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Given :ABCD is a rectangleED=16cmFG=FC=8cmBF=20cmFG is pependicular to BCCD=1/2 AD
To find : Area of shaded region
Solution :BC=BF+FC
=20+8
=28cm
CD=1/2 AD=1/2BC= 1/2 x 28=14m
AE=AD-ED =28-16=12cm
AREA OF SHADED REGION =AREA OF RECTANGLE - (AREA OF ΔAEB + AREA OF TRAPEZIUM)
=28X14 -(1/2 X12X14+1/2 X(8(8+14)
=392-(84+88)
=392-172
=220cm²
∴Area of shaded region is 220cm²
To find : Area of shaded region
Solution :BC=BF+FC
=20+8
=28cm
CD=1/2 AD=1/2BC= 1/2 x 28=14m
AE=AD-ED =28-16=12cm
AREA OF SHADED REGION =AREA OF RECTANGLE - (AREA OF ΔAEB + AREA OF TRAPEZIUM)
=28X14 -(1/2 X12X14+1/2 X(8(8+14)
=392-(84+88)
=392-172
=220cm²
∴Area of shaded region is 220cm²
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