The height of equilateral triangle is 6cm.Find area
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the height of triangle is 6cm
then base length is found by pythagoras theorem
![a^{2}= ( \frac{a}{2} )^{2}+ h^{2} a^{2}= ( \frac{a}{2} )^{2}+ h^{2}](https://tex.z-dn.net/?f=+a%5E%7B2%7D%3D+%28+%5Cfrac%7Ba%7D%7B2%7D+%29%5E%7B2%7D%2B+h%5E%7B2%7D+++)
![\frac{3a^2}{4}=36 \frac{3a^2}{4}=36](https://tex.z-dn.net/?f=+%5Cfrac%7B3a%5E2%7D%7B4%7D%3D36+)
![a^{2}=48 a^{2}=48](https://tex.z-dn.net/?f=+a%5E%7B2%7D%3D48+)
![a= \sqrt{48} a= \sqrt{48}](https://tex.z-dn.net/?f=a%3D+%5Csqrt%7B48%7D+)
The area of the triangle
![= \frac{1}{2}ah =3 \sqrt{48}=20.7846 cm^2 = \frac{1}{2}ah =3 \sqrt{48}=20.7846 cm^2](https://tex.z-dn.net/?f=%3D+%5Cfrac%7B1%7D%7B2%7Dah+%3D3+%5Csqrt%7B48%7D%3D20.7846+cm%5E2+)
then base length is found by pythagoras theorem
The area of the triangle
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Height of equilateral triangle is √3/2 X a, where 'a' is the side of the equilateral triangle. So √3/2 x a = 6.
Hence a = 12/√3.
Area of an equilateral triangle is 1/2 base x height
ie 1/2 x 12 ÷ √3 x 6 = 36 ÷√3
= 12√3.
Hence a = 12/√3.
Area of an equilateral triangle is 1/2 base x height
ie 1/2 x 12 ÷ √3 x 6 = 36 ÷√3
= 12√3.
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