Question

Plzz solve subpart 3

Answer 1

Step-by-step explanation:

We know Area of triangle = 1/2×Base × Height

i ) In ∆ ABC , Base = AB = 3 cm and Height = AC = 4 cm

So,

Area of ∆ ABC = (1/2)×3 × 4 = 3 × 2 = 6 cm2 ( Ans )

ii ) We apply Pythagoras theorem in ∆ ABC and get

BC^2 = AB^2 + AC°2 , Substitute all values we get

BC^2 = 3^2 + 4^2

BC^2 = 9 + 16

BC^2 = 25

BC = 5 cm ( Ans )

iii ) We apply Pythagoras theorem in ∆ ABD and get

AB^2 = AD^2 + BD^2 ,

AD^2 = AB^2 - BD^2 , Substitute values we get

AD^2 = 3^2 - BD^2

AD^2 = 9 - BD^2 --- ( 1 )

And apply Pythagoras theorem in ∆ ACD and get

AC^2 = AD^2 + CD^2 ,

AD^2 = AC^2 - CD^2

AD^2 = AC^2 - ( BC - BD )^2

AD^2 = AC^2 - ( BC^2 + BD^2 - 2 BC . BD ) , Substitute values we get

AD^2 = 3^2- ( 5^ 2 + BD^2 - 2 ( 5 ). BD )

AD^2 = 9 - ( 25 + BD^2 - 10 BD )

AD^2 = 9 - 25 - BD^2 + 10 BD

AD^2 = - 16 - BD^2 + 10 BD , Substitute value from equation 1 we get

9 - BD^2 = - 16 - BD^2 + 10 BD

10 BD = 25

BD = 2.5 , Substitute that value in equation 1 we get

AD2 = 9 - ( 2.5 )^2 = 9 - 6.25 = 2.75

AD = 1.65 cm

Therefore,

The length of altitude from A to BC = Ad = 1.65 cm