RBSE Solutions Class 9 Maths Chapter 10 Area of Triangles and Quadrilaterals Exercise 10.3

Chapter 10 Area of Triangles and Quadrilaterals Ex 10.3

Question 1. ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm², then ar (∆ABC) = 24 cm². Is this statement true or false. Give reason for your answer.
Answer: The given statement is False.
Given that ABCD is a parallelogram and X is the midpoint of AB.
We draw the diagonal AC. A diagonal of a parallelogram divides it into two triangles of equal areas.
So, the area of parallelogram ABCD is twice the area of triangle ADC.
\( \text{ar (parallelogram ABCD)} = 2 \times \text{ar (triangle ADC)} = 2 \times 24 \text{ cm}^2 = 48 \text{ cm}^2 \)
Now, CX is a median in triangle ABC. A median divides a triangle into two triangles of equal area.
So, the area of triangle BXC is half the area of triangle ABC.
\( \text{ar (triangle BXC)} = \frac{1}{2} \times \text{ar (triangle ABC)} = \frac{1}{2} \times 24 \text{ cm}^2 = 12 \text{ cm}^2 \)
The area of trapezoid AXCD can be found by subtracting the area of triangle BXC from the area of the parallelogram ABCD.
\( \text{ar (AXCD)} = \text{ar (parallelogram ABCD)} - \text{ar (triangle BXC)} = 48 \text{ cm}^2 - 12 \text{ cm}^2 = 36 \text{ cm}^2 \)
Since our calculated area of AXCD is 36 cm², and the question states it as 24 cm², the given statement is false. Understanding how medians and diagonals divide areas is key in geometry problems.
In simple words: The statement says two areas are equal, but when we calculate them using rules for parallelograms and medians, they turn out to be different. So, the original statement is wrong.

🎯 Exam Tip: Remember that a diagonal splits a parallelogram into two equal triangles, and a median splits a triangle into two equal areas. These are fundamental properties to use in area calculations.

Question 2. PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then ar (APAS) = 30 cm².
Answer: We aim to determine if the statement "ar (APAS) = 30 cm²" is true.
First, we find the length of side PQ of the rectangle using the Pythagorean theorem. In a rectangle inscribed in a quadrant, the diagonal of the rectangle is equal to the radius of the circle.
So, \( \text{PR} \), the diagonal of the rectangle, is \( 13 \text{ cm} \). Also, \( \text{PS} = 5 \text{ cm} \).
In the right-angled triangle PQR (where \( \angle \text{P} = 90^\circ \)):
\( \text{PQ}^2 + \text{PS}^2 = \text{PR}^2 \)
\( \text{PQ}^2 + 5^2 = 13^2 \)
\( \text{PQ}^2 + 25 = 169 \)
\( \text{PQ}^2 = 169 - 25 \)
\( \text{PQ}^2 = 144 \)
\( \text{PQ} = \sqrt{144} = 12 \text{ cm} \)
The area of triangle PQS (where A would be at Q) would be half the area of rectangle PQRS.
\( \text{Area of rectangle PQRS} = \text{PQ} \times \text{PS} = 12 \text{ cm} \times 5 \text{ cm} = 60 \text{ cm}^2 \)
\( \text{Area of triangle PQS} = \frac{1}{2} \times 60 \text{ cm}^2 = 30 \text{ cm}^2 \)
Now, A is any point on PQ. The area of triangle APS is calculated as \( \frac{1}{2} \times \text{base PS} \times \text{height AP} \).
For the area of triangle APS to be 30 cm², AP would need to be 12 cm (since PS is 5 cm). This condition only occurs if point A is exactly at point Q, making AP equal to PQ.
If A is anywhere else on PQ (other than Q), then AP will be less than 12 cm, and consequently, the area of triangle APS will be less than 30 cm².
Therefore, the statement "ar (APAS) = 30 cm²" is not always true; it is only true under a specific condition (when A coincides with Q). This problem shows how the area of a triangle changes based on its base and height, even when one side is fixed.
In simple words: We found the full side PQ is 12 cm. The area of triangle PQS (which is triangle APS if A is at Q) is 30 cm². But point A can be anywhere on PQ. So, unless A is exactly at Q, the area of triangle APS will be smaller than 30 cm². So, the statement that it's *always* 30 cm² is false.

🎯 Exam Tip: When a point can be anywhere on a segment, consider the extreme cases or how the area formula changes with the variable length. Don't assume the stated area holds for all positions.

Question 3. PQRS is a parallelogram, whose area is 180 cm² and A is any point on the diagonal QS. Then, the area of ∆ASR = 90 cm². This statement is true or false. Why?
Answer: The given statement is False.
Given: PQRS is a parallelogram with an area of 180 cm². QS is its diagonal.
A diagonal in a parallelogram divides it into two triangles of equal area. So, diagonal QS divides parallelogram PQRS into triangle PQS and triangle QRS, both having equal areas.
\( \text{Area of triangle QRS} = \frac{1}{2} \times \text{Area of parallelogram PQRS} = \frac{1}{2} \times 180 \text{ cm}^2 = 90 \text{ cm}^2 \)
Point A is located anywhere on the diagonal QS.
Consider triangle ASR. Its base is SR. The height of triangle ASR is the perpendicular distance from point A to the line containing SR.
If point A is exactly at point Q, then triangle ASR becomes triangle QSR, and its area is indeed 90 cm².
However, if A is any other point on QS (not Q), then the height of triangle ASR (with base SR) will be smaller than the height of triangle QSR (with base SR).
Therefore, the area of triangle ASR will be less than 90 cm² in most cases.
This means the statement "Area of triangle ASR = 90 cm²" is not always true; it is only true when A is at point Q. This illustrates that a point on a diagonal can create varying triangle areas unless it is a specific vertex.
In simple words: The diagonal QS splits the parallelogram into two triangles of 90 cm² each. Point A is on the diagonal. The area of triangle ASR will only be 90 cm² if point A is exactly at Q. If A is anywhere else on QS, the area of triangle ASR will be less than 90 cm². So, the statement is false because it's not always 90 cm².

🎯 Exam Tip: When a point's position on a line segment affects the area, remember to consider how the height of the triangle changes. The area formula depends directly on base and height.

Question 4. If ABC and BDE are two equilateral triangles such that D is the mid-point of BC, then ar (ABDE) = \( \frac{1}{4} \) ar (ΔABC). Is this statement true or false? Give reason for your answer.
Answer: The given statement is True.
Given: Triangle ABC. D is the midpoint of BC.
Since AD is a median of triangle ABC, it divides the triangle into two parts of equal area.
\( \text{ar (triangle ABD)} = \frac{1}{2} \text{ar (triangle ABC)} \) (Equation 1)
Now, consider triangle ABD. In the context of the solution, BE is referred to as a median of triangle ABD. This means E is the midpoint of AD.
Since BE is a median of triangle ABD, it also divides triangle ABD into two parts of equal area.
\( \text{ar (quadrilateral ABED)} = \frac{1}{2} \text{ar (triangle ABD)} \) (Equation 2)
Substitute Equation 1 into Equation 2:
\( \text{ar (quadrilateral ABED)} = \frac{1}{2} \times \left( \frac{1}{2} \text{ar (triangle ABC)} \right) \)
\( \text{ar (quadrilateral ABED)} = \frac{1}{4} \text{ar (triangle ABC)} \)
Therefore, the given statement is true. Medians are powerful tools in geometry for relating the areas of different parts of a triangle.
In simple words: Yes, the statement is true. A line from a corner to the middle of the opposite side (called a median) cuts a triangle into two equal halves. If you do this twice in a special way, the smaller area (quadrilateral ABED) becomes one-quarter of the big area (triangle ABC).

🎯 Exam Tip: Remember that a median divides a triangle into two triangles of equal area. This property is often used in area-related proofs, especially when dealing with nested areas.

Question 5. In figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD, then ar (ADPC) = \( \frac{1}{2} \) ar (EFGD). Is this statement true or false? Why?
Answer: The given statement is False.
Given: ABCD and EFGD are parallelograms. G is the mid-point of CD.
Consider triangle DPC and parallelogram ABCD. They share the same base DC and lie between the same parallel lines (DC and AB, assuming P is a point on AB).
Therefore, the area of triangle DPC is half the area of parallelogram ABCD.
\( \text{ar (triangle DPC)} = \frac{1}{2} \text{ar (parallelogram ABCD)} \) (Equation 1)
Next, consider parallelogram EFGD. Its base is DG. Since G is the midpoint of CD, DG is half the length of CD.
\( \text{DG} = \frac{1}{2} \text{CD} \)
Parallelogram EFGD and parallelogram ABCD share the same height (the perpendicular distance between the lines containing EF and DC).
Therefore, the area of parallelogram EFGD is half the area of parallelogram ABCD (because its base DG is half of CD).
\( \text{ar (parallelogram EFGD)} = \text{base DG} \times \text{height} = \frac{1}{2} \text{base CD} \times \text{height} = \frac{1}{2} \text{ar (parallelogram ABCD)} \) (Equation 2)
Comparing Equation 1 and Equation 2, we see that `\( \text{ar (triangle DPC)} = \text{ar (parallelogram EFGD)} \)` because both are half the area of parallelogram ABCD.
The original statement claims that `\( \text{ar (ADPC)} = \frac{1}{2} \text{ar (EFGD)} \)` (assuming ADPC refers to triangle DPC based on context).
Since `\( \text{ar (triangle DPC)} \)` is equal to `\( \text{ar (parallelogram EFGD)} \)`, the original statement is false. This problem highlights the relationship between the areas of triangles and parallelograms when they share bases and heights.
In simple words: The statement is false. We found that the area of triangle DPC is exactly the same as the area of parallelogram EFGD. But the question said the area of triangle ADPC is *half* the area of EFGD. Since they are equal, the statement is wrong.

🎯 Exam Tip: Remember the rule: the area of a triangle is half the area of a parallelogram if they share the same base and are between the same parallel lines. Also, scale the area of a parallelogram directly with its base length if height is constant.

Question 6. Prove that if L is the mid-point of BC, then ar (ABCD) = ar (APQD).
Answer: We aim to prove that the area of quadrilateral ABCD is equal to the area of quadrilateral APQD.
In the given figure, ABCD is a parallelogram, and L is the midpoint of BC. The line segment PQL passes through L, with P on the extension of line AB and Q on the extension of line DC.
Consider the two triangles, \( \triangle \text{ALPB} \) and \( \triangle \text{CLQ} \).
The angles \( \angle \text{PLB} \) and \( \angle \text{QLC} \) are vertically opposite angles, so they are equal.
Since AB is parallel to DC (property of a parallelogram) and PQL is a transversal, the alternate interior angles \( \angle \text{PBL} \) and \( \angle \text{QCL} \) are equal.
Also, since L is the midpoint of BC, \( \text{BL} = \text{LC} \).
Therefore, by ASA (Angle-Side-Angle) congruence rule, \( \triangle \text{PLB} \cong \triangle \text{QLC} \).
Congruent triangles have equal areas, so \( \text{ar (triangle PLB)} = \text{ar (triangle QLC)} \).
Now, let's consider the area of quadrilateral ABCD. It can be expressed as the sum of the area of triangle PLB and the area of the polygon ADCLP.
\( \text{ar (ABCD)} = \text{ar (triangle PLB)} + \text{ar (polygon ADCLP)} \) (Equation 1)
Similarly, the area of quadrilateral APQD can be expressed as the sum of the area of triangle QLC and the area of the polygon ADCLP.
\( \text{ar (APQD)} = \text{ar (triangle QLC)} + \text{ar (polygon ADCLP)} \) (Equation 2)
Since we have established that \( \text{ar (triangle PLB)} = \text{ar (triangle QLC)} \), we can substitute this into the equations.
Comparing Equation 1 and Equation 2, both `\( \text{ar (ABCD)} \)` and `\( \text{ar (APQD)} \)` are equal to the same sum, because the areas of the two triangles are equal and they share a common polygon area.
Therefore, `\( \text{ar (ABCD)} = \text{ar (APQD)} \)` (Hence proved). Area transformations like this can be used to compare complex shapes by breaking them into simpler, congruent parts.
In simple words: We showed that two small triangles (formed by the line PQL with the parallelogram's sides) are exactly the same size. Then, we saw that the original shape ABCD is like the first small triangle plus a common middle part. The new shape APQD is like the second small triangle plus the same common middle part. Since the two small triangles are equal, the big shapes ABCD and APQD must also have the same area.

🎯 Exam Tip: When proving area equality, look for congruent triangles or common regions that can simplify the comparison. Auxiliary lines and properties of parallel lines are often crucial.

Question 7. If the mid-points of the sides of a quadrilateral are joined in order, then prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral.
Answer: We are asked to prove a fundamental property related to quadrilaterals and midpoints.
Let ABCD be any quadrilateral. Let P, F, R, and S be the midpoints of sides AB, BC, CD, and DA respectively.
When these midpoints are joined in order, they form a parallelogram PFRS (or SFRP, depending on the order of joining).
According to the Mid-point Theorem applied to areas, the area of the parallelogram formed by joining the midpoints of the sides of any quadrilateral is half the area of the original quadrilateral.
Thus, \( \text{ar (parallelogram PFRS)} = \frac{1}{2} \text{ar (quadrilateral ABCD)} \).
This property is always true for any quadrilateral. This principle is a beautiful illustration of how midpoints create geometric relationships, leading to predictable area ratios.
In simple words: When you connect the middle points of all four sides of any shape with four straight sides, you always get a new shape inside that is a parallelogram. The space that this new parallelogram takes up is exactly half of the space that the original four-sided shape took up. This is a special rule in geometry.

🎯 Exam Tip: Remember the Mid-point Theorem for quadrilaterals: connecting midpoints forms a parallelogram whose area is half of the original quadrilateral. This is a key result often used in proofs and problem-solving.

Question 8. A man walks 10 m towards East and then he walks 30 m towards North. Find his distance from the starting point.
Answer: Let the starting point be O.
The man walks 10 m towards East to point A. So, the distance \( \text{OA} = 10 \text{ m} \).
Then he walks 30 m towards North from A to point B. So, the distance \( \text{AB} = 30 \text{ m} \).
Walking East and then North creates a path that forms two sides of a right-angled triangle, with the right angle at point A. The distance from the starting point to the final point (OB) is the hypotenuse of this triangle.
Using the Pythagorean theorem:
\( \text{OB}^2 = \text{OA}^2 + \text{AB}^2 \)
\( \text{OB}^2 = 10^2 + 30^2 \)
\( \text{OB}^2 = 100 + 900 \)
\( \text{OB}^2 = 1000 \)
\( \text{OB} = \sqrt{1000} \)
\( \text{OB} = \sqrt{100 \times 10} \)
\( \text{OB} = 10\sqrt{10} \text{ m} \)
So, the man's distance from the starting point is \( 10\sqrt{10} \text{ m} \). The Pythagorean theorem is very useful for finding distances when movement is in perpendicular directions.
In simple words: Imagine a map. The man walked East, then North, making a perfect square corner (right angle). His starting point, the East point, and the North point form a triangle. We use a special rule (Pythagoras theorem) to find the straight distance from where he started to where he finished. The answer is \( 10\sqrt{10} \text{ m} \).

🎯 Exam Tip: Always draw a simple diagram for navigation problems involving perpendicular directions; it makes applying the Pythagorean theorem much clearer and helps avoid errors.

Question 9. A ladder is placed in such a way that its foot is at a distance of 7 m from the wall. If its other end reaches a window, a height of 24 m, then find the length of the ladder.
Answer: We can visualize this scenario as a right-angled triangle.
Let the wall be represented by a vertical line, the ground by a horizontal line, and the ladder connecting the top of the wall to the ground. The right angle is formed where the wall meets the ground.
The distance from the foot of the ladder to the wall is the base of the triangle: \( \text{AB} = 7 \text{ m} \).
The height the ladder reaches on the wall (to the window) is the perpendicular height: \( \text{BC} = 24 \text{ m} \).
The length of the ladder is the hypotenuse of this right-angled triangle, let's call it AC.
Using the Pythagorean theorem (also known as Baudhayan theorem):
\( \text{AC}^2 = \text{AB}^2 + \text{BC}^2 \)
\( \text{AC}^2 = 7^2 + 24^2 \)
\( \text{AC}^2 = 49 + 576 \)
\( \text{AC}^2 = 625 \)
\( \text{AC} = \sqrt{625} \)
\( \text{AC} = 25 \text{ m} \)
So, the length of the ladder is 25 m. This is a classic example of applying the Pythagorean theorem to real-world scenarios involving heights and distances.
In simple words: The ladder, the wall, and the ground make a triangle with a square corner. The distance on the ground is 7 m, and the height on the wall is 24 m. We use a special rule (Pythagoras theorem) to find the length of the ladder, which is 25 m.

🎯 Exam Tip: Visualizing the problem with a simple sketch helps identify the right-angled triangle and its sides (hypotenuse, base, height). This is crucial for correctly applying the Pythagorean theorem.

Question 10. Two poles stand vertically on a level ground. Their heights are 7 m and 12 m respectively. If the distance between their feet is 12 m. Find the distance between their tops.
Answer: Let's imagine the two poles. One pole is 7 m tall, and the other is 12 m tall. The distance between their bases (feet) on the ground is 12 m.
To find the distance between their tops, we can construct a right-angled triangle.
Draw a horizontal line from the top of the shorter pole (7 m) to meet the taller pole (12 m). This horizontal line will be parallel to the ground.
The length of this horizontal line will be 12 m, which is the same as the distance between the poles' bases. This forms the base of our right-angled triangle.
The vertical side of this right-angled triangle will be the difference in the heights of the two poles: \( 12 \text{ m} - 7 \text{ m} = 5 \text{ m} \). This is the height of our right-angled triangle.
The distance between the tops of the poles is the hypotenuse of this new right-angled triangle.
Using the Pythagorean theorem:
\( (\text{Distance between tops})^2 = (\text{Base})^2 + (\text{Height})^2 \)
\( (\text{Distance between tops})^2 = 12^2 + 5^2 \)
\( (\text{Distance between tops})^2 = 144 + 25 \)
\( (\text{Distance between tops})^2 = 169 \)
\( \text{Distance between tops} = \sqrt{169} \)
\( \text{Distance between tops} = 13 \text{ m} \)
Therefore, the distance between the tops of the poles is 13 m. This problem demonstrates how to apply the Pythagorean theorem by creatively constructing a right-angled triangle from a given geometric setup.
In simple words: Picture two tall sticks (poles). One is 7m, the other is 12m. They are 12m apart at the bottom. To find the distance between their tops, imagine a horizontal line from the shorter pole's top to the taller pole. This makes a right-angle triangle. The bottom of this triangle is 12m. The height is the difference between the poles' heights (12-7=5m). Then, use Pythagoras to find the slanted distance (the hypotenuse), which is 13 m.

🎯 Exam Tip: When dealing with two vertical objects on level ground, always draw a horizontal line from the top of the shorter object to the taller one. This creates a right-angled triangle, simplifying the problem for the Pythagorean theorem.

Question 11. Find the length of the altitude and area of an equilateral triangle of side 'a'.
Answer: Let ABC be an equilateral triangle with each side equal to 'a'. So, \( \text{AB} = \text{BC} = \text{CA} = a \).
**1. Finding the length of the altitude (height):**
Draw an altitude AD from vertex A to side BC. In an equilateral triangle, the altitude also acts as a median, which means it bisects the base BC.
So, D is the midpoint of BC, and \( \text{BD} = \frac{1}{2} \text{BC} = \frac{a}{2} \).
Now, consider the right-angled triangle ADB (with the right angle at D).
Using the Pythagorean theorem:
\( \text{AB}^2 = \text{BD}^2 + \text{AD}^2 \)
\( a^2 = \left(\frac{a}{2}\right)^2 + \text{AD}^2 \)
\( a^2 = \frac{a^2}{4} + \text{AD}^2 \)
Now, we solve for AD (the altitude):
\( \text{AD}^2 = a^2 - \frac{a^2}{4} \)
\( \text{AD}^2 = \frac{4a^2 - a^2}{4} \)
\( \text{AD}^2 = \frac{3a^2}{4} \)
\( \text{AD} = \sqrt{\frac{3a^2}{4}} \)
\( \text{AD} = \frac{\sqrt{3}a}{2} \)
So, the length of the altitude of an equilateral triangle with side 'a' is \( \frac{\sqrt{3}}{2} a \) units.
**2. Finding the area:**
The formula for the area of any triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \).
For our equilateral triangle ABC, the base is BC = 'a', and the height (altitude) is AD = \( \frac{\sqrt{3}}{2} a \).
\( \text{Area} = \frac{1}{2} \times \text{BC} \times \text{AD} \)
\( \text{Area} = \frac{1}{2} \times a \times \frac{\sqrt{3}}{2} a \)
\( \text{Area} = \frac{\sqrt{3}}{4} a^2 \)
Therefore, the area of an equilateral triangle with side 'a' is \( \frac{\sqrt{3}}{4} a^2 \) square units. Knowing these standard formulas for equilateral triangles can save a lot of time in competitive exams.
In simple words: For an equilateral triangle (all sides equal, say 'a'): The altitude (height) from one corner to the middle of the opposite side is found using Pythagoras, and it is \( \frac{\sqrt{3}}{2} \) times the side length. The area is then found by \( \frac{1}{2} \) times base times height, which works out to \( \frac{\sqrt{3}}{4} \) times the side length squared.

🎯 Exam Tip: Memorize the formulas for the altitude and area of an equilateral triangle (\( \frac{\sqrt{3}}{2} a \) and \( \frac{\sqrt{3}}{4} a^2 \) respectively) as they are frequently used and can be directly applied in many problems.

Question 12. Find the length of the diagonal of a square whose each side is of 4 m.
Answer: Let the side of the square be 'a'. We are given that \( a = 4 \text{ m} \).
A diagonal of a square divides it into two right-angled triangles. The two sides of the square form the perpendicular sides of this right-angled triangle, and the diagonal itself is the hypotenuse.
Using the Pythagorean theorem:
\( (\text{diagonal})^2 = (\text{side})^2 + (\text{side})^2 \)
\( (\text{diagonal})^2 = a^2 + a^2 \)
\( (\text{diagonal})^2 = 2a^2 \)
To find the length of the diagonal, take the square root of both sides:
\( \text{diagonal} = \sqrt{2a^2} \)
\( \text{diagonal} = a\sqrt{2} \)
Now, substitute the given side length \( a = 4 \text{ m} \):
\( \text{diagonal} = 4\sqrt{2} \text{ m} \)
So, the length of the diagonal of the square is \( 4\sqrt{2} \text{ m} \). The relationship between a square's side and its diagonal is a direct application of the Pythagorean theorem, simplifying calculations for squares.
In simple words: For a square with sides of 4 m, the diagonal makes a right-angle triangle with two sides of the square. Using Pythagoras, the diagonal is \( \text{side} \times \sqrt{2} \). So, it's \( 4\sqrt{2} \text{ m} \).

🎯 Exam Tip: Remember the shortcut formula for a square's diagonal: \( d = a\sqrt{2} \), where 'a' is the side length. This formula saves time compared to re-deriving it with Pythagoras every time.

Question 13. In an equilateral triangle ABC, AD is perpendicular to BC, prove that \( 3\text{AB}^2 = 4\text{AD}^2 \).
Answer: Given: Triangle ABC is an equilateral triangle. AD is perpendicular to BC.
To prove: \( 3\text{AB}^2 = 4\text{AD}^2 \)
**Proof:**
In an equilateral triangle, the altitude (AD) drawn from a vertex to the opposite side also bisects that side. Therefore, D is the midpoint of BC.
This means \( \text{BD} = \frac{1}{2} \text{BC} \).
Since \( \triangle \text{ABC} \) is equilateral, all its sides are equal: \( \text{AB} = \text{BC} = \text{CA} \).
So, we can replace BC with AB in the expression for BD:
\( \text{BD} = \frac{1}{2} \text{AB} \)
Now, consider the right-angled triangle ADB (where the right angle is at D, because AD is perpendicular to BC).
Using the Pythagorean theorem:
\( \text{AB}^2 = \text{BD}^2 + \text{AD}^2 \)
Substitute the value of BD (\( \frac{1}{2} \text{AB} \)) into the equation:
\( \text{AB}^2 = \left(\frac{1}{2} \text{AB}\right)^2 + \text{AD}^2 \)
\( \text{AB}^2 = \frac{1}{4} \text{AB}^2 + \text{AD}^2 \)
To isolate \( \text{AD}^2 \), subtract \( \frac{1}{4} \text{AB}^2 \) from both sides:
\( \text{AB}^2 - \frac{1}{4} \text{AB}^2 = \text{AD}^2 \)
\( \frac{4\text{AB}^2 - \text{AB}^2}{4} = \text{AD}^2 \)
\( \frac{3}{4} \text{AB}^2 = \text{AD}^2 \)
Finally, multiply both sides by 4 to get the desired result:
\( 3\text{AB}^2 = 4\text{AD}^2 \)
Hence proved. This proof demonstrates a fundamental relationship between the side length and altitude of any equilateral triangle.
In simple words: In an equilateral triangle, if you draw a line straight down from a corner to the middle of the opposite side (this is called the altitude), you create a right-angle triangle. Using the Pythagoras rule for this triangle, we can show a special connection between the side length of the big triangle and the length of this altitude. We proved \( 3\text{AB}^2 = 4\text{AD}^2 \).

🎯 Exam Tip: For equilateral triangle proofs, remember that the altitude acts as a median and angle bisector. This property creates a right-angled triangle, which is crucial for applying the Pythagorean theorem to relate side lengths and altitudes.

Question 14. O is a point inside the rectangle ABCD. Prove that \( \text{OB}^2 + \text{OD}^2 = \text{OA}^2 + \text{OC}^2 \).
Answer: Given: ABCD is a rectangle. O is any point inside the rectangle.
To Prove: \( \text{OB}^2 + \text{OD}^2 = \text{OA}^2 + \text{OC}^2 \)
**Construction:** Draw a line segment LM passing through point O, parallel to sides AB and DC. Let this line meet AB at M and DC at L.
**Proof:**
Since ABCD is a rectangle, AB is parallel to DC, and AD is parallel to BC. By construction, LM is parallel to AB and DC.
This implies that the quadrilaterals AMLD and MBLC are also rectangles because their opposite sides are parallel and all angles are 90 degrees.
In rectangle AMLD, opposite sides are equal, so \( \text{AM} = \text{DL} \).
In rectangle MBLC, opposite sides are equal, so \( \text{MB} = \text{CL} \).
Now, consider the four right-angled triangles formed with O as a vertex: \( \triangle \text{OMB} \), \( \triangle \text{OLD} \), \( \triangle \text{OMA} \), and \( \triangle \text{OLC} \).
1. In right-angled \( \triangle \text{OMB} \) (with right angle at M):
By Pythagorean theorem: \( \text{OB}^2 = \text{OM}^2 + \text{MB}^2 \)
Substitute \( \text{MB} = \text{CL} \):
\( \text{OB}^2 = \text{OM}^2 + \text{CL}^2 \) (Equation 1)
2. In right-angled \( \triangle \text{OLD} \) (with right angle at L):
By Pythagorean theorem: \( \text{OD}^2 = \text{OL}^2 + \text{LD}^2 \)
Substitute \( \text{LD} = \text{AM} \):
\( \text{OD}^2 = \text{OL}^2 + \text{AM}^2 \) (Equation 2)
Add Equation 1 and Equation 2:
\( \text{OB}^2 + \text{OD}^2 = \text{OM}^2 + \text{CL}^2 + \text{OL}^2 + \text{AM}^2 \) (Equation 3)
3. In right-angled \( \triangle \text{OMA} \) (with right angle at M):
By Pythagorean theorem: \( \text{OA}^2 = \text{OM}^2 + \text{AM}^2 \) (Equation 4)
4. In right-angled \( \triangle \text{OLC} \) (with right angle at L):
By Pythagorean theorem: \( \text{OC}^2 = \text{OL}^2 + \text{CL}^2 \) (Equation 5)
Add Equation 4 and Equation 5:
\( \text{OA}^2 + \text{OC}^2 = \text{OM}^2 + \text{AM}^2 + \text{OL}^2 + \text{CL}^2 \) (Equation 6)
Comparing Equation 3 and Equation 6, we observe that their right-hand sides are identical.
Therefore, `\( \text{OB}^2 + \text{OD}^2 = \text{OA}^2 + \text{OC}^2 \)` (Hence proved). This elegant property shows that the sum of the squares of distances from an interior point to opposite vertices of a rectangle is constant.
In simple words: Imagine a point O inside a rectangle. If you draw lines from O to all four corners, this rule states that if you square the length of the line to one corner (like OB) and add it to the squared length of the line to the corner opposite it (like OD), this sum will be the same as if you did it for the other pair of opposite corners (OA squared plus OC squared). We prove this by drawing a helper line through O, which creates several right-angled triangles.

🎯 Exam Tip: For problems involving a point inside a rectangle or square, drawing auxiliary lines parallel to the sides often helps create right-angled triangles and apply the Pythagorean theorem effectively.

Question 15. In an obtuse angled triangle ABC, \( \angle \text{C} \) is an obtuse angle and AD is perpendicular to BC which meets BC produced at D. Then prove that \( \text{AB}^2 = \text{AC}^2 + \text{BC}^2 + 2\text{BC} \cdot \text{CD} \).
Answer: Given: Triangle ABC is an obtuse-angled triangle with \( \angle \text{C} \) being the obtuse angle. AD is perpendicular to BC, and D lies on BC produced.
To Prove: \( \text{AB}^2 = \text{AC}^2 + \text{BC}^2 + 2\text{BC} \cdot \text{CD} \)
**Proof:**
Consider the right-angled triangle ADB (with the right angle at D).
Using the Pythagorean theorem:
\( \text{AB}^2 = \text{AD}^2 + \text{DB}^2 \)
From the figure, the length of the segment DB is the sum of DC and CB:
\( \text{DB} = \text{DC} + \text{CB} \)
Substitute this expression for DB into the equation for \( \text{AB}^2 \):
\( \text{AB}^2 = \text{AD}^2 + (\text{DC} + \text{CB})^2 \)
Now, expand the term \( (\text{DC} + \text{CB})^2 \) using the algebraic identity \( (a+b)^2 = a^2 + b^2 + 2ab \):
\( \text{AB}^2 = \text{AD}^2 + \text{DC}^2 + \text{CB}^2 + 2\text{DC} \cdot \text{CB} \) (Equation 1)
Next, consider the right-angled triangle ADC (with the right angle at D).
Using the Pythagorean theorem:
\( \text{AC}^2 = \text{AD}^2 + \text{DC}^2 \) (Equation 2)
Now, substitute \( (\text{AD}^2 + \text{DC}^2) \) with \( \text{AC}^2 \) from Equation 2 into Equation 1:
\( \text{AB}^2 = \text{AC}^2 + \text{CB}^2 + 2\text{DC} \cdot \text{CB} \)
Finally, rearranging the terms and using BC for CB and CD for DC (as they represent the same lengths), we get:
\( \text{AB}^2 = \text{AC}^2 + \text{BC}^2 + 2\text{BC} \cdot \text{CD} \)
Hence proved. This theorem, often called a form of the Apollonius Theorem for obtuse triangles, relates the sides of a triangle when an altitude falls outside the base.
In simple words: When you have a triangle with one big angle (obtuse angle), and you draw a line straight down from the opposite corner to the side extended, you can find a special relationship. By using the Pythagoras rule in the two right-angle triangles formed, we can prove that the square of one side (\( \text{AB}^2 \)) equals the sum of the squares of the other two sides (\( \text{AC}^2 + \text{BC}^2 \)) plus twice the product of the base (\( \text{BC} \)) and the extended part (\( \text{CD} \)).

🎯 Exam Tip: For problems involving obtuse-angled triangles, remember that the altitude may fall outside the triangle, requiring the base to be extended. This changes how the Pythagorean theorem is applied by adding segments, often leading to results like this one.