# Another SA of Pyramid Question, please can you check where my mistake/s are: (1 Viewer)

##### Member

SA front = 1/2 x 12 x 5 = 30 cm^2

SA bottom = 1/2 x 9 x 12 = 54 cm^2

SA back = 1/2 x 15 x 5 = 37.5 cm^2 used Pythagoras to get 15 cm

SARHS = 1/2 x 9 x root 148.75 = 54.9 cm^2 used Pythagoras h^2 = 13^2 - 4.5^2, h = root 148.75

TOTAL SA = 176.4 cm^2

However the answer says 180 cm^2 (what am I doing wrong?)

#### Lith_30

##### New Member
You made a mistake in calculating the area of the top triangle. If you look at the pyramid from a birds eye view, the triangle at the very top is a right angle triangle as its vertices bend to the same degree as the triangle at the base.

##### Member
You made a mistake in calculating the area of the top triangle. If you look at the pyramid from a birds eye view, the triangle at the very top is a right angle triangle as its vertices bend to the same degree as the triangle at the base.
Thank you, but that will still be an SA of 175.5cm^2 the answer says 180 cm^2 (to one decimal place). So not too sure where the 4.5 cm^2 comes from

#### Lith_30

##### New Member
Thank you, but that will still be an SA of 175.5cm^2 the answer says 180 cm^2 (to one decimal place). So not too sure where the 4.5 cm^2 comes from
I think I might of explained it badly, the height of the triangle at the top would be the hypotenuse of the triangle at the front. Since they are both right angle triangles.

$\bg_white \\h=\sqrt{5^2+12^2}\\\\h=13cm$ find the height of the top triangle

$\bg_white \\A_{top}=\frac{1}{2}\times13\times9\\\\A_{top}=58.5cm^2$ the area of the top triangle is 58.5cm^2

Add all the other areas up and the answer should be 180cm^2

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$\bg_white \\h=\sqrt{5^2+12^2}\\\\h=13cm$ find the height of the top triangle
$\bg_white \\A_{top}=\frac{1}{2}\times13\times9\\\\A_{top}=58.5cm^2$ the area of the top triangle is 58.5cm^2