ab and dc are parallel sides of the trapezium abcd and angle adc = 90 degrees. given ab= 15 cm,cd=40 cm and diagonal ac= 41 cm, calculate the area of trapezium abcd.
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Answered by
233
ADC = 90°
Ab parallel to dc
ab = 15 cm
cd = 40cm
ac = 41 cm
Using Pythagoras Theorem
ad² + dc² = ac²
ad² + 40² = 41²
ad² + 1600 = 1681
ad = √1681- 1600
ad = √81
as = 9 cm
area of trapezium =1/2 × (base1 + base2) × h
1/2× ( 15 + 41) × 9
= 247.5 cm²
Ab parallel to dc
ab = 15 cm
cd = 40cm
ac = 41 cm
Using Pythagoras Theorem
ad² + dc² = ac²
ad² + 40² = 41²
ad² + 1600 = 1681
ad = √1681- 1600
ad = √81
as = 9 cm
area of trapezium =1/2 × (base1 + base2) × h
1/2× ( 15 + 41) × 9
= 247.5 cm²
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Answered by
38
in right angled ΔADC,
AD = √(41^2-40^2)= 9
area=(1/2)(15+40)(9)=247•5cm^2
AD = √(41^2-40^2)= 9
area=(1/2)(15+40)(9)=247•5cm^2
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