Question

The areas of two similar triangles ABC and DEF are respectively 100cm square and 49cm square. If the altitude of triangle ABC is 5cm, find the corresponding altitude of triangle DEF.

Answer 1

Hey Friends!!

Here is your answer↓⬇


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$\huge \bf \sf{It \: is \: given \: that:-)}$

$\bf{ar(abc) = 100 {cm}^{2} }$
$\bf{ar(def) = 49 {cm}^{2} }.$



$\bf{triangle \: ABC \sim \: triangle \: DEF}.$

$\bf{and \: AG = 5cm.}$

▶⏩
$\huge \boxed{We \: know \: that:-)}$
↪➡ The ratio of the areas of two similar triangles is equal to the square of their corresponding altitudes.


$\huge \bf = > \frac{ar(abc) }{ar(def)} = \frac{ ({ag})^{2} }{ (dh)^{2} } .$



$\huge \bf = > \frac{100}{49} = {( \frac{5}{dh} )}^{2} .$



$\huge \bf = > \sqrt{ \frac{100}{49} } = \frac{5}{dh} .$



$\huge \bf = > \frac{10}{7} = \frac{5}{dh} .$


$\huge \bf = > 10dh = 35.$


$\huge \bf = > dh = \frac{35}{10}.$



$\huge \boxed{ = > dh = 3.5cm \: .}$



✅✅ Hence, the corresponding side of∆DEF = 3•5cm.✔✔.



$\huge \boxed{THANKS}$



$\huge \bf \underline{Hope \: it \: is \: helpful \: for \: you}$

Answer 2

Answer: The corresponding altitude of Δ DEF is 3.5 cm.

Step-by-step explanation:

Since we have given that

Area of triangle ABC = 100 cm²

Area of triangle DEF = 49 cm^2

Length of altitude of ΔABC = 5 cm

Let the length of altitude of ΔDEF be 'x'.

According to "Area Similarity theorem" which states that the ratio of two similar triangles is equal to square of their corresponding sides of the two triangles.

$\dfrac{Ar(ABC)}{Ar(DEF)}=\dfrac{5^2}{x^2}\\\\\dfrac{100}{49}=(\dfrac{5}{x})^2\\\\\sqrt{\dfrac{100}{49}}=\dfrac{5}{x}\\\\\dfrac{10}{7}=\dfrac{5}{x}\\\\x=\dfrac{7\times 5}{10}\\\\x=3.5\ cm$

Hence, The corresponding altitude of Δ DEF is 3.5 cm.