answer this plzzzzz
Answers
Answer: 32.75cm sq.
Given that: i) MP=9cm
MD=7cm; AN=2.5cm
MC=6cm; OC=3cm
MB=4cm; BR=2.5cm
MA=2cm; QD=2cm
ii)NA, OC, QD and RB are perpendicular to diagonal MP.
Find: Area of polygon MNOPQR.
ar(MNOPQR)=ar(ΔMAN)+ar(ΔOCP)+ar(ΔQDP)+ar(ΔMBR)+ar(trap.ANOC)+ar(trap.RBDQ)
Required values:
i) CP= MP-MC
=> CP= 9cm-6cm
=> CP= 3cm
ii) DP= MP-MD
=> DP= 9cm-7cm
=> DP= 2cm
iii) AC= MC-MA
=> AC=6cm-2cm
=>AC=4cm
iv) BD= MD-MB
=> BD= 7cm-4cm
=> BD= 3cm
Step-by-step explanation:
i) ar(ΔMAN)= ½ of Base x Height
=> ar(ΔMAN)= ½ x MA x AN
= ½ x 2cm x 2.5cm
= ½ x 5cm sq.
= 2.5cm sq.
ii) ar(ΔOCP)= ½ of Base x Height
=> ar(ΔOCP)= ½ x CP x OC
= ½ x 3cm x 3cm
= ½ x 9cm sq.
= 4.5cm sq.
iii) ar(ΔQDP)= ½ of Base x Height
=> ar(ΔQDP)= ½ x DP x QD
= ½ x 2cm x 3cm
= ½ x 6cm sq.
= 3cm sq.
iv) ar(ΔMBR)= ½ of Base x Height
=> ar(ΔMBR)= ½ x MB x BR
= ½ x 4cm x 2.5cm
= ½ x 10cm sq.
= 5cm sq.
v) ar(trap.ANOC)= ½ x(Sum of parallel sides) x
Height
=>ar(trap.ANOC)= ½ x(AN+OC) x AC
= ½ x(2.5cm+3cm) x 4cm
= ½ x 5.5cm x 4cm
= ½ x 22cm sq.
= 11cm sq.
vi) ar(trap.RBDQ)= ½ x(Sum of parallel sides) x
Height
=>ar(trap.RBDQ)= ½ x(RB+DQ) x BD
= ½ x(2.5cm+2cm) x 3cm
= ½ x 4.5cm x 3cm
= ½ x 13.5cm sq.
= 6.75cm sq.
Therefore,
ar(MNOPQR)= (2.5+4.5+3+5+11+6.75)cm sq.
= 32.75cm sq .