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Question Number 49605 by Rio Michael last updated on 08/Dec/18

Commented by Rio Michael last updated on 08/Dec/18

$T h e f i g u r e a b o v e h a s △ A B C a n d △ M B D w h e r e A C i s p a r a l l e l$ $t o D M . F i n d t h e l e n g h t A M i f$ $a) B M = 6.5 m$ $b) G i v e n t h e t h e a r e a o f △ A B C = 120 m^{2} f i n d t h e a r e a$ $o f t h e T r a p e z i u m A C D M .$
Commented by afachri last updated on 08/Dec/18

$what is CD sir ? ? ?$ ${can}^{'} t see it sir .$
Commented by Rio Michael last updated on 08/Dec/18

$C D = 2 m$
	Answered by Kunal12588 last updated on 08/Dec/18
	$in △ABC and △MBD ∠ABC=∠MBD=90° consider AC∥MD and AB as transversal so ∠CAB=∠DMB {corosponding ∠s} in any two triangle if 2 angles are equal so their 3^(rd) were also equal. so △ABC ∼ △MBD AAA criteria (1) now, ((AB)/(MB))=((BC)/(BD)) ((AM+MB)/(MB))=((BD+CD)/(BD)) ((AM)/(MB))=((CD)/(BD)) AM=(2/6)×6.5=((13)/6) m 2)from (1) ((ar(△ABC))/(ar(△MBD)))=((AB^2 )/(MB^2 ))=((BC^2 )/(BD^2 ))=((CA^2 )/(DM^2 )) ((ar(△MBD))/(ar(△ABC)))=((BD^2 )/(BC^2 )) invertendo ((ar(△ABC)−ar(MBD))/(ar(△ABC)))=((BC^2 −BD^ )/(BC^2 )) compendo ((ar(AMDC))/(ar(△ABC)))=(((6+2)^2 −6^2 )/((6+2)^2 )) ar(AMDC)=((64−36)/(64))×120=((85)/2)m^2 first can also be solved using thales theorem$
	$i n △ A B C a n d △ M B D$ $∠ A B C = ∠ M B D = 90 °$ $c o n s i d e r A C ∥ M D a n d A B a s t r a n s v e r s a l$ $s o ∠ C A B = ∠ D M B {c o r o s p o n d i n g ∠ s}$ $i n a n y t w o t r i a n g l e i f 2 a n g l e s a r e e q u a l$ $s o t h e i r 3^{r d} w e r e a l s o e q u a l .$ $s o △ A B C \sim △ M B D A A A c r i t e r i a (1)$ $n o w,$ $\frac{A B}{M B} = \frac{B C}{B D}$ $\frac{A M + M B}{M B} = \frac{B D + C D}{B D}$ $\frac{A M}{M B} = \frac{C D}{B D}$ $A M = \frac{2}{6} \times 6.5 = \frac{13}{6} m$ $2) f r o m (1)$ $\frac{a r (△ A B C)}{a r (△ M B D)} = \frac{{A B}^{2}}{{M B}^{2}} = \frac{{B C}^{2}}{{B D}^{2}} = \frac{{C A}^{2}}{{D M}^{2}}$ $\frac{a r (△ M B D)}{a r (△ A B C)} = \frac{{B D}^{2}}{{B C}^{2}} i n v e r t e n d o$ $\frac{a r (△ A B C) - a r (M B D)}{a r (△ A B C)} = \frac{{B C}^{2} - {B D}^{}}{{B C}^{2}} c o m p e n d o$ $\frac{a r (A M D C)}{a r (△ A B C)} = \frac{{(6 + 2)}^{2} - 6^{2}}{{(6 + 2)}^{2}}$ $a r (A M D C) = \frac{64 - 36}{64} \times 120 = \frac{85}{2} m^{2}$ $f i r s t c a n a l s o b e s o l v e d u s i n g t h a l e s t h e o r e m$
	Commented by Rio Michael last updated on 08/Dec/18

	$t h a n k s s i r$
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