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One side other than the hypotenuse of the right angle isosceles triangle is 6 cm. The length of the perpendicular on the hypotenuse from the opposite vertex is:
- a)6 cm
- b)6√2 cm
- c)4 cm
- d)3√2 cm
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
One side other than the hypotenuse of the right angle isosceles trian...
Let ∠B be right angle in △ABC
and BD be perpendicular on hypotenuse AC then
1/p2 = 1/62 + 1/62
P2 = 62/2
⇒ p = 3√2
Most Upvoted Answer
One side other than the hypotenuse of the right angle isosceles trian...
Problem: One side other than the hypotenuse of the right angle isosceles triangle is 6 cm. The length of the perpendicular on the hypotenuse from the opposite vertex is:
Solution:
Given, the right angle isosceles triangle with one side other than the hypotenuse is 6 cm.
Let's assume that the length of the hypotenuse is 'x' cm.
Then, the length of the remaining side will also be 'x' cm.
Using Pythagoras theorem, we can write:
x^2 = 6^2 + x^2/4
=> 3x^2/4 = 36
=> x^2 = 48
=> x = 4√3
Now, let's draw the figure and label the perpendicular from the opposite vertex to the hypotenuse as 'h'.
We can see that the right-angled triangle formed by the perpendicular and the two halves of the hypotenuse is also an isosceles triangle.
Let's label the base of this triangle as 'b'.
Using Pythagoras theorem, we can write:
b^2 + h^2 = (x/2)^2
=> b^2 + h^2 = (4√3/2)^2
=> b^2 + h^2 = 12
But, we also know that b = x/2 = 2√3.
Substituting the value of b, we get:
(2√3)^2 + h^2 = 12
=> 12 + h^2 = 12
=> h^2 = 12 - 12
=> h^2 = 0
=> h = 0
Therefore, the length of the perpendicular on the hypotenuse from the opposite vertex is 0 cm.
Hence, the correct option is
a) 6 cm
b) 6√2 cm
c) 4 cm
d) 3√2 cm
Solution:
Given, the right angle isosceles triangle with one side other than the hypotenuse is 6 cm.
Let's assume that the length of the hypotenuse is 'x' cm.
Then, the length of the remaining side will also be 'x' cm.
Using Pythagoras theorem, we can write:
x^2 = 6^2 + x^2/4
=> 3x^2/4 = 36
=> x^2 = 48
=> x = 4√3
Now, let's draw the figure and label the perpendicular from the opposite vertex to the hypotenuse as 'h'.
We can see that the right-angled triangle formed by the perpendicular and the two halves of the hypotenuse is also an isosceles triangle.
Let's label the base of this triangle as 'b'.
Using Pythagoras theorem, we can write:
b^2 + h^2 = (x/2)^2
=> b^2 + h^2 = (4√3/2)^2
=> b^2 + h^2 = 12
But, we also know that b = x/2 = 2√3.
Substituting the value of b, we get:
(2√3)^2 + h^2 = 12
=> 12 + h^2 = 12
=> h^2 = 12 - 12
=> h^2 = 0
=> h = 0
Therefore, the length of the perpendicular on the hypotenuse from the opposite vertex is 0 cm.
Hence, the correct option is
d) 3√2 cm
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