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# The area of two similar triangles is $81\;c{m^2}\,and\,49\;c{m^2}$ respectively. If the altitude of the bigger training is 4.5 cm to find the corresponding altitude.

Last updated date: 15th Sep 2024
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Hint: 1. First of all we need to know about altitude, which is a line drawn from a vertex of a triangle and it is perpendicular to the base of that triangle.
2. In these types of questions, students should make a rough diagram of the data given to understand the demands of the question.
3. Students should consider the units of the given data.

Formula used:
The area of triangle is given as-
Area$= \dfrac{1}{2} \times base \times altitude$

Complete step by step solution:
Let us note down the given information.
We have been given The area of two similar triangles is $81c{m^2}\,and\,49c{m^2}$.
The question demands that If the altitude of the bigger training is 4.5cm to find the corresponding altitude.
In this question, we have to find out the ratio of altitude of similar triangles.
We know that in case of similar triangle, the ratio of areas is equal to squares of their altitudes
It is given that ${A_1} = 81c{m^2}\;,\;{A_2} = 49\;c{m^2}\;\& \;Altitude = 4.5\;cm$.
Let areas of two similar triangles are
${A_1} = 49c{m^2}$
${A_2} = 81c{m^2}$
And these corresponding altitudes
${h_1} = ?$
${h_2} = 4.5cm$
By a known theorem,
And properties of triangle we can say that,
$\dfrac{{h_1^2}}{{h_2^2}} = \dfrac{{{A_1}}}{{{A_2}}}$
$h_1^2 = \dfrac{{{A_1}}}{{{A_2}}} \times h_2^2$
$h_1^2 = \dfrac{{49}}{{81}} \times {4.5^2}$
$\Rightarrow{\dfrac{49}{81} \times \dfrac{45}{10} \times \dfrac{45}{10}}$
Cancel the terms by 9. So the equation becomes,
${h_1} = \sqrt {\dfrac{{49 \times 5 \times 5}}{{10 \times 10}}} = \dfrac{{7 \times 5}}{{10}}$
$\Rightarrow 3.5cm$
In addition to the above theory,
We know that if two triangles are similar then
$\dfrac{{{A_1}}}{{{A_2}}} = {\left( {\dfrac{{altitude\;1}}{{altitude\;2}}} \right)^2} \\ \left( {\dfrac{{altitude\;1}}{{altitude\;2}}} \right) = \sqrt {\dfrac{{81}}{{49}}} = \dfrac{9}{7} \\$
Further simplified as
$altitude\;2 = \dfrac{{4.5}}{{\left( {\dfrac{9}{7}} \right)}} = 3.5\;cm$
Hence the length of the smaller altitude is 3.5cm.

The altitude is 3.5cm.

Note: 1.When two triangle are similar then Ratio of square of their side = Ratio of square of their altitudes
2.Students are advised to go through the properties and theorems associated with the triangle for a better understanding of the question.
3. Students are advised to solve the question step by step to avoid any calculation mistake.