Question

plss answer this question with an appropriate explanation!


wrong answers or irralevant answers will be reported.. //

Answer 1

Answer:

In ∆ABC,

by phythagoras theorem

(AC)^2=(AB)^2+(BC)^2

(20)^2=(AB)^2+(12)^2

(AB)^2=400-144

AB=√356

Answer 2

Required Solution:

Given:

• AC ( Hypotenuse ) = 20cm

• BC ( Base ) = 12cm

To calculate:

• Unknown side ( AB or perpendicular )

Calculation:

Here, we'll apply pythagoras property to find out the measure of the perpendicular.

We know that,

• ${\underline {\boxed {\Large {\bf \gray { {H}^{2} = {B}^{2} + {P}^{2} } }}}}$

$\qquad$ $\sf { \longmapsto {AC}^{2} = {BC}^{2} + {AB}^{2} }$

$\qquad$

$\qquad$ $\sf { \longmapsto {20}^{2} = {12}^{2} + {AB}^{2} }$

$\qquad$

$\qquad$ $\sf { \longmapsto 400= 144 + {AB}^{2} }$

$\qquad$

$\qquad$ $\sf { \longmapsto {AB}^{2} = 400 - 144 }$

$\qquad$

$\qquad$ $\sf { \longmapsto {AB}^{2} = 256 }$

$\qquad$

$\qquad$ $\sf { \longmapsto AB = \sqrt{256} }$

$\qquad$

$\qquad$ $\sf { \longmapsto \boxed{ \mathfrak \pink {AB = 16cm}} }$

$\underline { \sf { \therefore Unknown \: side \: of \: this \: ∆ \: is \: \bold{16cm} }}$

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Verification:

• ${\underline {\boxed {\Large {\bf \gray { {H}^{2} = {B}^{2} + {P}^{2} } }}}}$

$\qquad$ $\sf { \longmapsto {20}^{2} = {12}^{2} + {16}^{2} }$

$\qquad$

$\qquad$ $\sf { \longmapsto 400 = 144 + 256 }$

$\qquad$

$\qquad$ $\sf { \longmapsto 400 = 400}$

$\underline { \sf { Hence, Verified!! } }$

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