find the area of a trapezium whose parallel sides are 25 cm and 13 cm and another sides are 15 cm and 15 cm
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area of a trapezium=h(a+b)÷2
area of the triangle formed is 18root 21 cm.
hence finding the height the area bcomes 45root21 cmsquare
area of the triangle formed is 18root 21 cm.
hence finding the height the area bcomes 45root21 cmsquare
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Given:- ABCD Is a trapezium
AB = 25 cm
DC = 13 cm
AD & BC = 15 cm
Construction:- Draw CE || AD
To Find :- Area of trapezium ABCD
Solution :- ADCE is a parallelogram ( AD || CE & AE || CD).
∴ AE = DC = 13 cm ( Opposite side of parallelogram are equal)
BE = AB - AE
BE = 25 - 13
BE = 12 cm
In ∆ BCE
S = a + b + c/2
S = 15 + 15 + 12 /2
S = 21
Area of ∆ BCE = √ s( s - a)(s - b)( s - c )
Area of ∆ BCE = √ 21(21-15)(21-15)(21-12)
Area of ∆ BCE = √ 21 × 6 × 6 × 9
Area of ∆ BCE = 18√21 cm^2 -----1
h is the height of ∆ BCE
Area of BCE = 1/2 ( Base × Height )
= 1/2(12)(h)
= 6h -----2
From 1 & 2
6h = 18√21
=> h = 3√21 cm
The height of trapezium ABCD is equal to height of ∆ BCE.
Area of trapezium = 1/2 ( AB + CD ) × h
= 1/2 (25 + 13) × 3√21cm^2
= 57√21 cm^2
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