Question

The difference of the parallel sides of a trapezoidal field is 20 m. Its area is 450 m' and its altitude is 15 m. Find the length of the parallel sides ​

Answer 1

Let ABCD be a trapezium with parallel sides AB and CD

Let ABCD be a trapezium with parallel sides AB and CDLet AB=20 (Given) and CD=x

Given) and CD=xA(ABCD)=480 m

Given) and CD=xA(ABCD)=480 m 2

Given) and CD=xA(ABCD)=480 m 2 ,height(h)=15 m (Given)

Given) and CD=xA(ABCD)=480 m 2 ,height(h)=15 m (Given)We know that, A(ABCD)=

Given) and CD=xA(ABCD)=480 m 2 ,height(h)=15 m (Given)We know that, A(ABCD)= 2

Given) and CD=xA(ABCD)=480 m 2 ,height(h)=15 m (Given)We know that, A(ABCD)= 2(AB+CCD

×h

×h⟹480=

×h⟹480= 2

×h⟹480= 220+x

×h⟹480= 220+x

×h⟹480= 220+x ×15

×h⟹480= 220+x ×15⟹

×h⟹480= 220+x ×15⟹ 15

×h⟹480= 220+x ×15⟹ 152×480

×h⟹480= 220+x ×15⟹ 152×480

×h⟹480= 220+x ×15⟹ 152×480 =20+x

×h⟹480= 220+x ×15⟹ 152×480 =20+x⟹64=20+x

×h⟹480= 220+x ×15⟹ 152×480 =20+x⟹64=20+x⟹x=44

×h⟹480= 220+x ×15⟹ 152×480 =20+x⟹64=20+x⟹x=44Hence, length of other parallel side CD is 44 m.

Answer 2

Answer:

Let one side be L1 and another side L2. It is given

L1-L2=20.

L1=20+L2. equation 1

lt is given

(1/2)(L1+L2)×h=450

(L1+L2)×15=450×2

L1+L2=450×2/15=60

Put the value of L1 in this equation, you will get

20+L2+L2=60

2L2=40

L2=40/2=20

L1=20+20=40