Question

the area of two similar triangles are 81 cm square and 49 CM square respectively the altitude of the triangle is 4.5 find the corresponding altitude ​

Answer 1

Answer:

3.5 cm

Step-by-step explanation:

In two similar triangles, the ratio of their areas is the square of the ratio of their sides, altitudes and medians are all in the same ratio. Therefore, the area ratio will be the square of any of these ratios too.

therefore we can say

area of 1st ∆ / area of 2nd ∆ = square of altitude of first triangle / square of altitude of 2nd triangle

$\frac{81}{49} = \frac{ {(4.5)}^{2} }{ {x}^{2} } \\ 81 {x}^{2} = 49 \times {(4.5)}^{2} \\ {x}^{2} = \frac{49 \times {(4.5)}^{2} }{81} \\ {x}^{2} = \frac{ {7}^{2} \times {(4.5)}^{2} }{ {9}^{2} } \\ {x}^{2} = {( \frac{7 \times 4.5}{9} )}^{2} \\ x = \sqrt{ {( \frac{7 \times 4.5}{9} )}^{2} } \\ x = \frac{7 \times 4.5}{9} \\ x = \frac{7}{2} \\ x = 3.5 \: cm$

the altitude of 2nd triangle corresponding to area 49 sq cm is 3.5 cm

hope you get your answer