how to calculate the area of a rhombus is in $cm^2$
How do I calculate the area of a rhombus is in $cm^2$?
Is the formula $\frac12 \times 17 \times 16$? Anyone can help me to solved this? I don't know the rhombus formula.
Based on my exercise book, the answer is $240 cm^2$
$\endgroup$ 13 Answers
$\begingroup$Hint: You can draw the vertical diagonal of the rhombus, which will cut the 16 cm diagonal in half and make the rhombus into four right triangles.
$\endgroup$ $\begingroup$You did some of the work by now.
You just have to be more careful.
If you noticed by cutting the $17$cm by $16$cm by $17$cm isoceles triangle in half you can figure out its height by using the pythagorean theorem. $a^2 + b^2 = c^2$.
This means that the height of the triangle, let's say $b$, is $(8\text{cm})^2 + b^2 = (17\text{cm})^2$
or $b = 15\text{cm}$.
Recall that the area of a triangle is $.5 \times b \times h$. So this means that the area of the large $17$ by $16$ by $17$ triangle is $.5 \times 16\text{cm} \times 15\text{cm}$ since the height of the triangle is $15$. Since two of these triangles make up the area of the rhombus however, we have $2 \times .5 \times 16\text{cm} \times 15\text{cm or }16\text{cm} \times 15\text{cm} = 240\text{cm}^2$.
One more thing I would like to add: because a rhombus is a particular type of parallelogram, the area of formula for a rhombus is the same as the area formula of a parallelogram: $ A = base \times height = b \times h$. In this case however, finding the height of the rhombus required more trigonometry than was needed to solve the problem. So the best method was decomposing the rhombus into two triangles.
$\endgroup$ $\begingroup$Your edited attempt is close, but not quite. After drawing the vertical diagonal, you have a right triangle with hypotenuse $17\,\text{cm}$ and base $8\,\text{cm}$.
To find the height, we must use the Pythagorean Theorem: $$b^2 + h^2 = c^2$$ Where $b$ is the base of the triangle, $h$ is the height, and $c$ is the hypotenuse.
See if you can find the height of the triangle... Let me know in a comment if you need more help.
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