28) A (-4,2), B (-3,-5), C (3,-2) and D (2, k) are the vertices of a quadrilateral ABCD. If the areaof the quadrilateral is 28 sq. units, find the value of ‘K’. About the author Ella
Let the points be A(4,2),B(3,5),C(3,2) and D(2,3) The quadrilateral ABCD can be divided into triangles ABC and ACD and hence the area of the quadrilateral is the sum of the areas of the two triangles. Area of a triangle with vertices (x 1 ,y 1 ) ; (x 2 ,y 2 ) and (x 3 ,y 3 ) is ∣ ∣ ∣ ∣ ∣ 2 x 1 (y 2 −y 3 )+x 2 (y 3 −y 1 )+x 3 (y 1 −y 2 ) ∣ ∣ ∣ ∣ ∣ Hence, substituting the points (x 1 ,y 1 )=(−4,−2) ; (x 2 ,y 2 )=(−3,−5) and (x 3 ,y 3 )=(3,−2) in the area formula, we get Area of triangle ABC = ∣ ∣ ∣ ∣ ∣ 2 (−4)(−5+2)+(−3)(−2+2)+3(−2+5) ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ 2 12+0+9 ∣ ∣ ∣ ∣ ∣ = 2 21 =10.5squnits And, substituting the points (x 1 ,y 1 )=(−4,−2) ; (x 2 ,y 2 )=(3,−2) and (x 3 ,y 3 )=(2,3) in the area formula, we get Area of triangle ACD = ∣ ∣ ∣ ∣ ∣ 2 −4(−2−3)+(3)(3+2)+2(−2+2 ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ 2 20+15+0 ∣ ∣ ∣ ∣ ∣ = 2 35 =17.5squnits Hence, Area of quadrilateral =10.5+17.5=28squnits Reply
[tex] \huge \fbox \purple{question}[/tex] A (-4,2), B (-3,-5), C (3,-2) and D (2, k) are the vertices of a quadrilateral ABCD. If the area of the quadrilateral is 28 sq. units, find the value of ‘K’. . [tex] \huge \fbox \green{solution}[/tex] A (-4,2), B (-3,-5), C (3,-2), D(2,k) Area of triangle ABCD = [tex] \frac{1}{2} ( – 4( – 5 + 2) – 3( – 2 + 2) + 3( – 2 + 5)) = \frac{21}{2} [/tex] sq.unit Area of Triangle ABC= [tex] \frac{1}{2} ( – 4( – 2 – k) + 3(k + 2) + 2( – 2 + 2)) = \frac{(7k + 14)}{2} [/tex] sq unit Area of quadrilateral ABCD =28sq. unit [tex]therefore \frac{21}{2} + \frac{7k + 14}{2} = 28 \\ \frac{7k + 35}{2} = 28 \\ 7k = 21 \\ therefore \: k = 3 [/tex] hope this helps u.. Reply
Let the points be A(4,2),B(3,5),C(3,2) and D(2,3)
The quadrilateral ABCD can be divided into triangles ABC and ACD and hence the
area of the quadrilateral is the sum of the areas of the two triangles.
Area of a triangle with vertices (x
1
,y
1
) ; (x
2
,y
2
) and (x
3
,y
3
) is
∣
∣
∣
∣
∣
2
x
1
(y
2
−y
3
)+x
2
(y
3
−y
1
)+x
3
(y
1
−y
2
)
∣
∣
∣
∣
∣
Hence, substituting the points (x
1
,y
1
)=(−4,−2) ; (x
2
,y
2
)=(−3,−5) and (x
3
,y
3
)=(3,−2) in the area formula, we get
Area of triangle ABC =
∣
∣
∣
∣
∣
2
(−4)(−5+2)+(−3)(−2+2)+3(−2+5)
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
2
12+0+9
∣
∣
∣
∣
∣
=
2
21
=10.5squnits
And, substituting the points (x
1
,y
1
)=(−4,−2) ; (x
2
,y
2
)=(3,−2) and (x
3
,y
3
)=(2,3) in the area formula, we get
Area of triangle ACD =
∣
∣
∣
∣
∣
2
−4(−2−3)+(3)(3+2)+2(−2+2
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
2
20+15+0
∣
∣
∣
∣
∣
=
2
35
=17.5squnits
Hence, Area of quadrilateral =10.5+17.5=28squnits
[tex] \huge \fbox \purple{question}[/tex]
A (-4,2), B (-3,-5), C (3,-2) and D (2, k) are the vertices of a quadrilateral ABCD. If the area
of the quadrilateral is 28 sq. units, find the value of ‘K’.
.
[tex] \huge \fbox \green{solution}[/tex]
A (-4,2), B (-3,-5), C (3,-2), D(2,k)
Area of triangle ABCD =
[tex] \frac{1}{2} ( – 4( – 5 + 2) – 3( – 2 + 2) + 3( – 2 + 5)) = \frac{21}{2} [/tex]
sq.unit
Area of Triangle ABC=
[tex] \frac{1}{2} ( – 4( – 2 – k) + 3(k + 2) + 2( – 2 + 2)) = \frac{(7k + 14)}{2} [/tex]
sq unit
Area of quadrilateral ABCD =28sq. unit
[tex]therefore \frac{21}{2} + \frac{7k + 14}{2} = 28 \\ \frac{7k + 35}{2} = 28 \\ 7k = 21 \\ therefore \: k = 3 [/tex]
hope this helps u..