Question

I want to understand ii question so please solve it​

Answer 1

Answer:

sorryyy i don't knoww thiss quessss....

this is veryy harddd

Answer 2

$\sf\Large{\underline{\underline{Answer:-}}}$

In the given polygon ABCDEF contains 4 triangles,

• ∆ABF$\Longrightarrow$ Base = 6.5cm, Height = 2cm
• ∆FBC$\Longrightarrow$ Base = 7cm, Height = 4cm
• ∆FDC $\Longrightarrow$ Base = 7cm, Height = 4cm
• ∆FED $\Longrightarrow$ Base = 5cm, Height = 2cm

⠀

We have to find the Area of all the triangles by the Formula,

$\large{\sf{\red{ \boxed{\sf area = \frac{1 }{2}\times base \times height}}}}$

⠀

First

• Area of ∆ABF

$\: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;area=\;\bf{\dfrac{1}{2} \times 6.5 \times 2\:}}\end{gathered}$

$\: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;area=\;\bf{\dfrac{1}{2} \times 13\:}}\end{gathered}$

$\: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;area=\;\bf{{\sf{\cancel\dfrac{13}{2}}}\:}}\end{gathered}$

$\: \: \: \: \: \: \: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;\bf{{\large {\sf { \blue{ \boxed{\sf area \: = {6.5^{2} } }}}}}}}\end{gathered}$

⠀

Now,

• Area of ∆FBC

$\: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;area=\;\bf{\dfrac{1}{2} \times 7 \times 4\:}}\end{gathered}$

$\: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;area=\;\bf{\dfrac{1}{2} \times 28\:}}\end{gathered}$

$\: \: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;area=\;\bf{{\sf{\cancel\dfrac{28}{2}}}\:}}\end{gathered}$

$\: \: \: \: \: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;\bf{{\large {\sf { \blue{ \boxed{\sf area \: = {14cm^{2} } }}}}}}}\end{gathered}$

⠀

Now,

• Area of ∆FDC

$\: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;area=\;\bf{\dfrac{1}{2} \times 7 \times 4\:}}\end{gathered}$

$\: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;area=\;\bf{\dfrac{1}{2} \times 28\:}}\end{gathered}$

$\: \: \: \: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;area=\;\bf{{\sf{\cancel\dfrac{28}{2}}}\:}}\end{gathered}$

$\: \: \: \: \: \: \: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;\bf{{\large {\sf { \blue{ \boxed{\sf area \: = {14cm^{2} } }}}}}}}\end{gathered}$

⠀

Now,

• Area of ∆FED

$\: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;area=\;\bf{\dfrac{1}{2} \times 5 \times 2\:}}\end{gathered}$

$\: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;area=\;\bf{\dfrac{1}{2} \times 10\:}}\end{gathered}$

$\: \: \: \: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;area=\;\bf{{\sf{\cancel\dfrac{10}{2}}}\:}}\end{gathered}$

$\: \: \: \: \: \: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;\bf{{\large {\sf { \blue{ \boxed{\sf area \: = {5cm^{2} } }}}}}}}\end{gathered}$

⠀

$\Longrightarrow$Now, add the areas of all the triangle.

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area=\;\bf{{∆ABF + ∆FBC +∆FDC + ∆FED }}}\end{gathered}$

$\: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;area=\;\bf{{6.5 + 14 +14 + 5}}}\end{gathered}$

$\: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;the\;required \: area=\;\bf{{39.5cm^{2} }}}\end{gathered}$

∴ The required area is 39.5cm².