plz solve this answer bro and sis
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Given in triangle xyz xp/py =qx/qz =3
py/xp +1 =qx/qz +1 =1/3+1
Py +xp /xp =qz +qx /qx =1+3 /3
xy /xp =xz/xq =4/3
since pq is parallel to yz
now from triangle xpq and xyz
xp/xy =qx/xz =3/4 from eq 1
angle x=angle x (common on both)
triangle xpq in triangle xyz (sas)
aera of xpq /aera of xyz =(xp/xy) 2
aera of xyz - aera of quadrilateral pxzq /32
(3/4)2
=32-aera of pyzq /32=9/16 32-aera of pqyz=18
Aera of a quadrilateral pyzq =32 - 12 =14 cm2
py/xp +1 =qx/qz +1 =1/3+1
Py +xp /xp =qz +qx /qx =1+3 /3
xy /xp =xz/xq =4/3
since pq is parallel to yz
now from triangle xpq and xyz
xp/xy =qx/xz =3/4 from eq 1
angle x=angle x (common on both)
triangle xpq in triangle xyz (sas)
aera of xpq /aera of xyz =(xp/xy) 2
aera of xyz - aera of quadrilateral pxzq /32
(3/4)2
=32-aera of pyzq /32=9/16 32-aera of pqyz=18
Aera of a quadrilateral pyzq =32 - 12 =14 cm2
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