Question

5. Find the area of trapezium where length of parallel sides are 15 cm and 25 cm and the third side measures 12 cm. no spamming allowed..​

Answer 1

Answer:

well you MAM and the guy you know who are the best biground in this number and I

Answer 2

Answer:

Answer:

240 cm² is the required area of Trapezium.

Step-by-step explanation:

According to the Question

It is given that,

Length of Parallel side 15 & 25 cm

Distance between them ,h = 12cm

we have to calculate the area of trapezium .

As we know that ,

Area of Trapezium = ½ × (sum of parallel side) × distance between them

On substituting the value we get

↠ Area of Trapezium = ½ × (15+25) × 12

↠ Area of Trapezium = ½ × (40) × 12

↠ Area of Trapezium = 20 × 12

↠ Area of Trapezium = 240 cm²

Hence, the area of trapezium is 240cm².

Additional Information !!

\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}

FormulasofAreas:−

⋆Square=(side)

2

⋆Rectangle=Length×Breadth

⋆Triangle=

2

1

×Breadth×Height

⋆Scalene△=

s(s−a)(s−b)(s−c)

⋆Rhombus=

2

1

×d

1

×d

2

⋆Rhombus=

2

1

d

4a

2

−d

2

⋆Parallelogram=Breadth×Height

⋆Trapezium=

2

1

(a+b)×Height

⋆EquilateralTriangle