Question

please solve please ​

Answer 1

Given

• △ABC is similar to △DEF AL and DM are altitudes drawn in △ABC and △DEF respectively.

To prove:

• $\sf\dfrac{Area \: of△ABC}{Area \: of△DEF}=\dfrac{AL²}{DM²}$

Proof:

In △ALB and △DEM

∠ALB=∠DME=90° (given)

∠B=∠E (∵△ABC is similar to △DEF)

∴∠BAL=∠EDM (Third angle)

∴ ∠ABL ~ ∠DEM (AA similarity criteria)

$\\\tt {\dfrac{AB}{ DE}= \dfrac{AL}{DM} ~~~~~----(1)}$

Now, △ABC ~ △DEF ( SAS Similarity criteria)

$\\{\sf\dfrac{Area \: of△ABC}{Area \: of△DEF}=\bigg(\dfrac{AB}{DE}\bigg)^{2} = \bigg(\dfrac{ AB}{DE} \bigg) ^2= \bigg( \dfrac{ AL}{DM}\bigg)^2 }$

$\\{\sf\dfrac{Area \: of△ABC}{Area \: of△DEF}= \bigg( \dfrac{ AL}{DY}\bigg)^2 }_{_{_{~~~~~~~~~~~~\bf Proved}}}$