ABCD is a quadrilateral with Ag = 20 cm, BC-15 cm,
CD = 25 cm 4 = 30 cm and ABC = 90" The area of
the quadrilateral ABCD
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What is the area of a quadrilateral ABCD in which AB = 10 cm, BC = 25 cm, CD = 8 cm, DA = 15 cm and ∠ADC = 90 degree?
ABCD is a quadrilateral and angle ADC is 90°. Draw a quadrilateral ABCD and join A to C.
In ∆ADC, Pythagoras can be used
AC2=AD2+DC2
AC2=152+82
AC=17cm
Area of ΔADC=1/2×base×height
Area of ADC=1/2×15×8
Area of ADC=60 sq cm
Now area of ΔABC
S=10+25+17/2
S=26cm
Area of ΔABC=√26(26–10)(26–25)(26–17)
Area of ΔABC=√26×16×1×9
Area of ΔABC=12√26cm2
Area of ABCD=(60+12√26) sq cm
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