Question

The roof of a house seen from the front is an isosceles triangle \Delta ABC ΔABC with base 14.5 m and sides of length 8.5 m. Find how high above line [BC] the vertex A is​

Answer 1

Answer:

4.4 m

Step-by-step explanation:

Draw the height of the triangle (perpendicular line from point A to line BC).  Since ABC is an isosceles triangle, the height is a perpendicular bisector, so it splits the triangle into two congruent right triangles.

Use Pythagorean theorem to find the height.

c² = a² + b²

(8.5 m)² = (14.5/2 m)² + h²

h ≈ 4.4 m

Answer 2

This  question is based on the Pythagorean theorem. Therefore, the high above line [BC] the vertex A is​ is 4.43 cm.

 

Given:

ΔABC with base 14.5 m and sides of length 8.5 m.  

We need to determined the high above line [BC] the vertex A is​.

According to the question,

Draw the height of the triangle (perpendicular line from point A to line BC).  So, ABC is an isosceles triangle, the height is a perpendicular bisector, so it splits the triangle into two congruent right triangles.

From the given figure, we have to considered the base  = 7.25cm.

By using Pythagorean theorem,

$a^2 +b^2 = c^2$

$a^2 + (7.25)^2 = (8.5)^2\\\\a^2 + 52.56 = 72.25$

Now, solve it further, we get,

$a^2 = 72.25 - 52.56\\\\a^2 = 19.69$

Taking both sides square root both sides.

We get,

$\sqrt{a^2} = \sqrt{19.69}$

a = 4.43 cm

Therefore, the high above line [BC] the vertex A is​ is 4.43 cm.

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https://brainly.com/question/24252852