Question

.......Please answer fast and get 15 points.


The parallel sides of an isoscles triangle ABCD are 14m and 8m . Also the distance between them in 6m . AM and BN are the perpindicular drawn to CD . Find the ratio of the area of trapezium ABCD and area of the triangle ADM .

Answer 1

$\underline{\large{\mathbf{Hey\: Mate\: Here\: Is\: your\:Answer}}}$

$\underline{\underline{\bold{Given\: That}}}$

ABCD Is An Isosceles Trapezium In Which The Measure Of The Parallel Sides are 14m & 8m , Also Given That AM And BN Are Perpendiculars Having Length Of 6m

, And In Question And ∆ADM Is drawn Through The

Perpendicular AM , Now According To Questions We Need To Find Ratio Of

$\bold{ar(ADM)\: And \:ar(ABCD)}$

So Now Let's Move For Solution

$\underline{\underline{\large{\bold{Solution..}}}}$

So Now To Find The Ratio .. Follow The Simple Steps..

$\underline{\bold{Step-1\: )\: D e fi n e\: All \:Required\: Values\: }}$

Now According To Question The Length Of

$\mathsf{\longrightarrow AB = 8m}$

$\mathsf{\longrightarrow CD = 14m}$

So Now In Question It's Said That The Perpendicular Distance Is

$\mathsf{\longrightarrow AM = 6m }$

$\underline{\textbf{Step-2 ) Find Area Of Trapezium ABCD }}$

SO We Know That Area Of Trapezium Is Equal To

$\boxed{\mathbf{ \: \frac{1}{2} (sum \ \: of \: \: parallel \: \: sides \: )(height)}}$

According To Question We Have

$\mathbf{Parallel\: Side \: AB\: =\: 8m}$

$\mathbf{Second\: Parallel\: Side\: CD\: =\:14m}$

And

$\mathbf{Height\:AM\:=\:6m}$

So now Putting Values Of Height And Parallel Sides in The

Formula We Have ...

$\mathbf{ar(abcd) = \frac{1}{2}(14m + 8m)(6m)}$

That Is...

$\mathbf{ar(ABCD)=\frac{1}{2}\:(22)(6)}$

That Is...

$\mathbf{ar(ABCD)=\frac{1}{2}(132m)}$

That Is Equal To ..

$\mathbf{ar(ABCD)\:= \:66m}$

So Now We Have

$\boxed{\bold{ar(ABCD)\: = 66m}}$

_______________________________

$\underline{\textbf{Step-3 Find Area Of Triangle ADM }}$

So We Know That Area Of A Triangle Is Equal To

$\boxed{\mathbf{\frac{1}{2}\:(Base)(Height)}}$

Now According To Question It's Given That Height Is

$\bold{AM\:=\:6m}$

And As , The Trapezium Is Isolated Therefore It's Symmetrical
So By This We Have ..

$\bold{DM(Base)\:=\:3m}$

Therefore Now By Putting This Values In The Formula We Have ...

$\mathbf{ar(ADM\:=\frac{1}{2}(3)(6)}$

That Is ..

$\mathbf{ar(ADM\:={1}{2}(18)}$

Therefore We Have Area Of

$\boxed{\bold{ADM\:=\:9{m}^{2}}}$

$\underline{\textbf{Step-4 ) Find Ratio of ADM And ABCD}}$

So Now Ratio of Area ADM And ABCD Is

$\mathbf{\frac{ar(ADM)}{ar(ABCD)}}$

That Is ..

$\bold{\frac{9{m}^{2}}{66{m}^{2}}}$

That Is ...

$\bold{\frac{3{m}^{2}}{22{m}^{2}}}$

$\underline{\textbf{HenceRequired Ratio Is ...}}$

$\boxed{\bold{3\: : \:22}}$

$\boxed{\boxed{\boxed{\bold{THANKS...}}}}$