Simple Geometry Question paper for Students. [New]

🏛️ Advanced Geometry Paper (Grade X) 📈

Challenge yourself with these problems covering key Grade X topics. Think critically, apply the theorems, and check your work. Best of luck! 🧠

Question 1: Similar Triangles 🔼

The areas of two similar triangles, \(\triangle ABC\) and \(\triangle PQR\), are 81 cm² and 144 cm² respectively. If the longest side of the smaller triangle, \(\triangle ABC\), is 27 cm, find the length of the longest side of the larger triangle, \(\triangle PQR\).

✅ Solution

We use the theorem which states that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$$ \frac{Area(\triangle ABC)}{Area(\triangle PQR)} = \left(\frac{\text{Corresponding Side of } \triangle ABC}{\text{Corresponding Side of } \triangle PQR}\right)^2 $$

Given information:

• Area(\(\triangle ABC\)) = 81 cm²
• Area(\(\triangle PQR\)) = 144 cm²
• Longest side of \(\triangle ABC\) = 27 cm

Let the longest side of \(\triangle PQR\) be \(x\).

$$ \frac{81}{144} = \left(\frac{27}{x}\right)^2 $$

Taking the square root of both sides:

$$ \sqrt{\frac{81}{144}} = \frac{27}{x} $$ $$ \frac{9}{12} = \frac{27}{x} $$

Now, we solve for \(x\):

$$ 9x = 27 \times 12 $$ $$ x = \frac{27 \times 12}{9} $$ $$ x = 3 \times 12 = 36 $$

Therefore, the length of the longest side of \(\triangle PQR\) is 36 cm. 💯

Question 2: Circles and Tangents 🔘

In the figure below, a circle with center O is inscribed in a quadrilateral ABCD such that it touches the sides AB, BC, CD, and DA at points P, Q, R, and S respectively. If AB = 29 cm, BC = 23 cm, CD = 20 cm, find the length of the side DA.

✅ Solution

The key property here is that the lengths of two tangents drawn from an external point to a circle are equal.

• From point A: \(AP = AS\)
• From point B: \(BP = BQ\)
• From point C: \(CQ = CR\)
• From point D: \(DR = DS\)

A property of a tangential quadrilateral (a quadrilateral with an inscribed circle) is that the sums of opposite sides are equal.

$$ AB + CD = BC + DA $$

We are given:

• AB = 29 cm
• BC = 23 cm
• CD = 20 cm

Substitute these values into the equation:

$$ 29 + 20 = 23 + DA $$ $$ 49 = 23 + DA $$ $$ DA = 49 - 23 $$ $$ DA = 26 \text{ cm} $$

So, the length of the side DA is 26 cm. 🎯

Question 3: Coordinate Geometry 🗺️

Find the ratio in which the y-axis divides the line segment joining the points A(5, -6) and B(-1, -4). Also, find the coordinates of the point of division.

✅ Solution

Let the ratio in which the y-axis divides the line segment be \(k:1\). Let the point of division on the y-axis be P.

Since the point P lies on the y-axis, its x-coordinate must be 0. So, \(P = (0, y)\).

Using the section formula for the x-coordinate:

$$ x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} $$

Here, \((x_1, y_1) = (5, -6)\), \((x_2, y_2) = (-1, -4)\), \(m_1 = k\), and \(m_2 = 1\).

$$ 0 = \frac{k(-1) + 1(5)}{k + 1} $$ $$ 0 = \frac{-k + 5}{k + 1} $$ $$ -k + 5 = 0 $$ $$ k = 5 $$

So, the ratio is \(k:1\), which is 5:1.

Now, let's find the y-coordinate of the point of division P using the section formula for y:

$$ y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} $$ $$ y = \frac{5(-4) + 1(-6)}{5 + 1} $$ $$ y = \frac{-20 - 6}{6} = \frac{-26}{6} = -\frac{13}{3} $$

Therefore, the point of division is (0, -13/3). 📍

Question 4: Surface Areas & Volumes ⚗️

A medicinal capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. (Use \(\pi = 22/7\))

✅ Solution

First, let's break down the dimensions of the capsule.

• Diameter = 5 mm, so Radius (r) = 2.5 mm
• Total length = 14 mm

The total length includes two hemispherical ends. The length of the cylindrical part is the total length minus the radii of the two hemispheres.

Height of cylinder (h) = Total length - 2 \(\times\) radius

$$ h = 14 - 2(2.5) = 14 - 5 = 9 \text{ mm} $$

The total surface area of the capsule is the sum of the curved surface area (CSA) of the cylinder and the CSA of the two hemispheres.

$$ \text{Total SA} = (\text{CSA of Cylinder}) + (2 \times \text{CSA of Hemisphere}) $$

Note: Two hemispherical CSAs make up the surface area of one sphere.

$$ \text{Total SA} = (2\pi rh) + (2 \times 2\pi r^2) = 2\pi rh + 4\pi r^2 $$ $$ \text{Total SA} = 2\pi r (h + 2r) $$

Let's substitute the values:

$$ \text{Total SA} = 2 \times \frac{22}{7} \times 2.5 \times (9 + 2 \times 2.5) $$ $$ \text{Total SA} = 2 \times \frac{22}{7} \times 2.5 \times (9 + 5) $$ $$ \text{Total SA} = 2 \times \frac{22}{7} \times 2.5 \times 14 $$

Simplify the calculation:

$$ \text{Total SA} = 2 \times 22 \times 2.5 \times \frac{14}{7} $$ $$ \text{Total SA} = 44 \times 2.5 \times 2 $$ $$ \text{Total SA} = 44 \times 5 = 220 \text{ mm}^2 $$

The surface area of the capsule is 220 mm². 💊

Excellent work! 🚀 Consistent practice is the highway to mastering geometry. 🌌