[tex]\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}[/tex]
the given figure is a composition of a rectangle as well as a right angled triangle !
we've been given the two sides of the rectangle and we're required to find out the height of the triangle , so as to find it's area ~
we know the the opposite sides of a rectangle are equal , therefore we can break the longest side ( length = 9.5 cm ) into two parts ! the first part of length = 7 cm which is the length of the rectangle and the rest 2.5 cm ( 9.5 - 7 = 2.5 ) will become the height of the triangle !
For perimeter of the figure -
[tex]perimeter \: of \: figure = perimeter \: of \: rectangle + perimeter \: of \: triangle \\ \\ [/tex]
now ,
perimeter of rectangle = 2 ( l + b )
where ,
l = length
b = breadth
[tex]\longrightarrow \: perimeter = 2(7 + 6) \\ \longrightarrow \: 2(13) \\ \longrightarrow \: 26 \: cm[/tex]
and ,
[tex]perimeter \: of \: \triangle = 6.5 + 2.5 + 6 \\ \longrightarrow \: 15 \: cm[/tex]
Perimeter of figure in total = 26 cm + 15 cm
thus ,
[tex]\qquad\quad\bold\red{perimeter \: = \: 41 \: cm}[/tex]
For area of the figure -
[tex]area \: of \: figure = area \: of \: rectangle + area \: of \: rectangle \\ [/tex]
now ,
area of rectangle = l × b
where ,
l = length
b = breadth
[tex]area \: of \: rectangle = 7 \times 6 \\ \longrightarrow \: 42 \: cm {}^{2} [/tex]
and ,
[tex]area \: of\triangle = \frac{1}{2} \times base \times height \\ \\ \longrightarrow \: \frac{1}{\cancel2} \times \cancel6 \times 2.5 \\ \\ \longrightarrow \: 3 \times 2.5 \\ \\ \longrightarrow \: 7.5 \: cm {}^{2} [/tex]
Area of figure in total = 42 cm² + 7.5 cm²
thus ,
[tex]\qquad\quad\bold\red{Area \: = \: 49.5 \: cm^{2}}[/tex]
hope helpful :)
