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The semi-perimeter of a right-angled triangle ABC is 12 cm, and the shortest median is 5 cm. What is the area of the triangle which has the largest median of triangle ABC as its longest side?
  • a)
    √(73) cm2
  • b)
    10 cm2
  • c)
    12 cm2
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The semi-perimeter of a right-angled triangle ABC is 12 cm, and the sh...
Given: S = 12 cm & BP = 5 cm
∴ AP = PC = BP = 5 cm
∴ AC = 10 cm
We know the ratio of Right triangle like (3:4:5)
then we assume the ratio, (6:8:10)
So, the other two sides = 8 cm, 6 cm
∴ Area of △QBC = ½ x QB x BC = 1/2 x 4 x 6 = 12cm2
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Most Upvoted Answer
The semi-perimeter of a right-angled triangle ABC is 12 cm, and the sh...
Let the sides of the right-angled triangle be a, b, and c, with c being the hypotenuse.
We know that a^2 + b^2 = c^2 (Pythagorean theorem).
The semi-perimeter is given by (a+b+c)/2 = 12 cm.
The shortest median is given by (1/2)*sqrt(2*(a^2 + b^2) - c^2) = 5 cm.
From the first equation, we can express c in terms of a and b as c = sqrt(a^2 + b^2).
Substituting this into the second equation, we get (1/2)*sqrt(2*(a^2 + b^2) - (a^2 + b^2)) = 5 cm.
Simplifying, we get sqrt(a^2 + b^2) = 10 cm.
Squaring both sides, we get a^2 + b^2 = 100.
Using the semi-perimeter equation, we can express c in terms of a and b as c = 24 - a - b.
To find the largest median as the longest side, we want to maximize c.
Since a^2 + b^2 = 100, we can rewrite c as c = 24 - sqrt(100) = 14 cm.
The area of the triangle is (1/2)*a*b = (1/2)*(10)*(10) = 50 cm^2.
Therefore, the area of the triangle with the largest median as its longest side is 50 cm^2. Answer: a) 50.
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The semi-perimeter of a right-angled triangle ABC is 12 cm, and the shortest median is 5 cm. What is the area of the triangle which has the largest median of triangle ABC as its longest side?a)√(73) cm2b)10 cm2c)12 cm2d)None of theseCorrect answer is option 'C'. Can you explain this answer?
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