Question

FOR 50 POINTS AND BRAINLIEST!

The sides of a parallelogram are 6 and 10 cm, and the longer diagonal is 14 cm. Find the area of the parallelogram.

Topic:
- Area of a triangle
- Law of sines/cosines​

Answer 1

Answer:

$30\sqrt{3} \, \textsf{or}\, 52 \, \textsf{cm}^2$.

Step-by-step explanation:

To find the area of the parallelogram, we need to first find the height of the parallelogram. Then we can use the formula for the area of a parallelogram, which:

$\Large\boxed{\boxed{\sf Area_{parallelogram}=base \times height }}$

Given that the sides of the parallelogram are 6 cm and 10 cm, and the longer diagonal is 14 cm, we can apply the law of cosines to find the angle between the two given sides. Let's denote this angle as $\theta$.

The law of cosines states:

$\Large\boxed{\boxed{\sf c^2 = a^2 + b^2 - 2ab \cdot \cos(\theta)}}$

Where:

• $c$ is the length of the side opposite the angle $\theta$ (the diagonal in this case)
• $a$ and $b$ are the lengths of the other two sides

Plugging in the values, we get:

$14^2 = 6^2 + 10^2 - 2 \cdot 6 \cdot 10 \cdot \cos(\theta)$

$196 = 36 + 100 - 120 \cdot \cos(\theta)$

$196 = 136 - 120 \cdot \cos(\theta)$

$120 \cdot \cos(\theta) = 136 - 196$

$120 \cdot \cos(\theta) = -60$

$\cos(\theta) = \dfrac{-60}{120}$

$\cos(\theta) = -\dfrac{1}{2}$

Now, we know that $\cos(\theta) = -\dfrac{1}{2}$ corresponds to an angle of $\theta = 120^\circ$ or $\theta = 240^\circ$.

Since the angle cannot be greater than $180^\circ$ in this context, we take $\theta = 120^\circ$.

Now, to find the height of the parallelogram, we use the sine of $\theta$:

$\sin(\theta) = \dfrac{\textsf{opposite side}}{\textsf{hypotenuse}}$

$\sin(120^\circ) = \dfrac{\textsf{height}}{10}$

$\textsf{height} = 10 \cdot \sin(120^\circ)$

$\textsf{height} = 10 \cdot \dfrac{\sqrt{3}}{2}$

$\textsf{height} = 5\sqrt{3} \, \textsf{cm}$

Now, the area of the parallelogram is given by:

$\textsf{Area} = \textsf{base} \times \textsf{height}$

$\textsf{Area} = 6 \times 5\sqrt{3}$

$\textsf{Area} = 30\sqrt{3} \, \textsf{cm}^2$

$\textsf{Area} = 51.961524227066\, \textsf{cm}^2$

$\textsf{Area} = 52 \, \textsf{cm}^2 \textsf{(in nearest whole number)}$

So, the area of the parallelogram is $30\sqrt{3} \, \textsf{or}\, 52 \, \textsf{cm}^2$.

Answer 2

Answer:

30√3 cm² ≈ 51.96 cm² (nearest tenth)

Step-by-step explanation:

The longer diagonal of a parallelogram divides it into two congruent triangles. Therefore, to find the area of the parallelogram, we need to determine the area of one of these triangles and then double it.

To find the area of one of the triangles, we can use the Sine Rule. This requires the measure of the included angle between the two sides of the parallelogram, which can be found using the Law of Cosines.

$\boxed{\begin{array}{l}\underline{\textsf{Cosine Rule (for finding angles)}} \\\\\cos(C)=\dfrac{a^2+b^2-c^2}{2ab}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{ C is the angle.}\\ \phantom{ww}\bullet\;\textsf{ a and b are the sides adjacent the angle.}\\ \phantom{ww}\bullet\;\textsf{ c is the side opposite the angle.}\end{array}}$

In this case:

• a = 6 cm
• b = 10 cm
• c = 14 cm

Substitute the values into the formula and solve for C:

$\cos(C)=\dfrac{6^2+10^2-14^2}{2(6)(10)}\\\\\\\cos(C)=\dfrac{36+100-196}{120}\\\\\\\cos(C)=\dfrac{-60}{120}\\\\\\\cos(C)=-\dfrac{1}{2}\\\\\\C= \cos^{-1}\left(-\dfrac{1}{2}\right)\\\\\\C=120^{\circ}$

Therefore, the larger angle between the sides of the parallelogram is 120°.

Now, we can use the Sine Rule to find the area of one triangle.

$\boxed{\begin{array}{l}\underline{\text{Area of a Triangle - Sine Rule}}\\\\A=\dfrac{1}{2}ab \sin C\\\\\text{where:}\\ \phantom{ww}\bullet \;\text{ C is the angle.} \\ \phantom{ww}\bullet \;\text{ a and b are the sides enclosing the angle.}\end{array}}$

In this case:

• a = 6 cm
• b = 10 cm
• C = 120°

Substitute the values into the formula and solve for the exact area (A):

$A=\dfrac{1}{2} \cdot 6 \cdot 10 \sin 120^{\circ}\\\\\\A=30\sin 120^{\circ}\\\\\\A=30 \cdot \dfrac{\sqrt{3}}{2}\\\\\\A=15\sqrt{3}$

Finally, to determine the area of the parallelogram, simply double the area of one triangle (A):

$\textsf{Area of parallelogram}=2A\\\\\\\textsf{Area of parallelogram}=2\cdot 15\sqrt{3}\\\\\\\textsf{Area of parallelogram}=30\sqrt{3}\\\\\\\textsf{Area of parallelogram}\approx 51.96\; \sf cm^2\; (nearest\;thousandth)$

Therefore, the area of a parallelogram with sides of 6 cm and 10 cm, and a longer diagonal measuring 14 cm, is exactly 30√3 cm², which is approximately 51.96 cm² rounded to the nearest thousandth.