What is the area of the parallelogram with vertices A(-3, 0), B(-1, 5), C(7, 4), and D(5, -1).

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un4t5o4v · Accepted Answer

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Jeffrey Jordon · Accepted Answer

Consider the parallelogram A=(-3;0), B(-1,5), C(7;4) and D=(5;-1)  The objective is to find the area of parallelogram  Find the are as follows:  The area of the parallelogram is the magnitude of the cross product of the adjacent edges.  That is, Area=|AB―×AD―|  Find the adjacent edges AB― and AD― as follows  AB―=B−A  =(−1;5)−(−3;0)  =(−1+3;5−0)  =(2;5)  AD―=D−A  =(5;−1)−(−3;0)  =(8;1)  Find AB―×AD― as,  AB―×AD―=[ijk2508−10]  =5(0)−0(−1)i−(2(0)−0(8))j+(2(−1)−5(8))k  =0i+0j−42k  The area of the parallelogram is,  Area =|AB×AD|  =|(0i+0j−42k|  =(0)2+(0)2+(−42)2  =(42)2  = 42 square unit  Hence, the area of parallelogram is 42

Jeffrey Jordon · Accepted Answer

The area of a parallelogram can be found by taking the cross product of its adjacent sides. Therefore, we can find the area of the parallelogram with vertices A(−3,0), B(−1,5), C(7,4), and D(5,−1) as follows:First, we need to find the two adjacent sides of the parallelogram. Let v→ be the vector from A to B and u→ be the vector from A to D. Then, we have:v→=[−1−(−3)5−0]=[25]u→=[5−(−3)−1−0]=[8−1]Next, we take the cross product of v→ and u→ to find the area of the parallelogram:v→×u→=|i→j→k→2508−10|=[00−42]Therefore, the area of the parallelogram is |−42|=42.

Vasquez · Accepted Answer

To find the area of the parallelogram with vertices A(−3,0), B(−1,5), C(7,4), and D(5,−1) is to use the distance formula to find the lengths of its adjacent sides.Let ABCD be the parallelogram with vertices A(−3,0), B(−1,5), C(7,4), and D(5,−1). Then, AB and AD are adjacent sides of the parallelogram.Using the distance formula, we have:AB=(xB−xA)2+(yB−yA)2=(−1−(−3))2+(5−0)2=20AD=(xD−xA)2+(yD−yA)2=(5−(−3))2+(−1−0)2=74The area of the parallelogram can then be found by taking the product of the lengths of its adjacent sides and the sine of the angle between them:Area=AB·AD·sin(θ)where θ is the angle between vectors AB→ and AD→. We can find θ using the dot product:AB→·AD→=(xB−xA)(xD−xA)+(yB−yA)(yD−yA)=(−1−(−3))(5−(−1))+(5−0)(−1−0)=−42Since AB→·AD→=AB·AD·cos(θ), we have:cos(θ)=AB→·AD→AB·AD=−422074=−21185Taking the sine of θ gives:sin(θ)=1−cos2(θ)=1−(−21185)2=8185Therefore, the area of the parallelogram is:Area=AB·AD·sin(θ)=20·74·8185=42

Find the area of the parallelogram with vertices A(-3, 0), B(-1, 5), C(7, 4), and D(5, -1).

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