And ∆ABC is similar to ∆DEF and it's corresponding sides are AB and DE.

Altitude of AB = 10

_______________ [ GIVEN ]

• We have to find DE.

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Now ..

We know that Ratio of two similar triangle is the ratio of the square of their sides (corresponding sides).

Let DE = a

Now..

$\dfrac{ 100}{ 49} \: = \: \dfrac{ {(AB)}^{2} }{ {(DE)}^{2} }$

→ $\dfrac{ 100}{ 49} \: = \: \dfrac{ {(10)}^{2} }{ {(a)}^{2} }$

→ $\dfrac{ 100}{ 49} \: = \: \dfrac{ 100}{ {(a)}^{2} }$

→ $\dfrac{ 100}{ 49\:\times\:100} \: = \: {a}^{2}$

→ $\dfrac{1}{49}$ = $a^{2}$

→ a² = 49

→ a = 7

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Altitude of DEF = 7

___________ [ ANSWER ]

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✡ Verification :

$\dfrac{ 100}{ 49} \: = \: \dfrac{ {(10)}^{2} }{ {(a)}^{2} }$

From above calculations we have DEF = 7

Put value of DEF above

=> $\dfrac{ 100}{ 49} \: = \: \dfrac{ {(10)}^{2} }{ {(7)}^{2} }$

=> $\dfrac{ 100}{ 49} \: =\:\dfrac{ 100}{ 49}$

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