GSEB Class 10 Maths Solutions Chapter 6 Triangle Exercise 6.5

Question 1. Sides of triangles are given below. Determine which of these are right triangle. Write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
Answer:
(i) Given sides: 7 cm, 24 cm, 25 cm.
Here, \( 7^2 = 49 \) cm
\( 24^2 = 576 \) cm
\( 25^2 = 625 \) cm
We observe that \( 49 + 576 = 625 \)
So, \( 7^2 + 24^2 = 25^2 \)
The given triangle is a right-angled triangle. Its hypotenuse is 25 cm.
In simple words: We check if the sum of squares of the two shorter sides equals the square of the longest side. If it does, it's a right triangle, and the longest side is the hypotenuse.

Exam Tip: Remember the Pythagorean theorem: \( a^2 + b^2 = c^2 \). The longest side is always the hypotenuse, \( c \).

Answer:
(ii) Given sides: 3 cm, 8 cm, 6 cm.
Here, \( 3^2 = 9 \) cm
\( 6^2 = 36 \) cm
\( 8^2 = 64 \) cm
We observe that \( 9 + 36 = 45 \), which is not equal to \( 64 \).
So, \( 3^2 + 6^2 \neq 8^2 \)
The given triangle is not a right-angled triangle.
In simple words: Since the squares of the two shorter sides don't add up to the square of the longest side, this triangle is not a right triangle.

Exam Tip: Always identify the longest side first, as it's the potential hypotenuse. The other two sides are the legs.

Answer:
(iii) Given sides: 50 cm, 80 cm, 100 cm.
Here, \( 50^2 = 2500 \) cm
\( 80^2 = 6400 \) cm
\( 100^2 = 10,000 \) cm
We observe that \( 2500 + 6400 = 8900 \), which is not equal to \( 10,000 \).
So, \( 50^2 + 80^2 \neq 100^2 \)
The given triangle is not a right-angled triangle.
In simple words: We check the numbers. If the two smaller squares don't sum up to the biggest square, then it's not a right-angled triangle.

Exam Tip: Squaring numbers correctly is crucial for these problems. Double-check your calculations.

Answer:
(iv) Given sides: 13 cm, 12 cm, 5 cm.
Here, \( 5^2 = 25 \) cm
\( 12^2 = 144 \) cm
\( 13^2 = 169 \) cm
We observe that \( 25 + 144 = 169 \)
So, \( 5^2 + 12^2 = 13^2 \)
The given triangle is a right-angled triangle. Its hypotenuse is 13 cm.
In simple words: Since the sum of the squares of the two shorter sides equals the square of the longest side, this is a right triangle, and the hypotenuse is 13 cm.

Exam Tip: Common Pythagorean triples (like 3-4-5, 5-12-13, 7-24-25, 8-15-17) can help you quickly identify right triangles.

Question 2. PQR is a triangle right angled at P and M is a point on QR such that \( PM \perp QR \). Show that \( PM^2 = QM \cdot MR \).
Answer:
In \( \triangle QMP \) and \( \triangle QPR \):
\( \angle QMP = \angle QPR \) (Both angles are \( 90^\circ \))
\( \angle PQM = \angle RQP \) (This angle is common to both triangles)
Therefore, \( \triangle QMP \sim \triangle QPR \) (by AA similarity criteria) .....(1)

Again, in \( \triangle PMR \) and \( \triangle QPR \):
\( \angle PMR = \angle QPR \) (Both angles are \( 90^\circ \))
\( \angle PRM = \angle QRP \) (This angle is common to both triangles)
Therefore, \( \triangle PMR \sim \triangle QPR \) (by AA similarity criteria) .....(2)

From (1) and (2), we can conclude that:
\( \triangle QMP \sim \triangle PMR \)

Since the triangles are similar, their corresponding sides are proportional:
\( \frac{QM}{PM} = \frac{PM}{RM} \)

Cross-multiplying gives:
\( PM^2 = QM \cdot RM \)
In simple words: When we drop a perpendicular from the right angle to the hypotenuse, it creates two smaller triangles. These smaller triangles are similar to the big triangle and also similar to each other. Because they are similar, their sides are proportional, which allows us to prove that the square of the perpendicular's length equals the product of the two segments it divides the hypotenuse into.

Exam Tip: This is a standard theorem regarding similar triangles in a right-angled triangle. Remember that dropping a perpendicular from the right angle vertex to the hypotenuse creates three similar triangles (the original, and the two smaller ones).

Question 3. In the figure ABD is a triangle, right angled at A and \( AC \perp BD \). Show that
(i) \( AB^2 = BC \cdot BD \)
(ii) \( AC^2 = BC \cdot DC \)
(iii) \( AD^2 = BD \cdot CD \) (CBSE 2010)
Answer:
(i) In \( \triangle ABC \) and \( \triangle BAD \):
\( \angle ACB = \angle BAD \) (Both angles are \( 90^\circ \))
\( \angle ABC = \angle ABD \) (This angle is common to both triangles)
Therefore, \( \triangle ABC \sim \triangle BAD \) (by AA similarity criteria)

Since the triangles are similar, their corresponding sides are proportional:
\( \frac{AB}{DB} = \frac{BC}{BA} \)
Cross-multiplying gives:
\( AB^2 = BC \cdot BD \)
In simple words: When we compare the small triangle ABC with the large triangle BAD, we find they are similar because they share a common angle and both have a 90-degree angle. This similarity means their sides are in proportion, leading to the formula \( AB^2 = BC \cdot BD \).

Exam Tip: To prove relationships involving squares of sides, think about similar triangles. Look for common angles and right angles to establish similarity.

Answer:
(ii) In \( \triangle ACB \) and \( \triangle DCA \):
We know \( \angle BAC + \angle CBA = 90^\circ \) (Sum of angles in right triangle \( \triangle ABC \)) .....(1)
Also, in \( \triangle ABD \), \( \angle BAD = 90^\circ \). Since \( AC \perp BD \), \( \angle BCA = 90^\circ \).
\( \angle BDA + \angle DBA = 90^\circ \)
From (1) and (2), we deduce \( \angle BAC = \angle BDA \).

Now, in \( \triangle ACB \) and \( \triangle DCA \):
\( \angle ACB = \angle DCA \) (Both angles are \( 90^\circ \))
\( \angle BAC = \angle BDA \) (Proved above)
Therefore, \( \triangle ACB \sim \triangle DCA \) (by AA similarity criteria)

Since the triangles are similar, their corresponding sides are proportional:
\( \frac{AC}{DC} = \frac{BC}{AC} \)
Cross-multiplying gives:
\( AC^2 = BC \cdot DC \)
In simple words: By showing that triangle ACB and triangle DCA are similar (again, using common angles and right angles), we can set up a proportion between their sides. This proportion then helps us prove the equation \( AC^2 = BC \cdot DC \).

Exam Tip: When dealing with multiple triangles within a larger one, try to identify all possible similar pairs. Sometimes, finding relationships between angles is the key to proving similarity.

Answer:
(iii) In \( \triangle ADB \) and \( \triangle CDA \):
\( \angle DAB = \angle DCA \) (Both angles are \( 90^\circ \))
\( \angle ADB = \angle CDA \) (This angle is common to both triangles)
Therefore, \( \triangle ADB \sim \triangle CDA \) (by AA similarity criteria)

Since the triangles are similar, their corresponding sides are proportional:
\( \frac{AD}{CD} = \frac{BD}{AD} \)
Cross-multiplying gives:
\( AD^2 = BD \cdot CD \)
In simple words: Looking at triangle ADB and triangle CDA, we can again establish similarity through shared angles and right angles. This similarity lets us set up another ratio of sides, proving that \( AD^2 \) is equal to the product of BD and CD.

Exam Tip: Practice drawing these complex figures and labeling all angles. It helps in visualizing which triangles are similar and why.

Question 4. ABC is an isosceles triangle right angled at C. Prove that \( AB^2 = 2AC^2 \). (CBSE 2012)
Answer:
We are given that \( \triangle ABC \) is a right-angled isosceles triangle, with the right angle at C. This means \( \angle C = 90^\circ \).
Since it is an isosceles triangle, two of its sides must be equal. As \( C \) is the right angle, the hypotenuse \( AB \) must be the longest side.
Therefore, the two equal sides are \( AC \) and \( BC \), so \( AC = BC \).

Applying the Pythagoras theorem to \( \triangle ABC \):
\( AB^2 = AC^2 + BC^2 \)
Since we know \( BC = AC \), we can substitute \( AC \) for \( BC \):
\( AB^2 = AC^2 + AC^2 \)
\( AB^2 = 2AC^2 \)
Thus, it is proven that \( AB^2 = 2AC^2 \).
In simple words: For an isosceles right-angled triangle, the two shorter sides are equal. Using the Pythagoras theorem, where the square of the hypotenuse equals the sum of the squares of the other two sides, we can replace one of the equal sides with the other, showing that the hypotenuse squared is twice the square of one of the equal sides.

Exam Tip: When a triangle is both right-angled and isosceles, the equal sides are always the two legs, not the hypotenuse. This understanding is key to correctly applying the Pythagorean theorem.

Question 5. ABC is an isosceles triangle with \( AC = BC \). If \( BC^2 = 2AC^2 \), prove that ABC is a right triangle.
Answer:
We are given that \( \triangle ABC \) is an isosceles triangle with \( AC = BC \).
We are also given the condition \( AB^2 = 2AC^2 \).

We can rewrite the given condition as:
\( AB^2 = AC^2 + AC^2 \)

Since we know that \( AC = BC \), we can substitute \( BC \) for one of the \( AC^2 \) terms:
\( AB^2 = AC^2 + BC^2 \)

According to the converse of the Pythagoras theorem, if the square of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the angle opposite the longest side is a right angle.
Here, \( AB^2 = AC^2 + BC^2 \), which means that the angle opposite side \( AB \) (which is \( \angle C \)) must be \( 90^\circ \).
Therefore, \( \angle C = 90^\circ \).

This shows that \( \triangle ABC \) is a right-angled triangle.
In simple words: We are given that two sides are equal and a special relationship between the sides. By rewriting the given equation, we find that it matches the Pythagorean theorem. This proves, using the opposite idea of Pythagoras's rule, that the angle opposite the longest side must be a right angle, making it a right triangle.

Exam Tip: The converse of the Pythagoras theorem is often used to prove that a triangle is right-angled. Make sure you state it clearly in your proof.

Question 6. ABC is an equilateral triangle of side \( 2a \). Find each of its altitudes.
Answer:
Given that \( \triangle ABC \) is an equilateral triangle with side length \( AB = AC = BC = 2a \).
Let's draw an altitude from vertex A to side BC, and label the intersection point as D. So, \( AD \perp BC \).

In an equilateral triangle, the altitude is also the median. This means D is the midpoint of BC.
Therefore, \( BD = CD = \frac{BC}{2} = \frac{2a}{2} = a \).

Now, consider the right-angled triangle \( \triangle ABD \) (or \( \triangle ACD \)).
Applying the Pythagoras theorem in \( \triangle ABD \):
\( AB^2 = BD^2 + AD^2 \)
Substitute the known values:
\( (2a)^2 = (a)^2 + AD^2 \)
\( 4a^2 = a^2 + AD^2 \)
Subtract \( a^2 \) from both sides:
\( AD^2 = 4a^2 - a^2 \)
\( AD^2 = 3a^2 \)
Take the square root of both sides to find \( AD \):
\( AD = \sqrt{3a^2} \)
\( AD = a\sqrt{3} \)

Since all altitudes in an equilateral triangle are equal in length, the length of each altitude is \( a\sqrt{3} \).
In simple words: For a triangle with all equal sides (equilateral), if you draw a line from the top corner straight down to the middle of the opposite side (this is called an altitude), that line also cuts the bottom side in half. Using Pythagoras's rule on one of the smaller right-angled triangles formed, we can calculate the length of this altitude. All three altitudes in an equilateral triangle are the same length.

Exam Tip: Remember the special properties of an equilateral triangle: all sides are equal, all angles are \( 60^\circ \), and an altitude bisects the opposite side and the vertex angle. The formula for the altitude of an equilateral triangle with side 's' is \( \frac{\sqrt{3}}{2}s \).

Question 7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. (CBSE 2012)
Answer:
Given: ABCD is a rhombus. This means all its sides are equal: \( AB = BC = CD = DA \).
The diagonals AC and BD intersect at point O.

To prove: \( AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2 \).

Proof: We know that the diagonals of a rhombus bisect each other at right angles.
Therefore, \( OA = OC = \frac{AC}{2} \) and \( OB = OD = \frac{BD}{2} \).
Also, all angles at the intersection O are \( 90^\circ \) (e.g., \( \angle AOB = 90^\circ \)).

Consider the right-angled triangle \( \triangle AOB \):
Applying the Pythagoras theorem:
\( AB^2 = OA^2 + OB^2 \) .....(1)

Similarly, for \( \triangle BOC \):
\( BC^2 = OB^2 + OC^2 \) .....(2)

For \( \triangle COD \):
\( CD^2 = OC^2 + OD^2 \) .....(3)

For \( \triangle DOA \):
\( DA^2 = OA^2 + OD^2 \) .....(4)

Now, let's add equations (1), (2), (3), and (4):
\( AB^2 + BC^2 + CD^2 + DA^2 = (OA^2 + OB^2) + (OB^2 + OC^2) + (OC^2 + OD^2) + (OA^2 + OD^2) \)
\( AB^2 + BC^2 + CD^2 + DA^2 = 2OA^2 + 2OB^2 + 2OC^2 + 2OD^2 \)

Substitute \( OA = \frac{AC}{2} \), \( OC = \frac{AC}{2} \), \( OB = \frac{BD}{2} \), and \( OD = \frac{BD}{2} \):
\( AB^2 + BC^2 + CD^2 + DA^2 = 2 \left( \frac{AC}{2} \right)^2 + 2 \left( \frac{BD}{2} \right)^2 + 2 \left( \frac{AC}{2} \right)^2 + 2 \left( \frac{BD}{2} \right)^2 \)
\( AB^2 + BC^2 + CD^2 + DA^2 = 2 \frac{AC^2}{4} + 2 \frac{BD^2}{4} + 2 \frac{AC^2}{4} + 2 \frac{BD^2}{4} \)
\( AB^2 + BC^2 + CD^2 + DA^2 = \frac{AC^2}{2} + \frac{BD^2}{2} + \frac{AC^2}{2} + \frac{BD^2}{2} \)
\( AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2 \)

Since \( AB = BC = CD = DA \), we can also write the left side as \( 4AB^2 \).
So, \( 4AB^2 = AC^2 + BD^2 \).
Thus, the sum of the squares of the sides of a rhombus equals the sum of the squares of its diagonals.
In simple words: A rhombus has four equal sides and its diagonals cut each other perfectly in half at a right angle. By using Pythagoras's theorem on the four small right-angled triangles formed at the center, and then adding their equations, we can show that the sum of the squares of all the sides is equal to the sum of the squares of the two diagonals.

Exam Tip: Key properties of a rhombus (equal sides, perpendicular bisecting diagonals) are essential for this proof. Breaking the rhombus into four right-angled triangles is the standard approach.

Question 8. In the figure, O is a point in the interior of a triangle ABC, \( OD \perp BC \), \( OE \perp AC \) and \( OF \perp AB \). Show that
(i) \( OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 = AF^2 + BD^2 + CE^2 \)
(ii) \( AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2 \)
Answer:
Given: O is a point inside \( \triangle ABC \). \( OD \perp BC \), \( OE \perp AC \) and \( OF \perp AB \).
Construction: Join OA, OB, and OC.

Proof for (i):
Consider the right-angled triangles formed by point O and the perpendiculars to the sides:
In \( \triangle AFO \): \( OA^2 = AF^2 + OF^2 \)
\( AF^2 = OA^2 - OF^2 \) .....(1)

In \( \triangle BDO \): \( OB^2 = BD^2 + OD^2 \)
\( BD^2 = OB^2 - OD^2 \) .....(2)

In \( \triangle CEO \): \( OC^2 = CE^2 + OE^2 \)
\( CE^2 = OC^2 - OE^2 \) .....(3)

Adding equations (1), (2), and (3):
\( AF^2 + BD^2 + CE^2 = (OA^2 - OF^2) + (OB^2 - OD^2) + (OC^2 - OE^2) \)
\( AF^2 + BD^2 + CE^2 = OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 \)
This proves the first part:
\( OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 = AF^2 + BD^2 + CE^2 \)
In simple words: By using the Pythagoras theorem in each of the right-angled triangles formed by point O and the perpendiculars, we can express the square of each segment of the sides in terms of the distance from O to the vertices and the perpendiculars. Adding these expressions together gives us the desired equality.

Exam Tip: When a point inside a triangle has perpendiculars to the sides, think about using Pythagoras theorem in the six small right-angled triangles formed (like \( \triangle AFO \), \( \triangle BDO \), etc.) to relate the sides and distances from the interior point.

Answer:
Proof for (ii):
From part (i), we have:
\( AF^2 + BD^2 + CE^2 = OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 \)

Now, let's look at the other set of segments. We can use the same Pythagoras relations but arranged differently:
In \( \triangle AEO \): \( OA^2 = AE^2 + OE^2 \)
\( AE^2 = OA^2 - OE^2 \) .....(4)

In \( \triangle CDO \): \( OC^2 = CD^2 + OD^2 \)
\( CD^2 = OC^2 - OD^2 \) .....(5)

In \( \triangle BFO \): \( OB^2 = BF^2 + OF^2 \)
\( BF^2 = OB^2 - OF^2 \) .....(6)

Adding equations (4), (5), and (6):
\( AE^2 + CD^2 + BF^2 = (OA^2 - OE^2) + (OC^2 - OD^2) + (OB^2 - OF^2) \)
\( AE^2 + CD^2 + BF^2 = OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 \)

Comparing this result with the result from part (i):
\( AF^2 + BD^2 + CE^2 = OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 \)
And \( AE^2 + CD^2 + BF^2 = OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 \)

Since both expressions are equal to the same quantity, they must be equal to each other:
\( AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2 \)
This proves the second part.
In simple words: After proving the first part, we can use a similar method by pairing different segments of the sides. We express the square of each of these new segments using Pythagoras theorem and then sum them up. Because both sums end up being equal to the same complex expression, we can conclude that the two sums of squares of segments are equal to each other.

Exam Tip: These types of proofs often involve algebraic manipulation of Pythagorean relations. Be organized with your equations and substitutions. Notice that both parts essentially relate to the same central expression \( OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 \).

Question 9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Answer:
Let AB represent the ladder, BC represent the wall, and AC represent the distance of the foot of the ladder from the base of the wall.
We are given:
Length of the ladder \( AB = 10 \) m.
Height of the window from the ground \( BC = 8 \) m.

The wall and the ground form a right angle, so \( \triangle ABC \) is a right-angled triangle with the right angle at C.
Applying the Pythagoras theorem:
\( AB^2 = AC^2 + BC^2 \)
Substitute the known values:
\( 10^2 = AC^2 + 8^2 \)
\( 100 = AC^2 + 64 \)
Subtract 64 from both sides:
\( AC^2 = 100 - 64 \)
\( AC^2 = 36 \)
Take the square root of both sides:
\( AC = \sqrt{36} \)
\( AC = 6 \) m

Hence, the distance of the foot of the ladder from the base of the wall is 6 m.
In simple words: Imagine a ladder leaning against a wall forms a right-angled triangle with the ground. The ladder is the longest side (hypotenuse), the wall is one side, and the distance from the wall to the bottom of the ladder is the other side. Using Pythagoras's rule, we can calculate this unknown distance.

Exam Tip: Visualize the scenario as a right-angled triangle. Clearly label the hypotenuse (ladder), perpendicular (wall height), and base (distance from wall). This ensures you apply Pythagoras's theorem correctly.

Question 10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut.
Answer:
Let BC represent the vertical pole, AB represent the guy wire, and AC represent the distance of the stake from the base of the pole.
We are given:
Height of the pole \( BC = 18 \) m.
Length of the guy wire \( AB = 24 \) m.

The pole stands vertically on the ground, so \( \triangle ABC \) is a right-angled triangle with the right angle at C.
Applying the Pythagoras theorem:
\( AB^2 = AC^2 + BC^2 \)
Substitute the known values:
\( 24^2 = AC^2 + 18^2 \)
\( 576 = AC^2 + 324 \)
Subtract 324 from both sides:
\( AC^2 = 576 - 324 \)
\( AC^2 = 252 \)
Take the square root of both sides:
\( AC = \sqrt{252} \)
To simplify \( \sqrt{252} \), we find its prime factors: \( 252 = 2 \times 2 \times 3 \times 3 \times 7 = 2^2 \times 3^2 \times 7 \)
\( AC = \sqrt{2^2 \times 3^2 \times 7} \)
\( AC = 2 \times 3 \sqrt{7} \)
\( AC = 6\sqrt{7} \) m

Hence, the stake should be driven \( 6\sqrt{7} \) m away from the base of the pole.
In simple words: We have a pole, a wire, and the ground forming a right-angled triangle. The wire is the hypotenuse, the pole is one leg, and the distance to the stake is the other leg. Using the Pythagoras theorem, we find the square root of the difference between the square of the wire's length and the square of the pole's height to get the required distance.

Exam Tip: Be careful with simplifying square roots. Factor out perfect squares to get the simplest radical form. Units (meters in this case) are also important in the final answer.

Question 11. An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will the two planes be after \( 1\frac{1}{2} \) hours?
Answer:
Let point A represent the airport.
The first aeroplane flies due North, and the second aeroplane flies due West. Since North and West are perpendicular directions, the paths of the two aeroplanes form a right angle at the airport. The distance between them will be the hypotenuse of the right-angled triangle formed.

Time \( t = 1\frac{1}{2} \) hours \( = \frac{3}{2} \) hours.

Distance travelled by the first aeroplane (North):
Speed \( = 1000 \) km/h
Distance \( AB = \text{Speed} \times \text{Time} = 1000 \times \frac{3}{2} = 1500 \) km.

Distance travelled by the second aeroplane (West):
Speed \( = 1200 \) km/h
Distance \( AC = \text{Speed} \times \text{Time} = 1200 \times \frac{3}{2} = 1800 \) km.

Now, \( \triangle ABC \) is a right-angled triangle with the right angle at A.
Applying the Pythagoras theorem to find the distance between the two planes (BC):
\( BC^2 = AB^2 + AC^2 \)
\( BC^2 = (1500)^2 + (1800)^2 \)
\( BC^2 = 2,250,000 + 3,240,000 \)
\( BC^2 = 5,490,000 \)
\( BC = \sqrt{5,490,000} \)
\( BC = \sqrt{549 \times 10,000} \)
\( BC = 100 \sqrt{549} \)
To simplify \( \sqrt{549} \): \( 549 = 9 \times 61 = 3^2 \times 61 \)
\( BC = 100 \sqrt{3^2 \times 61} \)
\( BC = 100 \times 3 \sqrt{61} \)
\( BC = 300\sqrt{61} \) km

The two planes will be \( 300\sqrt{61} \) km apart after \( 1\frac{1}{2} \) hours.
In simple words: Two planes fly away from an airport in directions that make a right angle (North and West). We calculate how far each plane travels using their speed and the time. Then, using Pythagoras's theorem, we find the direct distance between the two planes, which forms the longest side of the right triangle.

Exam Tip: Always draw a diagram for navigation problems. This helps correctly identify the right angle and which sides represent the legs and hypotenuse of the triangle. Ensure units are consistent throughout your calculations.

Question 12. Two poles of height 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Answer:
Let CD and AB be the two poles.
Height of pole CD \( = 6 \) m.
Height of pole AB \( = 11 \) m.
Distance between their feet \( AC = 12 \) m.

To find the distance between their tops (DB).

Draw a line DE parallel to AC from D to AB. This creates a rectangle CADE and a right-angled triangle \( \triangle DEB \).
Since CADE is a rectangle:
\( DE = AC = 12 \) m (Distance between bases)
\( CE = CD = 6 \) m (Height of smaller pole)

Now, find the length of BE:
\( BE = AB - AE \)
Since \( AE = CD = 6 \) m:
\( BE = 11 - 6 = 5 \) m.

Now, consider the right-angled triangle \( \triangle DEB \).
Applying the Pythagoras theorem:
\( DB^2 = DE^2 + BE^2 \)
Substitute the values:
\( DB^2 = 12^2 + 5^2 \)
\( DB^2 = 144 + 25 \)
\( DB^2 = 169 \)
Take the square root of both sides:
\( DB = \sqrt{169} \)
\( DB = 13 \) m.

Thus, the distance between the tops of the poles is 13 m.
In simple words: Imagine two poles of different heights standing on the ground. We draw a line from the top of the shorter pole straight across, parallel to the ground, to meet the taller pole. This creates a right-angled triangle. The horizontal distance between the poles is one side, and the difference in their heights is the other side. Using Pythagoras's theorem, we can find the distance between the tops of the poles.

Exam Tip: This problem involves creating an auxiliary right-angled triangle by drawing a line parallel to the ground. This is a common strategy for problems with non-right-angled shapes or irregular heights.

Question 13. D and E are points on the sides CA and CB respectively of a triangle ABC, right angled at C. Prove that \( AE^2 + BD^2 = AB^2 + DE^2 \).
Answer:
Given: \( \triangle ABC \) is right-angled at C (\( \angle C = 90^\circ \)). D is a point on AC, and E is a point on BC.

To prove: \( AE^2 + BD^2 = AB^2 + DE^2 \).

Proof:
Consider the right-angled triangles formed:

In \( \triangle ACE \):
\( AE^2 = AC^2 + CE^2 \) .....(1) (by Pythagoras theorem)

In \( \triangle BCD \):
\( BD^2 = BC^2 + CD^2 \) .....(2) (by Pythagoras theorem)

In \( \triangle ABC \):
\( AB^2 = AC^2 + BC^2 \) .....(3) (by Pythagoras theorem)

In \( \triangle DCE \):
\( DE^2 = CD^2 + CE^2 \) .....(4) (by Pythagoras theorem)

Add equations (1) and (2):
\( AE^2 + BD^2 = (AC^2 + CE^2) + (BC^2 + CD^2) \)
\( AE^2 + BD^2 = AC^2 + BC^2 + CE^2 + CD^2 \)

Now, from (3), we know \( AB^2 = AC^2 + BC^2 \).
From (4), we know \( DE^2 = CD^2 + CE^2 \).

Substitute \( AB^2 \) and \( DE^2 \) into the sum:
\( AE^2 + BD^2 = AB^2 + DE^2 \)
Thus, the proof is complete.
In simple words: In a right-angled triangle, if we pick points on the two shorter sides and connect them, we can form several other right-angled triangles. By writing out the Pythagoras theorem for each of these triangles and then cleverly adding their equations, we can show that the sum of the squares of certain lines is equal to the sum of the squares of other lines, including the hypotenuse.

Exam Tip: This type of problem often involves breaking down the main figure into multiple right-angled triangles. Write down the Pythagorean theorem for each relevant triangle and then look for algebraic substitutions to simplify and reach the desired result.

Question 14. The perpendicular from A on side BC at D of a \( \triangle ABC \) intersects BC at D such that \( DB = 3CD \) (see figure). Prove that \( 2AB^2 = 2AC^2 + BC^2 \).
Answer:
Given: \( \triangle ABC \) with \( AD \perp BC \). The point D on BC is such that \( DB = 3CD \).

To prove: \( 2AB^2 = 2AC^2 + BC^2 \).

Proof:
In the right-angled triangle \( \triangle ABD \):
\( AB^2 = AD^2 + BD^2 \) .....(1) (by Pythagoras theorem)

In the right-angled triangle \( \triangle ACD \):
\( AC^2 = AD^2 + CD^2 \) .....(2) (by Pythagoras theorem)

From the given information, we have \( DB = 3CD \).
We also know that \( BC = DB + CD \).
Substitute \( DB = 3CD \) into the equation for BC:
\( BC = 3CD + CD \)
\( BC = 4CD \)
This means \( CD = \frac{BC}{4} \).
And \( DB = 3CD = 3 \left( \frac{BC}{4} \right) = \frac{3BC}{4} \).

Subtract equation (2) from equation (1):
\( AB^2 - AC^2 = (AD^2 + BD^2) - (AD^2 + CD^2) \)
\( AB^2 - AC^2 = BD^2 - CD^2 \)

Substitute the expressions for BD and CD in terms of BC:
\( AB^2 - AC^2 = \left( \frac{3BC}{4} \right)^2 - \left( \frac{BC}{4} \right)^2 \)
\( AB^2 - AC^2 = \frac{9BC^2}{16} - \frac{BC^2}{16} \)
\( AB^2 - AC^2 = \frac{9BC^2 - BC^2}{16} \)
\( AB^2 - AC^2 = \frac{8BC^2}{16} \)
\( AB^2 - AC^2 = \frac{BC^2}{2} \)

Multiply the entire equation by 2:
\( 2(AB^2 - AC^2) = BC^2 \)
\( 2AB^2 - 2AC^2 = BC^2 \)

Rearrange the terms to get the required proof:
\( 2AB^2 = 2AC^2 + BC^2 \)
The proof is complete.
In simple words: We start with two right-angled triangles created by the perpendicular line from A. We apply Pythagoras's theorem to each. We use the given relationship between the segments of the base to express them in terms of the whole base. Subtracting the two Pythagoras equations and substituting these expressions helps us simplify the equation. After some algebraic steps, we arrive at the desired proof.

Exam Tip: For problems involving segments of a base with a given ratio (like \( DB = 3CD \)), always express the segments in terms of the total base length. This will help simplify algebraic expressions during the proof.

Question 15. In an equilateral triangle ABC, D is a point on side BC such that \( BD = \frac{1}{3} BC \). Prove that \( 9AD^2 = 7AB^2 \).
Answer:
Given: \( \triangle ABC \) is an equilateral triangle. Let each side be 's', so \( AB = BC = AC = s \).
D is a point on side BC such that \( BD = \frac{1}{3} BC = \frac{s}{3} \).

To prove: \( 9AD^2 = 7AB^2 \).

Construction: Draw an altitude \( AP \perp BC \).

Proof:
In an equilateral triangle, the altitude AP bisects the base BC.
So, \( BP = PC = \frac{BC}{2} = \frac{s}{2} \).

Now, we need to find the length of DP.
\( DP = BP - BD \)
\( DP = \frac{s}{2} - \frac{s}{3} \)
\( DP = \frac{3s - 2s}{6} = \frac{s}{6} \).

Consider the right-angled triangle \( \triangle APB \):
\( AB^2 = AP^2 + BP^2 \) (by Pythagoras theorem)
\( s^2 = AP^2 + \left( \frac{s}{2} \right)^2 \)
\( s^2 = AP^2 + \frac{s^2}{4} \)
\( AP^2 = s^2 - \frac{s^2}{4} \)
\( AP^2 = \frac{4s^2 - s^2}{4} = \frac{3s^2}{4} \).

Now, consider the right-angled triangle \( \triangle APD \):
\( AD^2 = AP^2 + DP^2 \) (by Pythagoras theorem)
Substitute the values of \( AP^2 \) and \( DP \):
\( AD^2 = \frac{3s^2}{4} + \left( \frac{s}{6} \right)^2 \)
\( AD^2 = \frac{3s^2}{4} + \frac{s^2}{36} \)
To add these fractions, find a common denominator (36):
\( AD^2 = \frac{3s^2 \times 9}{4 \times 9} + \frac{s^2}{36} \)
\( AD^2 = \frac{27s^2}{36} + \frac{s^2}{36} \)
\( AD^2 = \frac{27s^2 + s^2}{36} \)
\( AD^2 = \frac{28s^2}{36} \)
Simplify the fraction:
\( AD^2 = \frac{7s^2}{9} \)

Now, we want to prove \( 9AD^2 = 7AB^2 \).
Multiply both sides of \( AD^2 = \frac{7s^2}{9} \) by 9:
\( 9AD^2 = 7s^2 \)
Since \( AB = s \), we can substitute \( AB^2 \) for \( s^2 \):
\( 9AD^2 = 7AB^2 \)
The proof is complete.
In simple words: For an equilateral triangle with a point D on one side dividing it in a 1:2 ratio, we draw an altitude. We use Pythagoras's rule multiple times: first to find the length of the altitude, then to find the length of the segment DP, and finally, using the altitude and DP, to find AD. After getting AD squared in terms of the side length, we perform a few algebraic steps to prove the required relationship.

Exam Tip: Problems involving specific points on the base of an equilateral triangle often require drawing an altitude. This creates right-angled triangles, enabling the use of the Pythagoras theorem. Clearly define all segments, especially those involving ratios, to avoid errors.

Question 16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Answer:
Given: \( \triangle ABC \) is an equilateral triangle. Let each side be 's', so \( AB = BC = AC = s \).

To prove: \( 3AB^2 = 4AD^2 \), where AD is an altitude.

Construction: Draw an altitude \( AD \perp BC \).

Proof:
In an equilateral triangle, the altitude AD bisects the base BC.
So, \( BD = CD = \frac{BC}{2} = \frac{s}{2} \).

Consider the right-angled triangle \( \triangle ADB \):
Applying the Pythagoras theorem:
\( AB^2 = AD^2 + BD^2 \)
Substitute the known values:
\( s^2 = AD^2 + \left( \frac{s}{2} \right)^2 \)
\( s^2 = AD^2 + \frac{s^2}{4} \)

Now, we want to relate \( AD^2 \) and \( s^2 \). Rearrange the equation to solve for \( AD^2 \):
\( AD^2 = s^2 - \frac{s^2}{4} \)
\( AD^2 = \frac{4s^2 - s^2}{4} \)
\( AD^2 = \frac{3s^2}{4} \)

To get rid of the fraction, multiply both sides by 4:
\( 4AD^2 = 3s^2 \)

Since \( s = AB \) (the side of the equilateral triangle), we can substitute \( AB^2 \) for \( s^2 \):
\( 4AD^2 = 3AB^2 \)
This proves that three times the square of one side is equal to four times the square of one of its altitudes.
In simple words: If you take any equilateral triangle and draw a height line (altitude), it splits one of its sides in half and creates a right-angled triangle. Using Pythagoras's rule on this new right triangle, we can find a relationship between the side length and the height. After a little rearranging, we discover that three times the side squared is exactly equal to four times the altitude squared.

Exam Tip: This is a fundamental property of equilateral triangles often tested. Knowing the relationship between side and altitude (either deriving it or remembering \( h = \frac{\sqrt{3}}{2}s \)) is very useful.

Question 17. Tick the correct answer and justify: In AB = \( 6\sqrt{3} \) cm, AC = 12 cm, and BC = 6 cm. The angle B is
(a) 120°
(b) 60°
(c) 90°
(d) 45°
Answer: (c) 90°
We have AB equal to \( 6\sqrt{3} \) cm, AC is \( 12 \) cm, and BC is \( 6 \) cm.
Squaring AB, we get \( \text{AB}^2 = (6\sqrt{3})^2 = 36 \times 3 \).
So, \( \text{AB}^2 = 108 \).
For AC, \( \text{AC}^2 = 12^2 = 144 \).
And for BC, \( \text{BC}^2 = 6^2 = 36 \).
Since \( 144 = 108 + 36 \), which implies \( \text{AC}^2 = \text{AB}^2 + \text{BC}^2 \). By the converse of the Pythagoras theorem, the given triangle is a right-angled triangle. The right angle is at B, so \( \angle B = 90^\circ \). Therefore, the correct choice is (c).
In simple words: First, square each side length. Then, check if the square of the longest side (hypotenuse) equals the sum of the squares of the other two sides. If it does, the triangle is a right-angled triangle, and the angle opposite the longest side is 90 degrees.

Exam Tip: Remember the converse of the Pythagoras theorem: if in a triangle, the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.