The diagonal AC of a quadrilateral ABCD is 6 cm and perpendiculars , BM from the vertex B is 3 cm and DN is 5 cm , dropped on the same dropper on the same diagonal . Find the area of the quadrilateral ABCD
Answers
Step-by-step explanation:
Question:
The diagonal AC of a Quadrilateral ABCD is 6cm and the perpendiculars, BM from the vertex B is 3cm and DN from D is 5cm, dropped on the same diagonal. Find the area of the quadrilateral ABCD.
Given:
ABCD is a Quadrilateral.
Length of Diagonal AC is 6cm.
Perpendicular BM from Vertex B is 3cm.
Perpendicular DN from Vertex D is 5cm.
To Find:
Area of Quadrilateral ABCD.
Concept:
Here, we are given a Quadrilateral which has a diagonal and two perpendiculars of different lengths are dropped on the same diagonal. The diagonal cuts the quadrilateral into two different triangles. Finding the area of both triangles and adding them will give us the Area of Quadrilateral needed.
Formula Used:
\sf{Area\: of\: Triangle=\large{\frac{1}{2}}\times b\times h}AreaofTriangle=
2
1
×b×h
Where,
b is Base of Triangle.
h is Height of Triangle.
________________________________
Solution:
Let's firstly find out the area of triangle ABC.
» Base of ∆ABC= 6cm.
» Height of ∆ABC= 3cm.
Equating the values in Formula. We get,
\rightarrow\mathsf{Area\: of\: Triangle=\large{\frac{1}{2}}\times b\times h}→AreaofTriangle=
2
1
×b×h
\rightarrow\mathsf{Area\: of\: \triangle ABC=\large{\frac{1}{2}}\times 6\times 3}→Areaof△ABC=
2
1
×6×3
\rightarrow\mathsf{Area\: of\: \triangle ABC=\large{\frac{1}{2}}\times18}→Areaof△ABC=
2
1
×18
\rightarrow\mathsf{Area\: of\: \triangle ABC=\large{\frac{18}{2}}}→Areaof△ABC=
2
18
\rightarrow\bold{Area\: of\: \triangle ABC=9}→Areaof△ABC=9
This gives us the Area of Triangle ABC as 9 cm².
Now, Let's find the Area of triangle ADC.
» Base of ∆ADC= 6cm.
» Height of ∆ADC= 5cm.
Equating the values in Formula. We get,
\rightarrow\mathsf{Area\: of\: Triangle=\large{\frac{1}{2}}\times b\times h}→AreaofTriangle=
2
1
×b×h
\rightarrow\mathsf{Area\: of\: \triangle ADC=\large{\frac{1}{2}}\times 6\times 5}→Areaof△ADC=
2
1
×6×5
\rightarrow\mathsf{Area\: of\: \triangle ADC=\large{\frac{1}{2}}\times 30}→Areaof△ADC=
2
1
×30
\rightarrow\mathsf{Area\: of\: \triangle ADC=\large\frac{30}{2}}→Areaof△ADC=
2
30
\rightarrow\bold{Area\: of\: \triangle ADC=15}→Areaof△ADC=15
This gives us the Area of Triangle ADC as 15cm².
Now, adding the Areas of ∆ABC and ∆ADC gives Area of Quadrilateral ABCD. So,
\begin{gathered}\\ \implies\mathsf{Area\: of\: Quadrilateral\: ABCD=Area\: of\: \triangle ABC+ Area\: of\: \triangle ADC} \end{gathered}
⟹AreaofQuadrilateralABCD=Areaof△ABC+Areaof△ADC
\implies\mathsf{Area\: of\: Quadrilateral\: ABCD=9+15}⟹AreaofQuadrilateralABCD=9+15
\begin{gathered}\\ \implies\bold{Area\: of\: Quadrilateral\: ABCD\: =24} \end{gathered}
⟹AreaofQuadrilateralABCD=24
Required Answer:
The Area of Quadrilateral ABCD is 24cm².
________________________________
More to know:
\leadsto\mathsf{Area\: of\: Square={a}^{2}}⇝AreaofSquare=a
2
\leadsto\mathsf{Area\: of\: Rectangle=l\times b}⇝AreaofRectangle=l×b
\leadsto\mathsf{Area\: of\: Parallelogram=B\times H}⇝AreaofParallelogram=B×H
\leadsto\mathsf{Area\: of\: Kite=\large{\frac{1}{2}}\times {d}_{1}\times {d}_{2}}⇝AreaofKite=
2
1
×d
1
×d
2
a =Side of Square
l = Length of Rectangle
b = Breadth of Rectangle
B = Base of Parallelogram
H = Height of Parallelogram
d1 = First Diagonal of Kite
d2 = Second Diagonal of Kite
Answer: 24 sq cm
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Step-by-step explanation:
Area of quad. ABCD = 1/2 * diagonal AC * ( BM +DN)
= 1/2 * 6 * (3+5 )
= 3* 8
= 24 sq cm