Question

Please help it’s just only two questions

Answer 1

Question-6;

The value of
`x=7.5`
`in`

Question-7;

So the Answer is 'A'

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* DC`

`30=4* x`

`x=(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* AC)+(CD* CF)+(FG* FI)`

Plug all values in above equation,

Area of Rectangle 'ABIJ'
`=(9* 5)+(9* 4)+(9* 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* Area of \bigtriangleup CEF`

Total Area of this two triangle
`=2*(1)/(2)* BASE * HIGHT`

Total Area of this two triangle
`=2*(1)/(2)* EF * CF`

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

Total Area of this two triangle
`=2* (1)/(2) *3 * 4`

Total Area of this two triangle
`=12`
`cm^(2)`

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