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As shown in the figure, the middle angle ACB of triangle ABC is equal to 90 degrees, CA is equal to 15cm, CB is equal to 20cm, and the circle C with Ca as the radius intersects AB at d to find the length of AD
In RT Δ ABC, ∠ C = 90 ° and 〈 AB = √ (AC & # 178; + BC & # 178;) = 25,
If e is used as de ⊥ AC, then de ∥ BC,
∴ΔADE∽ΔABC,∴AE/DE=AC/BCD=3/4,
Let AE = 3x (x > 0), then de = 4x, CE = 15-3x,
In RT Δ CDE, CD & # 178; = CE & # 178; + de & # 178;,
225=(15-3X)²+16X²
25X²-90X=0
X = 18 / 5 or x = 0 (rounding off), AE = 3x = 54 / 5
∴AD/AB=AE/AC=(54/5)/15=18/25,
∴AD=18.
I hope I can help you
The side length of a right triangle is 15cm, 10cm, and the largest square is drawn. The area of the square is
It is easy to know that when the four vertices of a square fall on the three sides of a triangle, the square has the largest area
Let the side length of a square be x, then the two triangles above and below are similar
(15-x) / 15 = x / 10 gives x = 6
So the maximum square area is: 6 * 6 = 36cm ^ 2
I hope my answer is useful to you!
Draw the largest circle in a square with a side length of 10 cm, and calculate the area of the shadow part
10×10-3.14×10/2×10/2
=100-78.5
=21.5 square centimeters
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