find the area of the rectangle whose length is 40cm and a diagonal is 41 cm.
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Answered by
5
In Δ ACD
AB = CD = 40 CM
AD=41 CM
∠ACD = 90° ( RECTANGLE HAS 90° ON ALL THE BOTH SIDES)
By Pythagoras theorem
AD² = AC² + CD²
AD² - CD² = AC²
(41)² - (40)² = AC²
1681 - 1600 = 81 = AC²
√81 = AC
AC = 9 CM
Perimeter of Rectangle ABCD = 2 (l+b)
= 2(40 + 9) = 2(49) = 98 CM
Hope It Helps U.......
Itzharsh★
Answered by
2
Answer:
Step-by-step explanation:
Diagonal square = L square + B square
41 square = 40 Square + B Square
1681 = 1600 + Bsquare
81 = B Square
B = 9
area = LB = 40 x 9 = 360
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