Question

The areas of two similar triangles are 100 cm² and 49 cm² respectively. If the altitude of the larger triangle is 5 cm, find the corresponding altitude of the other.

Answer 1

here is your answer by Sujeet,


We know that when areas square is equal to the their altitude....

then,

Follow the instructions,

Let to be ∆ABC

Let to be ∆DEF

Let to be AB and DE Altitude of ∆ABC and ∆DEF. respectively
then,
ar(∆ABC)=100cm²
ar(∆DEF)=48cm²

Again,
ar(∆ABC)²=AB²
ar{∆DEF)²=DE²

100cm²=5²
49cm²=DE²

100=[5]
49=[DE)
√100/49=5/DE
10/7=5/DE
10DE=7*5
10DE=35
DE=35/10
DE=3.5 cm

Again,
The corresponding altitude of other ∆ is 3.5 cm





I'm giving you fully on full solution..........


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