Question

Please help it’s just only two questions

Answer 1

Answer:

Question-6;

The value of $x=7.5$ $in$

Question-7;

So the Answer is 'A'

Step-by-step explanation:

Question-6;

From the portion 'ABCD';

Given the area of rectangle 'ABCD' is $30$ $in^{2}$ , $AD=4$ $in$ and $AD=x$ $in$

We have the formula to find the area of rectangle;

Area of rectangle 'ABCD'$=AD\times DC$

$30=4\times x$

$x=\frac{30}{4}$

$x=7.5$ $in$

So the value of $x=7.5$ $in$.

Question-7;

From figure;

Given;

All values $AB=CD=FG=IJ=9$ $cm$ , $AC=BD=5$ $cm$ , $CF=DG=4$ $cm$ , $FI=GJ=3$ $cm$ and $EF=GH=3$ $cm$

In Rectangle 'ABIJ' have three sub-rectangle which are 'ABCD' , 'CDFG' and 'FGIJ'

Area of Rectangle 'ABIJ'  equal to sum of area of Rectangle 'ABCD' , area of Rectangle 'CDFG' and area of Rectangle 'FGIJ'

Area of Rectangle 'ABIJ'$=(AB\times AC)+(CD\times CF)+(FG\times FI)$

Plug all values in above equation,

Area of Rectangle 'ABIJ'$=(9\times 5)+(9\times 4)+(9\times 3)$

Area of Rectangle 'ABIJ'$=108$ $cm^{2}$

Also from figure;

$\bigtriangleup CEF$ and $\bigtriangleup DGH$ have same Area.

Total Area of this two triangle$=2\times Area of \bigtriangleup CEF$

Total Area of this two triangle$=2\times\frac{1}{2}\times BASE \times HIGHT$

Total Area of this two triangle$=2\times\frac{1}{2}\times EF \times CF$

Plug 'EF' and 'CF' values in above equation,

Total Area of this two triangle$=2\times \frac{1}{2} \times3 \times 4$

Total Area of this two triangle$=12$ $cm^{2}$

∴ Total surface area of the building block is $(108+12)=120$ $cm^{2}$