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The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, the other two sides of the triangle are equal to:
- a)Base = 10cm and Altitude = 5cm
- b)Base = 12cm and Altitude= 5cm
- c)Base = 14cm and Altitude = 10cm
- d)Base = 12cm and Altitude = 10cm
Correct answer is option 'B'. Can you explain this answer?
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The altitude of a right triangle is 7 cm less than its base. If the h...
Let the base be x cm.
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Altitude = (x – 7) cm
In a right triangle,
Base2 + Altitude2 = Hypotenuse2 (From Pythagoras theorem)
∴ x2 + (x – 7)2 = 132
By solving the above equation, we get;
⇒ x = 12 or x = – 5
Since the side of the triangle cannot be negative.
Therefore, base = 12cm and altitude = 12 - 7 = 5cm
Most Upvoted Answer
The altitude of a right triangle is 7 cm less than its base. If the h...
Given:
- Altitude of a right triangle is 7 cm less than its base.
- Hypotenuse is 13 cm.
To find:
- The other two sides of the triangle.
Solution:
Let the base of the triangle be 'x' cm.
Then, the altitude of the triangle will be 'x - 7' cm.
Using the Pythagorean theorem, we know that in a right triangle, the sum of the squares of the two legs (base and altitude) is equal to the square of the hypotenuse.
Therefore, we can write the equation as follows:
x^2 + (x - 7)^2 = 13^2
Simplifying the equation:
Expanding the equation:
x^2 + (x - 7)(x - 7) = 169
Simplifying:
x^2 + (x^2 - 14x + 49) = 169
Combining like terms:
2x^2 - 14x + 49 = 169
Subtracting 169 from both sides:
2x^2 - 14x + 49 - 169 = 0
2x^2 - 14x - 120 = 0
Dividing both sides by 2:
x^2 - 7x - 60 = 0
Factoring the quadratic equation:
(x - 12)(x + 5) = 0
Setting each factor equal to zero:
x - 12 = 0 or x + 5 = 0
Solving for x:
x = 12 or x = -5
Since the base of a triangle cannot be negative, we discard the solution x = -5.
Therefore, the base of the triangle is 12 cm.
Calculating the altitude:
The altitude is 7 cm less than the base, so it will be:
x - 7 = 12 - 7 = 5 cm
Therefore, the other two sides of the triangle are:
- Base = 12 cm
- Altitude = 5 cm
Hence, the correct answer is option 'B': Base = 12 cm and Altitude = 5 cm.
- Altitude of a right triangle is 7 cm less than its base.
- Hypotenuse is 13 cm.
To find:
- The other two sides of the triangle.
Solution:
Let the base of the triangle be 'x' cm.
Then, the altitude of the triangle will be 'x - 7' cm.
Using the Pythagorean theorem, we know that in a right triangle, the sum of the squares of the two legs (base and altitude) is equal to the square of the hypotenuse.
Therefore, we can write the equation as follows:
x^2 + (x - 7)^2 = 13^2
Simplifying the equation:
Expanding the equation:
x^2 + (x - 7)(x - 7) = 169
Simplifying:
x^2 + (x^2 - 14x + 49) = 169
Combining like terms:
2x^2 - 14x + 49 = 169
Subtracting 169 from both sides:
2x^2 - 14x + 49 - 169 = 0
2x^2 - 14x - 120 = 0
Dividing both sides by 2:
x^2 - 7x - 60 = 0
Factoring the quadratic equation:
(x - 12)(x + 5) = 0
Setting each factor equal to zero:
x - 12 = 0 or x + 5 = 0
Solving for x:
x = 12 or x = -5
Since the base of a triangle cannot be negative, we discard the solution x = -5.
Therefore, the base of the triangle is 12 cm.
Calculating the altitude:
The altitude is 7 cm less than the base, so it will be:
x - 7 = 12 - 7 = 5 cm
Therefore, the other two sides of the triangle are:
- Base = 12 cm
- Altitude = 5 cm
Hence, the correct answer is option 'B': Base = 12 cm and Altitude = 5 cm.
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The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, the other two sides of the triangle are equal to:a)Base = 10cm and Altitude = 5cmb)Base = 12cm and Altitude= 5cmc)Base = 14cm and Altitude = 10cmd)Base = 12cm and Altitude = 10cmCorrect answer is option 'B'. Can you explain this answer?
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Solutions for The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, the other two sides of the triangle are equal to:a)Base = 10cm and Altitude = 5cmb)Base = 12cm and Altitude= 5cmc)Base = 14cm and Altitude = 10cmd)Base = 12cm and Altitude = 10cmCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 10.
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