Area of the given triangle is equals to [tex]27.6 in^{2}[/tex](round to nearest tenth).
" Triangle is defined as two dimensional geometrical shape with three vertices, three sides and three angles enclosed in it."
Formula used
Pythagoras theorem
(Hypotenuse)² = (Base)² + (Height)²
Area of a right triangle = [tex]\frac{1}{2} \times b\times h[/tex]
[tex]b[/tex] represents the base of the triangle
[tex]h[/tex] represents the height of the triangle
According to the question,
In a given triangle,
Hypotenuse [tex]= 11in[/tex]
Height [tex]= 6in[/tex]
Substitute the value in Pythagoras theorem to get the measure of base,
[tex](11)^{2} = b^{2} + 6^{2}\\\\\implies b^{2}= 121 -36\\\\\implies b =\sqrt{85}[/tex]
Substitute the value in the formula to find area of a triangle,
Area of a triangle [tex]= \frac{1}{2}\times \sqrt{85} \times 6[/tex]
[tex]= \frac{1}{2} \times 9.22 \times 6\\\\= 9.22 \times 3\\\\= 27.66[/tex]
≈ [tex]27.6in^{2}[/tex] ( round to nearest tenth)
Hence, area of the given triangle is equals to [tex]27.6 in^{2}[/tex](round to nearest tenth).
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