in the following figure, ABCD is a rectangle with sides 8cm and 4cm . find areas of a ∆ CDA
hey guys give me answer in step by step
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Answered by
1
Hey mates here is your answer.
Cda =4^2+8^2
CDA = 16+64
CDA=root of 80
Cda =4^2+8^2
CDA = 16+64
CDA=root of 80
crawat7016:
A(CDA)=A(abc)
ok
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Answered by
12
first find the line ac bu Pythagoras theorem
(H)^2= (B)^2 + (P)^2
h^2 = 8^2+ 4^2
h^2 = 64 + 16
h^2 = 80
h = 9
AC = 9
now, let find the area of cda
as it is a rectangle so opposite sides are equal
AD = BC
AB = CD
Thus ,
a of triangle = (1÷2) × b × h
= (1÷2) × 4 ×8
= 1÷2 × 32
= 16
hence area of triangle is 16
hope you had got it and plz click on thanks
(H)^2= (B)^2 + (P)^2
h^2 = 8^2+ 4^2
h^2 = 64 + 16
h^2 = 80
h = 9
AC = 9
now, let find the area of cda
as it is a rectangle so opposite sides are equal
AD = BC
AB = CD
Thus ,
a of triangle = (1÷2) × b × h
= (1÷2) × 4 ×8
= 1÷2 × 32
= 16
hence area of triangle is 16
hope you had got it and plz click on thanks
Your answer is correct but can you explain how root of 80 is 9 ?
as root of 80 is 8.9 so if we will round it of it will be 9
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