two prallel sides of a trapazium are 60m and 77m, and the nonparallel sides are 25m and 26m. find the area of the trapazium.
Answers
Answer:
The area of the trapezium is 1644 m².
Step-by-step-explanation:
NOTE: Refer to the attachment for the diagram.
In figure, □ABCD is a trapezium.
AB ∥ CD
- AB = 60 m
- CD = 77 m
- AD = 25 m
- BC = 26 m
Now,
Draw BP ∥ AD and BQ ⊥ PC.
∴ □ABPD is a parallelogram.
∴ AB = PD = 60 m
AD = BP = 25 m
Now,
CD = PD + PC - - - [ C - P - D ]
⇒ PC = CD - PD
⇒ PC = 77 - 60
⇒ PC = 17 m
Now, in △BPC,
- BP ( a ) = 25 m
- PC ( b ) = 17 m
- BC ( c ) = 26 m
Semi perimeter = ( Sum of sides ) / 2
⇒ s = ( a + b + c ) / 2
⇒ s = ( 25 + 17 + 26 ) / 2
⇒ s = ( 25 + 25 + 17 + 1 ) / 2
⇒ s = ( 50 + 18 ) / 2
⇒ s = 68 ÷ 2
⇒ s = 34 m
Now, by Heron's formula,
Area of triangle = √[ s ( s - a ) ( s - b ) ( s - c ) ]
⇒ A ( △BPC ) = √[ 34 ( 34 - 25 ) ( 34 - 17 ) ( 34 - 26 ) ]
⇒ A ( △BPC ) = √( 34 * 9 * 17 * 8 )
⇒ A ( △BPC ) = √( 17 * 2 * 9 * 17 * 8 )
⇒ A ( △BPC ) = √( 17 * 17 * 3 * 3 * 16 )
⇒ A ( △BPC ) = √( 17 * 17 * 3 * 3 * 4 * 4 )
⇒ A ( △BPC ) = 17 * 3 * 4
⇒ A ( △BPC ) = 204 m²
Also,
Area of triangle = ( Base * Height ) / 2
⇒ A ( △BPC ) = ( PC * BQ ) / 2
⇒ 204 = ( 17 * BQ ) / 2
⇒ 17 * BQ = 204 * 2
⇒ BQ = ( 204 * 2 ) / 17
⇒ BQ = 204 ÷ 17 * 2
⇒ BQ = 12 * 2
⇒ BQ = 24 m
Now, we know that,
Area of trapezium = ( Sum of parallel sides * Height ) / 2
⇒ A ( □ABCD ) = [ ( AB + CD ) * BQ ] / 2
⇒ A ( □ABCD ) = [ ( 60 + 77 ) * 24 ] / 2
⇒ A ( □ABCD ) = ( 60 + 77 ) * 24 ÷ 2
⇒ A ( □ABCD ) = ( 60 + 77 ) * 12
⇒ A ( □ABCD ) = 12 * 60 + 12 * 77
⇒ A ( □ABCD ) = 720 + 12 * ( 70 + 7 )
⇒ A ( □ABCD ) = 720 + 840 + 84
⇒ A ( □ABCD ) = 1560 + 84
⇒ A ( □ABCD ) = 1644
∴ Area of trapezium = 1644 m²
∴ The area of the trapezium is 1644 m².
Step-by-step explanation:
Given :
- Two prallel sides of a trapazium are 60m and 77m, and the nonparallel sides are 25m and 26m. find the area of the trapazium.
Solution :
Draw a line BC parallel to AD
Draw a perpendicular line BE on DF
ABCD is a parallelogram.
- BC=AD=25cm
- CD=AB=60cm
- CF=77-CD=17cm
For Δ BCF
perimeter of triangle=(25+26+17)/2
= 68/2=34
By Heron's Formula of triangle= √s(s - a)(s - b)(s - c)
= √34(34 - 25)(34 - 26)(34 - 17)
= 204cm²
Now area of ΔBCF = 1/2 × base × height
= 1/2 BExCF204cm²
= 1/2x BEx17BE
= 408/17
= 24cm
Area of Trapezium = 1/2 (AB + DF) × BE
= 1/2(60 + 77) × 24
= 1644cm²