Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. if one of its diagonal is 8 cm long, find the length of another diagonal.

Question

If I solve for diagonal by using area = 1/2 (D1 x D2) and area = base x height and if I solve for diagonal by using Pythagoras theorem, why the answers are different?

Answer 1

The thing is, if a side of a rhombus is $6$ and one of the diagonals is $8$, the area cannot be $24$. The area will consist of four right triangles with hypothenuse $6$ and one of the legs $4$. $A=4\cdot(2\sqrt{36-16})=16\sqrt{5}\approx 35.8$

Answer 2

I agree with you

Using area = 1/2 (D1 x D2) we get $$area = 4 * D_2$$

and with area = base x height

$$area = 6 * 4 = 24$$

so $$D_2 = 24/4 = 6$$

Now Pythagorus' Theorem: taking $$ a = 4 b = D_2/2 c = 6$$

we get $$D_2 = 4*\sqrt(6^2-4^2)$$ so $$D_2 = 4*\sqrt(5)$$

I believe There may be a typo in the original question. If you take side length to be 5 then the answers agree. A rhombus with the dimensions you give doesn't exist