Question

help me to solve this question


http://im41.gulfup.com/ODADc.jpg

go to the link to see the question

Answer 1

Question 1. In the quadrilateral $ABCD$:

$CA\perp DB,\ \ BC=6cm,\ \ CD=3cm,DA=5cm.$

Find $AB$.

Solution. Let $O$ be the intersection of $CA$ and $DB$. By Pythagorean theorem for $\triangle AOB$, $\triangle BOC$, $\triangle COD$, $\triangle DOA$ we have

$OA^{2}+OB^{2}=AB^{2},$

$OB^{2}+OC^{2}=BC^{2},$

$OC^{2}+OD^{2}=CD^{2},$

$OA^{2}+OD^{2}=AD^{2}.$

Adding the first 3 equalities we get

$OA^{2}+OD^{2}+2(OB^{2}+OC^{2})=AB^{2}+BC^{2}+CD^{2}.$

But $OA^{2}+OD^{2}=AD^{2}$ and $OB^{2}+OC^{2}=BC^{2}$, so

$AD^{2}+2BC^{2}=AB^{2}+BC^{2}+CD^{2},$

whence

$AB^{2}=AD^{2}+BC^{2}-CD^{2}=25+36-9=52.$

Answer: $AB=\sqrt{52}=2\sqrt{13}$ cm.

$\square$