Geometry Practice Paper for Students 9th Grade [New]

📐 Geometry Challenge (Grade IX) 📚

Strengthen your geometry fundamentals with these problems. Take your time, draw diagrams, and think logically before checking the solution. Let's begin! 🚀

Question 1: Lines and Angles 📏

In the figure below, line \(l\) is parallel to line \(m\) (\(l \parallel m\)), and line \(t\) is a transversal. Find the value of \(x\).

✅ Solution

The angles given, \((3x - 10)°\) and \((2x + 15)°\), are consecutive interior angles.

A property of parallel lines states that consecutive interior angles are supplementary, meaning their sum is 180°.

$$ (3x - 10) + (2x + 15) = 180 $$

Combine like terms:

$$ 5x + 5 = 180 $$

Subtract 5 from both sides:

$$ 5x = 175 $$

Divide by 5:

$$ x = \frac{175}{5} $$ $$ x = 35 $$

Therefore, the value of \(x\) is 35. 💯

Question 2: Heron's Formula 🌳

A triangular park has sides of length 40 m, 24 m, and 32 m. Find the area of the park using Heron's formula.

✅ Solution

Heron's formula is used to find the area of a triangle when the lengths of all three sides are known.

First, we find the semi-perimeter (\(s\)) of the triangle.

$$ s = \frac{a + b + c}{2} $$

Given sides: \(a = 40\) m, \(b = 24\) m, \(c = 32\) m.

$$ s = \frac{40 + 24 + 32}{2} = \frac{96}{2} = 48 \text{ m} $$

Now, we apply Heron's formula:

$$ Area = \sqrt{s(s-a)(s-b)(s-c)} $$

Substitute the values:

$$ Area = \sqrt{48(48-40)(48-24)(48-32)} $$ $$ Area = \sqrt{48 \times 8 \times 24 \times 16} $$

To simplify, let's break down the numbers into factors:

$$ Area = \sqrt{(16 \times 3) \times 8 \times (3 \times 8) \times 16} $$ $$ Area = \sqrt{16^2 \times 8^2 \times 3^2} $$

Now, take the square root:

$$ Area = 16 \times 8 \times 3 = 384 $$

The area of the park is 384 square meters. 🌿

Question 3: Quadrilaterals 🔷

The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find the measure of all four angles.

✅ Solution

The sum of the interior angles of any quadrilateral is always 360°.

Let the angles be \(3x\), \(5x\), \(9x\), and \(13x\) based on the given ratio.

Set up the equation:

$$ 3x + 5x + 9x + 13x = 360 $$

Sum the terms with \(x\):

$$ 30x = 360 $$

Solve for \(x\):

$$ x = \frac{360}{30} = 12 $$

Now, find each angle by substituting \(x = 12\):

• First angle = \(3x = 3 \times 12 = \textbf{36°}\)
• Second angle = \(5x = 5 \times 12 = \textbf{60°}\)
• Third angle = \(9x = 9 \times 12 = \textbf{108°}\)
• Fourth angle = \(13x = 13 \times 12 = \textbf{156°}\)

Check: \(36 + 60 + 108 + 156 = 360°\). The solution is correct. ✅

Question 4: Circles and Chords ⚪

A chord of length 16 cm is drawn in a circle of radius 10 cm. Find the distance of the chord from the center of the circle.

✅ Solution

Let the circle have center O, the chord be AB, and M be the midpoint of AB. The line segment OM is the perpendicular distance from the center to the chord.

The perpendicular from the center of a circle to a chord bisects the chord. So:

$$ AM = MB = \frac{1}{2} \times AB = \frac{1}{2} \times 16 = 8 \text{ cm} $$

Now, consider the right-angled triangle \(\triangle OMA\). The hypotenuse is the radius OA.

• Radius (OA) = 10 cm
• Half-chord length (AM) = 8 cm
• Distance from center (OM) = ?

Using the Pythagorean theorem (\(a^2 + b^2 = c^2\)):

$$ OM^2 + AM^2 = OA^2 $$ $$ OM^2 + 8^2 = 10^2 $$ $$ OM^2 + 64 = 100 $$ $$ OM^2 = 100 - 64 = 36 $$ $$ OM = \sqrt{36} = 6 \text{ cm} $$

The distance of the chord from the center is 6 cm. 🎯

Great job completing the challenge! Keep practicing to build a strong foundation in geometry. 🌟