If the sides of rectangle are 3.2m and 2.4 m and the side of triangle be 2 m then the height of the figure is 3.6m.
Given the sides of rectangle are 3.2m and 2.4 m and the side of triangle be 2 m.
We are required to find the height of the figure.
Draw a perpendicular on the corner of triangle on the side of rectangle.
Now We have to give the names of the vertiex of the figure as A,B,C,D,E,F.
let AB=x,in this way DB=3.2-x.
In triangle ABC,
AB=[tex]\sqrt{4-x^{2} }[/tex]---------1 [From pythagoras theorem]
In triangle ADB,
AB=[tex]\sqrt{4-(3.2-x)^{2} }[/tex]-----------2 [From pythagoras theorem]
From 1&2
[tex]\sqrt{4-x^{2} }[/tex]=[tex]\sqrt{4-(3.2-x)^{2} }[/tex]
Taking square root both sides.
4-[tex]x^{2}[/tex]=4-[tex](3.2-x)^{2}[/tex]
[tex]x^{2}[/tex]=[tex](3.2-x)^{2}[/tex]
x=3.2-x
x+x=3.2
2x=3.2
x=1.6
Put the value of x in 1
AB=[tex]\sqrt{4-1.6^{2} }[/tex]
=[tex]\sqrt{4-2.56}[/tex]
=[tex]\sqrt{1.44}[/tex]
=1.2
Total height=AB+CF
=1.2+2.4
=3.6
Hence if the sides of rectangle are 3.2m and 2.4 m and the side of triangle be 2 m then the height of the figure is 3.6m.
Learn more about pythagoras theorem at https://brainly.com/question/343682
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