Work out the height of the following triangles

The height of first (a) triangle is 11.31 cm and the height of the second (b) triangle is 15.49 cm.
Step-by-step explanation:
The given is,
Triangle (a) and (b)
Step:1
For first (a) triangle
(Ref the attachment (a))
b = 4 cm
hyp = 12 cm
From Pythagoras theorem,
[tex]Hyp^{2} = b^{2} +h^{2}[/tex]...........................(1)
Equation becomes,
[tex]12^{2}[/tex] = [tex]4^{2}[/tex] + [tex]h^{2}[/tex]
144 = 16 + [tex]h^{2}[/tex]
h = [tex]\sqrt{128}[/tex]
h = 11.314 cm
Step:2
For first (b) triangle
(Ref the attachment (b))
b = 7 cm
hyp = 17 cm
From Pythagoras theorem,
[tex]Hyp^{2} = b^{2} +h^{2}[/tex]...........................(1)
Equation becomes,
[tex]17^{2}[/tex] = [tex]7^{2}[/tex] + [tex]h^{2}[/tex]
289 = 49 + [tex]h^{2}[/tex]
h = [tex]\sqrt{240}[/tex]
h = 15.492 cm
Result:
The height of first (a) triangle is 11.314 cm and the height of the second (b) triangle is 15.492 cm.
A) height = [tex]11.3cm[/tex] .
B) height = [tex]15.48cm[/tex] .
Step-by-step explanation:
Here , We need to calculate height of triangles let's do this :
A)
We draw a perpendicular from vertex to base line (8 cm ) , so length becomes 4 cm of base of triangle .
By Pythagoras Theorem we have ,
[tex]Hypotenuse^2 = Perpendicular^2+Base^2[/tex]
⇒ [tex]Hypotenuse^2 = Perpendicular^2+Base^2[/tex]
⇒ [tex]12^2 = Perpendicular^2+4^2[/tex]
⇒ [tex]144 = Perpendicular^2+16[/tex]
⇒ [tex]Perpendicular^2= 144-16[/tex]
⇒ [tex]Perpendicular= \sqrt{128}[/tex]
⇒ [tex]Perpendicular= \sqrt{2^6(2)}[/tex]
⇒ [tex]Perpendicular= 2^3\sqrt{2}[/tex]
⇒ [tex]Perpendicular= 8\sqrt{2}cm[/tex]
Therefore , height = [tex]11.3cm[/tex] .
B)
We draw a perpendicular from vertex to base line (14 cm ) , so length becomes 7 cm of base of triangle .
By Pythagoras Theorem we have ,
[tex]Hypotenuse^2 = Perpendicular^2+Base^2[/tex]
⇒ [tex]Hypotenuse^2 = Perpendicular^2+Base^2[/tex]
⇒ [tex]17^2 = Perpendicular^2+7^2[/tex]
⇒ [tex]289= Perpendicular^2+49[/tex]
⇒ [tex]Perpendicular^2=289-49[/tex]
⇒ [tex]Perpendicular= \sqrt{240}[/tex]
⇒ [tex]Perpendicular= 4\sqrt{15}cm[/tex]
Therefore , height = [tex]15.48cm[/tex] .