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If we know that the sum of the right sides of a right triangle is 20cm, then the maximum area of the right triangle is
Let the length of one right angle be a cm, and then the length of the other right angle be (20-a) cm
The area of right triangle s = a × (20-a) △ 2 = 10a-0.5a = - 0.5 (a-20a) = - 0.5 (A-2 × 10 × a + 10-10) = - 0.5 (A-2 × 10 × a + 10) + 50 = - 0.5 (A-10) + 50
When a = 10 cm, the largest area of right triangle is s = - 0.5 (A-10) + 50 = - 0.5 (10-10) + 50 = 50 cm
A: the maximum area of this right triangle is 50 square centimeters
The difference between the two right sides of a right triangle is 1cm, and the area is 20cm square. Find the length x of the longer right side
1/2*x（x-1）=20
x^2-x-40=0
x=［1±√（1+4*40）］/2
=（1±√161）/2
x1=（1+√161）/2
X2 = (1 - √ 161) / 2
∴x=（1+√161）/2
It is known that the area of a right triangle is 10 pieces of 3cm square, and the length of one right side is 3 pieces of 2cm
(10 root 3 * 2) / 3 root 2 = 10 root 6 / 3
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