Question

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

Answer 1

Hey there !!

→ Let the given triangles be ∆ABC and ∆DEF such that ar(∆ABC) = 100cm² and ar(∆DEF) = 49cm².

→ Let AL and DL be the corresponding altitudes of ∆ABC and ∆DEF respectively.

Then, AL = 5cm.

▶ We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding altitudes.

=> $\therefore \frac{ ar ( \triangle ABC ) }{ ar ( \triangle DEF ) } = \frac{ {AL}^{2} }{ {DM}^{2} } .$

=> $\frac{100}{49} = { ( \frac{AL}{DL} ) }^{2} .$

=> $\frac{100}{49} = { ( \frac{5}{DL} ) }^{2} .$

$= > \frac{5}{DL} = \sqrt{ \frac{100}{49} } .$

$= > \frac{5}{DL} = \frac{10}{7} .$

=> DL × 10 = 7 × 5.

=> 10DL = 35.

=> DL = $\frac{35}{10} .$


$\huge \boxed{ \boxed{ \bf => DL = 3.5 cm. }}$


✔✔ Hence, it is solved ✅✅.

____________________________________


$\huge \boxed{ \boxed{ \boxed { \mathbb{THANKS}}}}$


$\huge \bf{ \# \mathbb{B}e \mathbb{B}rainly.}$

Answer 2

Question;-

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

Method of Solution:-

Firstly, Let to ∆ABC and ∆DEF and also their areas 100cm² and 49cm² respectively.

Also Takes their Corresponding Side's , altitude of Bigger Triangle (∆ABC)=5 cm

Let to be Smaller Corresponding side of Triangle =x

Here, "We know that The ratio of areas of two Triangle is equal to their Corresponding Side's altitude of Given Triangle."

According to the above statement we have to written;-

$\implies \: \frac{ar( \triangle{ABC)} }{ar( \triangle{def})} = \frac{ {MN}^{2} }{ {OP}^{2} }$

Substitute the value of Required terms in Above Formula;-

$\frac{ar( \triangle{ABC)} }{ar( \triangle{def})} = \frac{ {MN}^{2} }{ {OP}^{2} } \\ \\ \\ \\ \: \implies \: \frac{100}{49} = \frac{ {5 \: cm}^{2} }{ {x}^{2} } \\ \\ \\ \\ \implies \sqrt{ \frac{100}{49} } = \frac{5 \: cm}{ \: x \: cm} \\ \\ \\ \implies \: \frac{10}{7} = \frac{5cm}{x \: cm} \\ \\ \\ \implies \: 10x = 35 \\ \\ \implies \: x = \frac{35}{10} = 3.5cm \\ \\ \implies \: x = \: 3.5cm$

Hence,The areas of two similar triangles are 100 cm square and 49 cm square respectively. the altitude of the bigger triangle is 5 cm and altitude of the smaller Triangle is 3.5 cm.