In the given figure shows a quadrilateral ABCD. A line is drawn from A parallel to BD intersects CD produced at Q. BP is a line parallel to CQ such that QP perpendicular to BP. If CQ =22 cm and PQ =5 cm. Find the area of quadrilateral ABCD.
Answers
55 cm² is Area
Step-by-step explanation:
CQ = 22 cm
PQ = 5 cm
BP ║ CQ
QP ⊥ BP
Ar (ΔBDA) = Ar (ΔBDN) Triangles formed between the same pair of parallel lines and with same base are equal in areas ,
BD is the base and BD || AQ]
BNQD is a parallelogram and ND is its diagonal.
Ar(ΔBDN) = 1/2*area (BNQD)
therefore
Ar(ΔBDA) = 1/2*area(BNQD)
Ar(ΔBDA) = 1/2*DQ*(height between the parallel lines BN and DQ]
Ar(ΔBDA) = 1/2*DQ*PQ
Ar(ΔBDA) = 1/2*DQ*5
Ar(ΔBCD) = 1/2*CD*(the length of perpendicular from B to CD)
Ar(ΔBCD) = 1/2*CD*5
Ar(ΔBDA)+ Ar(ΔBCD) = 1/2*DQ*5+1/2*CD*5
Ar(ABCD) = 1/2*(DQ+CD)*5
= 1/2*CQ*5
=1/2*22*5
=55 sq cm
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