Question

In figure $△ABC\sim △PQR$ then y+z is

• A. $5+\sqrt{3}$
• B. $4+3\sqrt{3}$
• C. $2+\sqrt{3}$
• D. $3+4\sqrt{3}$

Answer 1

Answer: (B)

$4+3\sqrt{3}$

$△PQR$ is a right angle triangle then

$(QR{)}^2=(PR{)}^2+(PO{)}^2$

$(6{)}^2=(y{)}^2+(3{)}^2$

$y=\sqrt{36-9}$

$y=\sqrt{27}$

$y=\sqrt{9\times 3}$

$y=3\sqrt{3}$

$△ABC$ is a right angle triangle then

$(BC{)}^2=(AB{)}^2+(AC{)}^2$

$(8{)}^2=(z{)}^2+(4\sqrt{3}{)}^2$

$z=\sqrt{64-48}$

$z=\sqrt{16}$

$z=4$

$\therefore y+z=4+3\sqrt{3}$

Or

This can be done using SSS criteria as both triangles are given to be similar.