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14 Apr 2020, 01:48
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71% (01:49) correct
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Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then what is the maximum possible length of AP, in cm?
A. 5
B. 5√2
C. 6√2
D. 10
E. 8√2
Are You Up For the Challenge: 700 Level Questions
Let ABC be a right-angled triangle with hypotenuse BC of length 20 cm. If AP is perpendicular on BC, then what is the maximum possible length of AP, in cm?
A. 5
B. 5√2
C. 6√2
D. 10
E. 8√2
For BC to be a hypotenuse, triangle ABC is right angled at A. For AP to be maximum triangle ABC has to be an isosceles triangle.
Thus,
Hypotenuse BC = √2*a where a = AB = AC
20 = √2 * a
a = 10√2
Area of triangle = \(\frac{1}{2}\)*AB * AC = \(\frac{1}{2}\) * AP * BC
10√2 * 10√2 = AP * 20
AP = 10
Answer D.
Note: For a better understanding you need to consider that the right angled triangle ABC is inside a circle where BC is diameter of the circle. So, in the half the circle with triangle ABC, the largest perpendicular distance from the circumference to side BC is from center of the circle(here P being the center of the circle).
A. 5
B. 5√2
C. 6√2
D. 10
E. 8√2
For BC to be a hypotenuse, triangle ABC is right angled at A. For AP to be maximum triangle ABC has to be an isosceles triangle.
Thus,
Hypotenuse BC = √2*a where a = AB = AC
20 = √2 * a
a = 10√2
Area of triangle = \(\frac{1}{2}\)*AB * AC = \(\frac{1}{2}\) * AP * BC
10√2 * 10√2 = AP * 20
AP = 10
Answer D.
Note: For a better understanding you need to consider that the right angled triangle ABC is inside a circle where BC is diameter of the circle. So, in the half the circle with triangle ABC, the largest perpendicular distance from the circumference to side BC is from center of the circle(here P being the center of the circle).
General Discussion
14 Apr 2020, 02:10
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AP Max is when we have an isosceles triangle
Thus we have 10/10 for the base and 10*sqrt(2) for the sides of the triangle
We use AB^2+BP^2=AP^2
(10*sqrt(2))^2=10^2+AP^2
AP=10
Answer D
Thus we have 10/10 for the base and 10*sqrt(2) for the sides of the triangle
We use AB^2+BP^2=AP^2
(10*sqrt(2))^2=10^2+AP^2
AP=10
Answer D
