Question

The dimensions of the squares ABMD, CDEF and EGHI are indicated in the given figure. Find the area of trapezium DJKE.​

Answer 1

mark me in brainliest answer

Answer 2

Solution :-

In ∆AJD and ∆AHI we have,

→ ∠ADJ = ∠AIJ { Each 90° because of square }

→ ∠JAD = ∠HAI { common }

So,

→ ∆AJD ~ ∆AHI { By AA similarity }

then,

→ JD/AD = HI/AI {when ∆'s are similar their corresponding sides are in same ratio. }

→ JD/2 = 5/10 { AI = AD + DE + EI = 2 + 3 + 5 = 10 units }

→ JD/2 = 1/2

→ JD = 2/2

→ JD = 1 unit .

similarly,

→ ∆AKE ~ ∆AHI { By AA similarity }

then,

→ KE/AE = HI/AI

→ KE/5 = 1/2

→ KE = (5/2) unit .

therefore,

→ Area trapezium (DJKE) = (1/2) * (sum of parallel sides) * Height

→ Area trapezium (DJKE) = (1/2) * (JD + KE) * DE

→ Area trapezium (DJKE) = (1/2) * (1 + 5/2) * 3

→ Area trapezium (DJKE) = (1/2) * (7/2) * 3

→ Area trapezium (DJKE) = (21/4)

→ Area trapezium (DJKE) = 5.25 unit² ≈ 5.2 unit² (B) (Ans.)

Learn more :-

Diagonals AC and BD of quadrilateral ABCD meet at E. IF AE = 2 cm, BE = 5 cm, CE = 10 cm, 8 सेमी.

https://brainly.in/question/27312677

Let abcd be a rectangle. the perpendicular bisector of line segment bd intersect ab and bc in point e and f respectively...

https://brainly.in/question/27813873