Quadratic Equations | Exercise 4.2

Question 5

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution
Understand the Question
  • Let the base of the right-angled triangle be x cmx\text{ cm}.
  • The altitude is 7 cm7\text{ cm} less than the base, so Altitude=(x−7) cm\text{Altitude} = (x - 7)\text{ cm}.
  • According to the Pythagoras theorem: Base2+Altitude2=Hypotenuse2\text{Base}^2 + \text{Altitude}^2 = \text{Hypotenuse}^2
  • Substitute the given values to form a quadratic equation, solve for xx, and reject any negative value since length must be positive.

Step 1 · Form the Quadratic Equation

Let the base of the right-angled triangle be x cmx\text{ cm}. Then, Altitude=(x−7) cm\text{Altitude} = (x - 7)\text{ cm} and Hypotenuse=13 cm\text{Hypotenuse} = 13\text{ cm}.Diagram 1

By Pythagoras theorem

x2+(x−7)2=132x2+(x2−14x+49)=1692x2−14x+49=1692x2−14x+49−169=02x2−14x−120=0\begin{aligned} x^2 + (x - 7)^2 &= 13^2 \\ x^2 + (x^2 - 14x + 49) &= 169 \\ 2x^2 - 14x + 49 &= 169 \\ 2x^2 - 14x + 49 - 169 &= 0 \\ 2x^2 - 14x - 120 &= 0 \end{aligned}

Dividing the entire equation by 22 x2−7x−60=0x^2 - 7x - 60 = 0

Step 2 · Solve for the Sides of the Triangle

Factorising the quadratic equation

x2−12x+5x−60=0x(x−12)+5(x−12)=0(x−12)(x+5)=0\begin{aligned} x^2 - 12x + 5x - 60 &= 0 \\ x(x - 12) + 5(x - 12) &= 0 \\ (x - 12)(x + 5) &= 0 \end{aligned}

  ⟹  x=12orx=−5\implies x = 12 \quad \text{or} \quad x = -5

Since side length cannot be negative, x≠−5x \ne -5.

Therefore, Base=x=12 cm\text{Base} = x = 12\text{ cm} Altitude=x−7=12−7=5 cm\text{Altitude} = x - 7 = 12 - 7 = 5\text{ cm}

Answer

5 cm5\text{ cm} and 12 cm12\text{ cm}

Common Mistakes
  • Accepting Negative Roots: Side lengths represent physical distances and cannot be negative, so x=−5x = -5 must be discarded.
  • Expansion Error: Incorrectly expanding (x−7)2(x - 7)^2 as x2−49x^2 - 49 or x2+49x^2 + 49 instead of (x2−14x+49)(x^2 - 14x + 49) using the identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2.

More questions in Exercise 4.2

Q1

Find the roots of the following quadratic equations by factorisation:

(i) x2−3x−10=0x^2 - 3x - 10 = 0

(ii) 2x2+x−6=02x^2 + x - 6 = 0

(iii) 2x2+7x+52=0\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0

(iv) 2x2−x+18=02x^2 - x + \dfrac{1}{8} = 0

(v) 100x2−20x+1=0100x^2 - 20x + 1 = 0

Q2

Solve the problems given in Example 1.

Q3

Find two numbers whose sum is 27 and product is 182.

Q4

Find two consecutive positive integers, sum of whose squares is 365.

Q5

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Q6

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹ 90, find the number of articles produced and the cost of each article.

← Back to Quadratic Equations