Question

ABC is a triangle in which AD is the altitude from A such that AD= 12m, BD= 9m, and DC=16m.show that triangle BAC is a right -angled triangle, right -angled at A.

Answer 1

in right triangle ADB
AB^2=AD^2+BD^2
=>AB=√12^2+9^2
=√144+81
=√225
=15. (i)
now,in right triangle ADC
AC^2=AD^2+DC^2
=>AC=√AD^2+DC^2
=√12^2+16^2
=√144+256
=√400
=20. (ii)

BC=BD+DC
=9+16
=25
so,we have
AB=15=>AB^2=225
AC=20=>AC^2=400
BC=25=>BC^2=625

here, we found
AB^2+AC^2=BC^2
thus,by the converse of Pythagoras theorem
∆ABC is a right angled triangle.

BC is hypotenuse here then,
angle A is 90°(angle opposite to hypotenuse is right angle)

Answer 2

Step-by-step explanation:

in right triangle ADB

AB^2=AD^2+BD^2

=>AB=√12^2+9^2

=√144+81

=√225

=15.

(i)

now,in right triangle ADC

AC^2=AD^2+DC^2 =>AC=VAD^2+DC^2

=√12^2+16^2

=√144+256

=√400

=20.

(ii)

BC=BD+DC

=9+16

=25

so, we have

AB=15->AB^2=225

AC-20=>AC^2=400

BC=25=>BC^2=625

here, we found AB^2+AC^2=BC^2

thus, by the converse of Pythagoras theorem AABC is a right angled triangle.

BC is hypotenuse here then, angle A is 90°(angle opposite to hypotenuse is right angle)