Question

The hypotenuse of a right angled triangle is 25 cm. If the length of the base is 7cm, what is the length of altitude of the triangle?

Answer 1

Solutioñ

from the Pythagorean theorem...

$= > hypotenuse {}^{2} = base {}^{2} + altitude {}^{2}$

now.....

hypotenuse=25

base=7

therefore...

$= > 25 {}^{2} = 7 {}^{2} + altitude {}^{2} \\ = > altitude = \sqrt{25 {}^{2} - 7 {}^{2} } \\ = > altitue = \sqrt{(25 + 7)(25 - 7)} \\ = > altitude = \sqrt{32 \times 18} \\ = > altitude = \sqrt{16 \times 2 \times 9 \times 2} \\ = > altitude = 4 \times 2 \times 3 = 24 \: cm$

Answer 2

Step-by-step explanation:

altitude²=hypotenuse²- base²

altitude= √hypoteneous²-base²

altitude=√(25²-7²)

=√(625-49)

=√576

=24cm

hope it will help