It's from R.S Aggarwal class 9 Plz solve it...
Answers
Mark me as brainliest if this helps please
From the point C construct CE || DA
We know that ADCE is a parallelogram having AE || DC and AD || EC with AD = 13m and D = 11m
It can be written as
AE = DC = 11m and EC = AD = 13m
So we get
BE = AB - AE
By substituting the values
BE = 25 –-11 = 14m
Consider Δ BCE
We know that BC = 15m, CE = 13m and BE = 14m
Take a = 15m, b = 13m and c = 14m
So we get
s=2a+b+c
By substituting the values
s=215+13+14
So we get
s=21m
We know that
Area=s(s−a)(s−b)(s−c)
By substituting the values
Area=21(21−15)(21−13)(21−14)
So we get
Area=21×6×8×7
It can be written as
Area=7×3×2×3×4×2×7
We get
Area=7×3×2×2
So we get
Area=84cm2
We know that
Area of △ BCE = 21×BE×CL
By substituting the values
84=21×14×CL
On further calculation
84=7×CL
By division
CL = 12m
We know that
Area of trapezium ABCD =21× sum of parallel sides × height
It can be written as
Area of trapezium ABCD = 21×(AB+CD)×CL
By substituting the values
Area of trapezium ABCD = 21×(11+25)×12
On further calculation
Area of trapezium ABCD = 36×6
By multiplication
Area of trapezium ABCD = 216m2