Question

please answer this question it's urgent​

Answer 1

Answer:

A is 90 degree angle

ao is 9 cm

Answer 2

Given that

AQ//BP

OP= 6cm

QO= 9cm

A and B is equal to 90%

Find <QOA

In OAQ and OBP, we have

<A and <B [Each are equal to 90^]

<AOQ = <BOP

So, by AA- criterion of similarity, we have

AOQ ~ BOP

$\frac{area(tri \: aoq)}{area(tri \: bop)} = \frac{ {oq}^{2} }{ {op}^{2} }$

$\frac{area(tri \: aoq)}{120} = \frac{ {9}^{2} }{ {6}^{2} } \\ area(tri \: aoq) = \frac{81}{36} \times 120 {cm}^{2} \\ area(tri \: aoq) = 405 {cm}^{2}$

So, the <QOA or <AOQ = 405cm^ ...answer

I hope it is helpful answer for you.