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3)
Given:-
Find:-
Solution:-
In ∆ABC
▶H² = P² + B² [Pythogoras Theorem]
▶H² = 5² + 4²
▶H² = 25 + 16
▶H² = 41
▶H = √(41)
▶H = 6.40cm
we, know that
[tex] \large{ \underline{\boxed{ \sf Area \: of \: Trapezium =area \: of \: rectangle + 2(area \: of \: triangle)}}}[/tex]
[tex] \large{ \underline{\boxed{ \sf Area \: of \: rectangle = l \times b}}}[/tex]
[tex] \large{ \underline{\boxed{ \sf Area \: of \: triangle = \dfrac{1}{2} \times b \times h}}}[/tex]
So,
[tex] \dashrightarrow\sf Area \: of \: Trapezium =area \: of \: rectangle + 2(area \: of \: triangle) \\ \\ [/tex]
[tex] \dashrightarrow\sf Area \: of \: Trapezium =l \times b+ 2 \bigg( \dfrac{1}{2} \times b \times h\bigg) \\ \\ [/tex]
where,
So,
[tex] \dashrightarrow\sf Area \: of \: Trapezium =19\times6.40+ 2 \bigg( \dfrac{1}{2} \times 4 \times 6.40\bigg) \\ \\ [/tex]
[tex] \dashrightarrow\sf Area \: of \: Trapezium =121.6+ 2 \bigg( \dfrac{1}{2} \times 4 \times 6.40\bigg) \\ \\ [/tex]
[tex] \dashrightarrow\sf Area \: of \: Trapezium =121.6+ 2 \bigg( \dfrac{1}{2} \times25.6\bigg) \\ \\ [/tex]
[tex] \dashrightarrow\sf Area \: of \: Trapezium =121.6+ 2 \times 12.8 \\ \\ [/tex]
[tex] \dashrightarrow\sf Area \: of \: Trapezium =121.6+ 25.6 \\ \\ [/tex]
[tex] \dashrightarrow\sf Area \: of \: Trapezium =147.2 {cm}^{2} \\ \\ [/tex]
Hence, Area of given Trapezium = 147.2cm²
____________________________
4)
Given:-
Find:-
Solution:-
AD = BC = 10cm
DE = 8cm
➟Perimeter = Sum of all sides
➟Perimeter = AB + BC + CD + DA
where,
Perimeter = 48cm
BC = 10cm = DA = 10cm
So,
➟ 48 = AB + 10 + CD + 10
➟ 48 = AB + CD + 10+10
➟ 48 = AB + CD + 20
➟ 48 - 20 = AB + CD
➟ AB + CD = 28cm
Now, we know that
[tex] \large{ \underline{\boxed{ \sf Area \: of \: Trapezium = \dfrac{1}{2} \times (sum \: of \: parallel \: sides) \times h}}}[/tex]
where,
So,
[tex] \implies\sf Area \: of \: Trapezium = \dfrac{1}{2} \times (AB + CD) \times DE \\ \\ [/tex]
[tex] \implies\sf Area \: of \: Trapezium = \dfrac{1}{2} \times (28) \times 8 \\ \\ [/tex]
[tex] \implies\sf Area \: of \: Trapezium = \dfrac{1}{2} \times 28 \times 8 \\ \\ [/tex]
[tex] \implies\sf Area \: of \: Trapezium = \dfrac{1}{2} \times224\\ \\ [/tex]
[tex] \implies\sf Area \: of \: Trapezium = \dfrac{224}{2}\\ \\ [/tex]
[tex] \implies\sf Area \: of \: Trapezium = 112 {cm}^{2} \\ \\ [/tex]
Hence, Area of Trapezium = 112cm²
____________________________
5)
Given:-
Find:-
Solution:-
we, know that
[tex] \large{ \underline{\boxed{ \sf Area \: of \: Trapezium = \dfrac{1}{2} \times (sum \: of \: parallel \: sides) \times h}}}[/tex]
where,
So,
[tex] \dashrightarrow\sf Area \: of \: Trapezium = \dfrac{1}{2} \times (sum \: of \: parallel \: sides) \times h \\ \\ [/tex]
[tex] \dashrightarrow\sf Area \: of \: Trapezium = \dfrac{1}{2} \times (34 + 10) \times 5 \\ \\ [/tex]
[tex] \dashrightarrow\sf Area \: of \: Trapezium = \dfrac{1}{2} \times (44) \times 5 \\ \\ [/tex]
[tex] \dashrightarrow\sf Area \: of \: Trapezium = \dfrac{1}{2} \times 44 \times 5 \\ \\ [/tex]
[tex] \dashrightarrow\sf Area \: of \: Trapezium = \dfrac{1}{2} \times 220 \\ \\ [/tex]
[tex] \dashrightarrow\sf Area \: of \: Trapezium = \dfrac{220}{2} \\ \\ [/tex]
[tex] \dashrightarrow\sf Area \: of \: Trapezium = 110m^2\\ \\ [/tex]
Hence, Area of Trapezium = 110m²