Question

Use the information given in the adjoining figure to find:
(i) the length of AC.
(ii) the are a of $\Delta ABC$
(iii) the length of BD, correct to one decimal place.

Answer 1

REF.Image

$A{C}^2=A{B}^2+B{C}^2$

$A{C}^2={24}^2+7^2$

AC = 26

area $\Delta ABC=\frac{1}{2}\times 24\times 7=84$

$B{D}^2=(24{)}^2-(AD{)}^2$

$(BC{)}^2=B{D}^2+(CD{)}^2$

$(BC{)}^2=(24{)}^2-(AD{)}^2+(CD{)}^2$

$49=(24{)}^2-(AD{)}^2+(AC-AD{)}^2$

$49=(24{)}^2-A{D}^2+(AC-AD{)}^2$

$49=(24{)}^2-A{D}^2+A{C}^2+A{D}^2-2ACAD$

$49=576+676-26AD$

$26\times 2AD=1252-49=1203$

$AD=\frac{1203}{52}=23.13$

$B{D}^2={24}^2-(23.13{)}^2$

$B{D}^2=40.9$

BD= 6.4