ABCD is a parallelogram, AC=9cm, DC=11cm, angle DAC=100'.
Calculate the are of the parallelogram
i think you use the cosine rule but im not sure, please help!
+23254 There is a formula for the area of a parallelogram: Area = a·b·sin(∠C)
where a and b are adjacent sides and ∠C is the angle included between those two sides:
Unfortunately, we have ∠DAC, not ∠DCA. So, let's find ∠DCA.
Look at triangle ACD: We can find ∠DCA if we know both ∠DAC and ∠ADC.
To find the size of ∠ADC, let's use the Law of Sines:
sin(∠CDA) / AC = sin(∠CAD) / CD
---> sin(∠CDA) / 9 = sin(100°) / 11 ---> sin(∠CDA) = 0.80575 ---> ∠CDA = 53.683°
Since ∠CAD = 100° and ∠CDA = 53.683° , then ∠ACD = 26.3°
---> Area = 9·11·sin(26.3°) = 43.86
+23254 There is a formula for the area of a parallelogram: Area = a·b·sin(∠C)
where a and b are adjacent sides and ∠C is the angle included between those two sides:
Unfortunately, we have ∠DAC, not ∠DCA. So, let's find ∠DCA.
Look at triangle ACD: We can find ∠DCA if we know both ∠DAC and ∠ADC.
To find the size of ∠ADC, let's use the Law of Sines:
sin(∠CDA) / AC = sin(∠CAD) / CD
---> sin(∠CDA) / 9 = sin(100°) / 11 ---> sin(∠CDA) = 0.80575 ---> ∠CDA = 53.683°
Since ∠CAD = 100° and ∠CDA = 53.683° , then ∠ACD = 26.3°
---> Area = 9·11·sin(26.3°) = 43.86
