First, the bigger square's area is 20*20= 400 cm^2
second, the circle's area is ¶r^2 = ¶(20/2)^2 = ¶(10)^2 = 100¶ cm^2
third, smaller square-- if we imagine it as 4 parts as 4 right triangles-- 1 part will be 1/2*(20/2)(20/2) = 1/2 (10)(10) = 50 cm^2 then the area of the smaller square is 4*50 = 200 cm^2
so the shaded parts = circle - smaller square =100¶ - 200 cm^2
the answer is 2
Answer:
The Proof is given below.
Step-by-step explanation:
Given:
LN⊥KM,
KL≅ML
To Prove:
ΔKLN≅ΔMLN
Proof:
In Δ KLN and Δ MLN
KL ≅ ML ....……….{Given i.e Hypotenuse }
LN ≅ LN …………..{Reflexive Property}
∠ LNK ≅ ∠ LNM ……….{ LN ⊥ KM i.e Measure of each angle is 90° given}
Δ KLN ≅ Δ MLN ….{By Hypotenuse Leg Theorem}
....Proved
Answer:
25
Step-by-step explanation:
Let n be the first integer.
Then the second integer will be (n + 1).
And the third will be (n + 2).
The sum is 78. Therefore:
Solve for n. Combine like terms:
So:
Therefore:
Therefore, the first integer is 25.
So our sequene is 25, 26, and 27.
The integer closest to zero will thus be 25.
