The altitude of a right-angled triangle is 7 cm less than its base. If its hypotenuse is 17 cm long, then (a) represent the above information in the form of a quadratic equation; (b) find the length of the sides of the triangle.

(a) x² - 7x - 120 = 0; (b) sides 15 cm, 8 cm and 17 cm

MathsCBSE Class 10 · Basic · Board exam 2024, set 430/3/1 · 3 marks · Short answer

CBSE Class 10 Mathematics Basic 2024: question 27

Set in the CBSE Class 10 board exam of 2024, paper 430/3/1, section C, question 27, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q27 · 3 marks · Short answerQuadratic Equations

The altitude of a right-angled triangle is $7$ cm less than its base. If its hypotenuse is $17$ cm long, then

(a)
represent the above information in the form of a quadratic equation;
1 mark
(b)
find the length of the sides of the triangle.
2 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 430/3/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Let the base be $x$ ; write the altitude in terms of $x$ .

Hint 2

Use (hypotenuse) $^{2}$ = (base) $^{2}$ + (altitude) $^{2}$ and simplify.

Hint 3

Split $- 7 x$ into two terms whose product is $- 120 x^{2}$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(a) Let the base be $x$ cm; then the altitude is $(x - 7)$ cm. By Pythagoras theorem: $x^{2} + (x - 7)^{2} = 1 7^{2}$
½ mark
(a) $x^{2} + x^{2} - 14 x + 49 = 289 \Rightarrow 2 x^{2} - 14 x - 240 = 0 \Rightarrow x^{2} - 7 x - 120 = 0$ , the required quadratic equation.
½ mark
(b) $x^{2} - 15 x + 8 x - 120 = 0 \Rightarrow (x - 15) (x + 8) = 0 \Rightarrow x = 15$ or $x = - 8$
½ mark
(b) Length cannot be negative, so $x \neq = - 8$ ; hence $x = 15$ .
½ mark
(b) Base = 15 cm, altitude = $15 - 7 = 8$ cm, hypotenuse = 17 cm.
1 mark

Answer: (a) $x^{2} - 7 x - 120 = 0$ ; (b) sides 15 cm, 8 cm and 17 cm

Where the marks are

Takes base $x$ and altitude $x - 7$ and applies Pythagoras theorem (½ mark)
Obtains $x^{2} - 7 x - 120 = 0$ (½ mark)
Factorises as $(x - 15) (x + 8) = 0$ (½ mark)
Rejects negative root, $x = 15$ (½ mark)
States the sides 15 cm and 8 cm (with hypotenuse 17 cm) (1 mark)

Revise the chapter

Chapter: Quadratic Equations (notes, formulas, where marks are lost and every past question)
Topic: Standard Form of a Quadratic Equation

Practise this chapterBoard-style questions from Quadratic Equations, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 430/3/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2024: question 27

Hints

Step-by-step solution

Revise the chapter

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