Question

Please help me solve 5 and 6 for my homework

Answer 1

Answer:

(5) The perpendicular height of the triangle is 8.48 mm.

(6) The 1184.86 cm far up the wall the ladder reach.

Step-by-step explanation:

Part (5):

When a perpendicular is drawn in an isosceles triangle then the perpendicular divided the base into 2 equal parts.

The figures is shown below.

In ΔABC,

CB = 6 mm

So,

CD = BD = 3 mm

Now calculating the perpendicular height of the triangle.

Using Pythagoras theorem in ΔADC:

$(AC)^2=(CD)^2+(AD)^2\\\\(9)^2=(3)^2+(AD)^2\\\\81=9+(AD)^2\\\\(AD)^2=81-9\\\\(AD)^2=72\\\\AD=8.48mm$

Thus, the perpendicular height of the triangle is 8.48 mm.

Part (6):

The figure is shown below.

Converting meter to centimeter:

1 m = 100 cm

So,

12 m = 1200 cm

Using Pythagoras theorem in ΔXYZ:

$(XZ)^2=(YZ)^2+(XY)^2\\\\(1200)^2=(190)^2+(XY)^2\\\\1440000=36100+(XY)^2\\\\(XY)^2=1440000-36100\\\\(XY)^2=1403900\\\\XY=1184.86cm$

Thus, 1184.86 cm far up the wall the ladder reach.