Question

please anyone tell me the answer of this question find the area​

Answer 1

Step-by-step explanation:

Area of rectangle=l×b

5*4=20cm

Area of rectangle = 20cm

Mark the mid point of BD as F

Area of triangle BDC = 2( area of right triangle )

Area of BFC = (hypotenuse )^2 = (side) ^2+ (side) ^2

(hy) ^2= 9+4

(hy) ^2=13

(hy) =√13

BC=3.60555128

Area of triangle BDC= 2(3.60555128)

= 7.21110256

Area of given figure = 20+ 7.21110256

= 27.21110256

Answer 2

To find:

• Area of the given figure

Solution :

Let do it parts for better understanding:

Part 1:

To find Area of rectangle ABDE

Given :

• BD=4cm
• DE= 5cm

We know:

$\blue{ \boxed{ \boxed{ \rm{}Area \: of \: rectangle = length \times breadth}}}$

By using this formula we can find value of Area of rectangle ABDE

$: \implies\sf{}Area \: of \: rectangle = length \times breadth$

$: \implies\sf{}Area \: of \: rectangle = 4 \times 5$

$: \implies \underline{ \underline{\sf{}Area \: of \: rectangle = 20 {cm}^{2} }}$

part 2:

To find area of triangle BCD

Given:

• CF = 3cm
• BD = 4cm

We know:

$\blue{ \boxed{ \boxed{ \rm{}Area \: of \: triangle = \frac{1}{2} \times length \times breadth}}}$

By using this formula we can find value of Area of triangle BCD.

$: \implies \sf{}Area \: of \: triangle = \dfrac{1}{2} \times length \times breadth$

$: \implies \sf{}Area \: of \: triangle = \dfrac{1}{2} \times 3\times 4$

$: \implies \sf{}Area \: of \: triangle = \dfrac{1}{ \cancel2} \times 3\times \cancel 4$

$: \implies \sf{}Area \: of \: triangle = 3 \times 4$

$: \implies \underline{ \underline{\sf{}Area \: of \: triangle = 12 {cm}^{2} }}$

Part 3:

To find area of Given figure:

• Area of figure = Area of rectangle + Area of triangle
• Area of figure = 20+12
• Area of figure = 32 cm²

.°. Area of figure= 32 cm²

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$\begin{array}{|c|c|c|}\cline{1-3}\bf Shape&\bf Volume\ formula&\bf Surface\ area formula\\\cline{1-3}\sf Cube&\tt l^3}&\tt 6l^2\\\cline{1-3}\sf Cuboid&\tt lbh&\tt 2(lb+bh+lh)\\\cline{1-3}\sf Cylinder&\tt {\pi}r^2h&\tt 2\pi{r}(r+h)\\\cline{1-3}\sf Hollow\ cylinder&\tt \pi{h}(R^2-r^2)&\tt 2\pi{rh}+2\pi{Rh}+2\pi(R^2-r^2)\\\cline{1-3}\sf Cone&\tt 1/3\ \pi{r^2}h&\tt \pi{r}(r+s)\\\cline{1-3}\sf Sphere&\tt 4/3\ \pi{r}^3&\tt 4\pi{r}^2\\\cline{1-3}\sf Hemisphere&\tt 2/3\ \pi{r^3}&\tt 3\pi{r}^2\\\cline{1-3}\end{array}$