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Question (1 Viewer)
- Thread starter highshill
- Start date
Let the length of diagonal AC be x cm.
AD = CD = 9cm (equal sides of rhombus ABCD)
∠DCB = 2*16° = 32° (diagonal AC bisects vertex ∠DCB through which it passes)
AD||BC (opposite sides of rhombus ABCD parallel)
Therefore, ∠D + ∠DCB = 180° (co-interior angles supplementary on AD||BC)
Therefore, ∠D = 180° - 32° = 148°
Now, x2 = 92 + 92 - 2*9*9cos148° (cosine rule)
x2 = 81*2 - 162cos148°
x2 = 299.3837916
x = 17.30271053cm
Therefore, x = 17.30cm (2 dec. pl.)
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Last edited:
I did it the long way, lol.}{9} \\ l &= 2 \times 9\cos{16\degree} \\ &\approx 17.3 \text{ cm} \end{align*} \\ The length of the other diagonal can be found using Pythagoras' Theorem.)
pikachu975
Premium Member
The diagonal is 9 cm not the side I think
I'm pretty sure it's talking about the side being 9cm, otherwise it wouldn't make much sense to find the "diagonals" when one is already given. If the diagonal length is 9cm, then it should ask you to find the "other diagonal", not the "diagonals". Check the answer to see if you can tell which one is 9cm.
pikachu975
Premium Member
A diagonal of a rhombus with the size 9 cm makes a angle of 16 degrees with the side Find the length of the diagonals
I think it's talking about the diagonal because it's saying the diagonal makes an angle of 16 degrees with the side so the 9 cm should be talking about the diagonal still. But yeah check the answers just in case.