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Question

The sides (in cm) of a right triangle are $(x – 13)$, $(x – 26)$ and $x$. Its area (in cm²) is:

The correct answer is
1014

Understanding the Right Triangle Properties

The problem describes a right triangle with side lengths given in terms of '$x$': $(x – 13)$ cm, $(x – 26)$ cm, and $x$ cm. In a right triangle, the longest side is the hypotenuse, and the other two sides are the legs (base and height). Since $x$ is the largest value among the three expressions (assuming $x > 26$), $x$ represents the hypotenuse. The other two sides, $(x – 13)$ and $(x – 26)$, are the legs of the right triangle.

Applying the Pythagorean Theorem

The Pythagorean theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (legs). We can write this relationship as:

$$ (\text{leg}_1)^2 + (\text{leg}_2)^2 = (\text{hypotenuse})^2 $$

Substituting the given side lengths:

$$ (x – 13)^2 + (x – 26)^2 = x^2 $$

Solving the Equation for 'x'

Now, we need to solve this equation for $x$. First, expand the squared terms:

  • $ (x – 13)^2 = x^2 - 2(x)(13) + 13^2 = x^2 - 26x + 169 $
  • $ (x – 26)^2 = x^2 - 2(x)(26) + 26^2 = x^2 - 52x + 676 $

Substitute these expanded forms back into the equation:

$$ (x^2 - 26x + 169) + (x^2 - 52x + 676) = x^2 $$

Combine like terms on the left side:

$$ 2x^2 - 78x + 845 = x^2 $$

Rearrange the equation to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$$ 2x^2 - x^2 - 78x + 845 = 0 $$

$$ x^2 - 78x + 845 = 0 $$

We can solve this quadratic equation by factoring. We need two numbers that multiply to 845 and add up to -78. These numbers are -13 and -65.

So, the equation factors as:

$$ (x – 13)(x – 65) = 0 $$

This gives two possible solutions for $x$:

  • $ x - 13 = 0 \implies x = 13 $
  • $ x - 65 = 0 \implies x = 65 $

Validating the Value of 'x'

We must check if these values of $x$ result in valid side lengths for a triangle (all sides must be positive).

  • If $x = 13$:
    • Side 1: $x - 13 = 13 - 13 = 0$ cm. A side length cannot be zero.
    • Side 2: $x - 26 = 13 - 26 = -13$ cm. A side length cannot be negative.
    Therefore, $x = 13$ is not a valid solution.
  • If $x = 65$:
    • Side 1: $x - 13 = 65 - 13 = 52$ cm.
    • Side 2: $x - 26 = 65 - 26 = 39$ cm.
    • Side 3 (Hypotenuse): $x = 65$ cm.
    All side lengths are positive, so $x = 65$ is the correct value. Let's verify the Pythagorean theorem: $39^2 + 52^2 = 1521 + 2704 = 4225$. And $65^2 = 4225$. The sides form a valid right triangle.

Calculating the Area

The area of a right triangle is calculated using the formula:

$$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$

Using the lengths of the two legs ($39$ cm and $52$ cm):

$$ \text{Area} = \frac{1}{2} \times 39 \, \text{cm} \times 52 \, \text{cm} $$

$$ \text{Area} = 39 \, \text{cm} \times \frac{52}{2} \, \text{cm} $$

$$ \text{Area} = 39 \, \text{cm} \times 26 \, \text{cm} $$

$$ \text{Area} = 1014 \, \text{cm}^2 $$

The area of the right triangle is 1014 cm².

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Important Questions from Triangles

  1. What is the foot of the altitude from the vertex A of the triangle ABC?

  2. In ΔABC, D is a point on BC such that ∠ADB = 2∠DAC, ∠BAC = 70° and ∠B = 56°. What is the measure of ∠ADC?

  3. In ΔABC, ∠A = 66° and ∠B = 50 °. If the bisectors of ∠B and ∠C meet at P, then ∠BPC – ∠PCA = ?

  4. In a triangle ABC, points P and Q are on AB and AC, respectively, such that AP = 4 cm, PB = 6 cm, AQ = 5 cm and QC = 7.5 cm. If PQ = 6 cm, then find BC (in cm).

  5. The perimeters of two similar ΔABC and  Δ PQR are 48.4 cm and 12.1 cm, respectively. What is the ratio of the areas of  Δ ABC and  Δ PQR?

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