Samacheer Kalvi Class 8 Maths Solutions Chapter 5 Geometry Exercise 5.2

Question 1. Fill in the blanks:
(i) If in a \( \Delta \) PQR, \( \mathrm{PR}^2 = \mathrm{PQ}^2 + \mathrm{QR}^2 \), then the right angle of \( \Delta \) PQR is at the vertex _________.
Answer: Q
In simple words: The Pythagorean theorem states that in a right-angled triangle, the square of the longest side (hypotenuse) equals the sum of the squares of the other two sides. Here, PR is the hypotenuse, and the right angle is always found opposite to it. So, the right angle is at vertex Q.

🎯 Exam Tip: For Pythagoras theorem questions, remember the right angle is always opposite the longest side (hypotenuse).

Question 1. Fill in the blanks:
(ii) If 'l' and 'm' are the legs and 'n' is the hypotenuse of a right angled triangle then, \( \mathrm{l}^2 = \) _________.
Answer: \( \mathrm{n}^2 - \mathrm{m}^2 \)
In simple words: In a right-angled triangle, if 'n' is the hypotenuse and 'l' and 'm' are the other two sides (legs), then according to Pythagoras theorem, \( \mathrm{n}^2 = \mathrm{l}^2 + \mathrm{m}^2 \). To find \( \mathrm{l}^2 \), we simply rearrange the formula. This shows how the square of one leg can be found by subtracting the square of the other leg from the square of the hypotenuse.

🎯 Exam Tip: Always recall the basic Pythagoras theorem \( \mathrm{a}^2 + \mathrm{b}^2 = \mathrm{c}^2 \) and then rearrange it as needed to find any unknown side.

Question 1. Fill in the blanks:
(iii) If the sides of a triangle are in the ratio 5:12:13 then, it is _________.
Answer: a right angled triangle
In simple words: The numbers 5, 12, and 13 form a special group called a Pythagorean triplet because \( 5^2 + 12^2 = 25 + 144 = 169 = 13^2 \). When the squares of the two shorter sides of a triangle add up to the square of the longest side, it means the triangle has a right angle, following the converse of Pythagoras theorem.

🎯 Exam Tip: Recognize common Pythagorean triplets like (3, 4, 5) or (5, 12, 13) to quickly identify right-angled triangles.

Question 1. Fill in the blanks:
(iv) The medians of a triangle cross each other at _________.
Answer: Centroid
In simple words: The centroid is a special point inside a triangle where all three medians meet. A median is a line segment drawn from a vertex to the midpoint of the opposite side. This point is often thought of as the triangle's "center of mass."

🎯 Exam Tip: Understand the definitions of key triangle points: centroid (intersection of medians), orthocenter (altitudes), incenter (angle bisectors), circumcenter (perpendicular bisectors).

Question 1. Fill in the blanks:
(v) The centroid of a triangle divides each medians in the ratio _________.
Answer: 2:1
In simple words: The centroid divides each median into two parts. The part closer to the vertex is always twice as long as the part closer to the midpoint of the opposite side. This 2:1 ratio is a fixed property of the centroid.

🎯 Exam Tip: Remember the 2:1 ratio for the centroid, with the larger part always being from the vertex side.

Question 2. Say True or False.
(i) 8, 15, 17 is a Pythagorean triplet.
Answer: True
In simple words: To check if 8, 15, and 17 form a Pythagorean triplet, we see if the square of the largest number (17) equals the sum of the squares of the other two numbers (8 and 15). We calculate \( 8^2 = 64 \), \( 15^2 = 225 \), and \( 17^2 = 289 \). Since \( 64 + 225 = 289 \), the statement is true. This property is fundamental to right-angled triangles.

🎯 Exam Tip: Always check if \( \mathrm{a}^2 + \mathrm{b}^2 = \mathrm{c}^2 \) for given numbers to determine if they form a Pythagorean triplet, where 'c' is the largest number.

Question 2. Say True or False.
(ii) In a right angled triangle, the hypotenuse is the greatest side.
Answer: True
In simple words: The hypotenuse is the side opposite the right angle in a right-angled triangle. It is always the longest side of the triangle because it spans the greatest angle. This is a basic property of right triangles.

🎯 Exam Tip: Visualize a right triangle to remember that the side opposite the 90-degree angle (the hypotenuse) is always the longest.

Question 2. Say True or False.
(iii) In any triangle the centroid and the incentre are located inside the triangle.
Answer: True
In simple words: The centroid, which is the intersection of medians, always lies inside the triangle. Similarly, the incenter, which is the intersection of angle bisectors, also always lies inside the triangle. Both are internal points for any type of triangle.

🎯 Exam Tip: Recall that the centroid and incenter are always *inside* a triangle, while the orthocenter and circumcenter can sometimes be outside (for obtuse triangles) or on the triangle (for right triangles).

Question 2. Say True or False.
(iv) The centroid, orthocentre, and incentre of a triangle are collinear.
Answer: True
In simple words: The statement says that the centroid, orthocenter, and incenter lie on the same line. This is true for special kinds of triangles, such as isosceles or equilateral triangles, where these points align due to symmetry.

🎯 Exam Tip: Remember that for most general triangles, only the centroid, orthocenter, and circumcenter are collinear (forming the Euler line). This specific statement holds true only for particular types of triangles like isosceles ones.

Question 2. Say True or False.
(v) The incentre is equidistant from all the vertices of a triangle.
Answer: False
In simple words: The incenter is the point where the angle bisectors of a triangle meet. It is equidistant from all *sides* of the triangle, not from its vertices. The circumcenter is the point that is equidistant from all vertices.

🎯 Exam Tip: Clearly distinguish between the incenter (equidistant from sides) and the circumcenter (equidistant from vertices).

Question 3. Check whether given sides are the sides of right-angled triangles, using Pythagoras theorem.
(i) 8, 15, 17
Answer:
Take \( a = 8 \), \( b = 15 \) and \( c = 17 \).
Now \( a^2 + b^2 = 8^2 + 15^2 = 64 + 225 = 289 \).
And \( c^2 = 17^2 = 289 \).
Since \( a^2 + b^2 = c^2 \), by the converse of Pythagoras theorem, the triangle with given measures is a right-angled triangle. This set of numbers is a common Pythagorean triplet.
In simple words: We check if the sum of squares of the two smaller sides equals the square of the largest side. Since \( 8^2 + 15^2 = 17^2 \), these sides form a right-angled triangle.

🎯 Exam Tip: Always identify the longest side first and set it as 'c' (potential hypotenuse) when applying Pythagoras theorem.

Question 3. Check whether given sides are the sides of right-angled triangles, using Pythagoras theorem.
(ii) 12, 13, 15
Answer:
Take \( a = 12 \), \( b = 13 \) and \( c = 15 \).
Now \( a^2 + b^2 = 12^2 + 13^2 = 144 + 169 = 313 \).
And \( c^2 = 15^2 = 225 \).
Since \( a^2 + b^2 \neq c^2 \), the triangle with given measures is not a right-angled triangle.
In simple words: We check if the sum of squares of the two smaller sides equals the square of the largest side. Since \( 12^2 + 13^2 \neq 15^2 \), these sides do not form a right-angled triangle.

🎯 Exam Tip: Be careful with calculations; even a small arithmetic error can lead to a wrong conclusion about the triangle type.

Question 3. Check whether given sides are the sides of right-angled triangles, using Pythagoras theorem.
(iii) 30, 40, 50
Answer:
Take \( a = 30 \), \( b = 40 \) and \( c = 50 \).
Now \( a^2 + b^2 = 30^2 + 40^2 = 900 + 1600 = 2500 \).
And \( c^2 = 50^2 = 2500 \).
Since \( a^2 + b^2 = c^2 \), by the converse of Pythagoras theorem, the triangle with given measures is a right-angled triangle. This is a scaled version of the (3, 4, 5) triplet.
In simple words: We check if the sum of squares of the two smaller sides equals the square of the largest side. Since \( 30^2 + 40^2 = 50^2 \), these sides form a right-angled triangle.

🎯 Exam Tip: Recognize scaled Pythagorean triplets (e.g., (30, 40, 50) is 10 times (3, 4, 5)) as they also form right-angled triangles.

Question 3. Check whether given sides are the sides of right-angled triangles, using Pythagoras theorem.
(iv) 9, 40, 41
Answer:
Take \( a = 9 \), \( b = 40 \) and \( c = 41 \).
Now \( a^2 + b^2 = 9^2 + 40^2 = 81 + 1600 = 1681 \).
And \( c^2 = 41^2 = 1681 \).
Since \( a^2 + b^2 = c^2 \), by the converse of Pythagoras theorem, the triangle with given measures is a right-angled triangle.
In simple words: We check if the sum of squares of the two smaller sides equals the square of the largest side. Since \( 9^2 + 40^2 = 41^2 \), these sides form a right-angled triangle.

🎯 Exam Tip: Practice squaring numbers efficiently to quickly test for Pythagorean triplets.

Question 3. Check whether given sides are the sides of right-angled triangles, using Pythagoras theorem.
(v) 24, 45, 51
Answer:
Take \( a = 24 \), \( b = 45 \) and \( c = 51 \).
Now \( a^2 + b^2 = 24^2 + 45^2 = 576 + 2025 = 2601 \).
And \( c^2 = 51^2 = 2601 \).
Since \( a^2 + b^2 = c^2 \), by the converse of Pythagoras theorem, the triangle with given measures is a right-angled triangle.
In simple words: We check if the sum of squares of the two smaller sides equals the square of the largest side. Since \( 24^2 + 45^2 = 51^2 \), these sides form a right-angled triangle.

🎯 Exam Tip: If the numbers are large, sometimes dividing by a common factor first can simplify calculations. For instance, 24, 45, 51 can all be divided by 3, resulting in 8, 15, 17, which is a known triplet.

Question 4. Find the unknown side in the following triangles.
(i)
Answer:
In the given right-angled triangle ABC, we have sides AB = 9, AC = 40, and the hypotenuse BC = x. According to the Pythagoras theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
So, \( \mathrm{BC}^2 = \mathrm{AB}^2 + \mathrm{AC}^2 \).
\( x^2 = 9^2 + 40^2 \)
\( x^2 = 81 + 1600 \)
\( x^2 = 1681 \)
\( x = \sqrt{1681} \)
\( x = 41 \)
The unknown side is 41.
In simple words: We use the Pythagoras theorem where \( x^2 = 9^2 + 40^2 \). This calculates to \( x^2 = 1681 \), so \( x \) is 41.

🎯 Exam Tip: Clearly label the sides (hypotenuse and legs) before applying the Pythagoras theorem to avoid mistakes.

Question 4. Find the unknown side in the following triangles.
(ii)
Answer:
In the right-angled triangle PQR, PR is the hypotenuse (34), QR is one leg (30), and PQ is the other leg (y).
By the Pythagoras theorem, \( \mathrm{PR}^2 = \mathrm{PQ}^2 + \mathrm{QR}^2 \).
\( 34^2 = y^2 + 30^2 \)
\( 1156 = y^2 + 900 \)
\( y^2 = 1156 - 900 \)
\( y^2 = 256 \)
\( y = \sqrt{256} \)
\( y = 16 \)
In simple words: We find the unknown leg 'y' using Pythagoras theorem: \( y^2 = 34^2 - 30^2 \). This simplifies to \( y^2 = 256 \), so \( y \) is 16.

🎯 Exam Tip: Remember to rearrange the Pythagoras formula correctly when finding a leg instead of the hypotenuse: \( \mathrm{leg}^2 = \mathrm{hypotenuse}^2 - \mathrm{other \, leg}^2 \).

Question 4. Find the unknown side in the following triangles.
(iii)
Answer:
In the right-angled triangle XYZ, YZ is the hypotenuse (39), XZ is one leg (36), and XY is the other leg (z).
Using the Pythagoras theorem, \( \mathrm{YZ}^2 = \mathrm{XY}^2 + \mathrm{XZ}^2 \).
\( 39^2 = z^2 + 36^2 \)
\( 1521 = z^2 + 1296 \)
\( z^2 = 1521 - 1296 \)
\( z^2 = 225 \)
\( z = \sqrt{225} \)
\( z = 15 \)
In simple words: We use the Pythagoras theorem to find the unknown leg 'z': \( z^2 = 39^2 - 36^2 \). This calculation gives \( z^2 = 225 \), so \( z \) is 15.

🎯 Exam Tip: Double-check your squares and subtractions, especially when dealing with larger numbers, to ensure accuracy.

Question 5. An isosceles triangle has equal sides each 13 cm and a base 24 cm in length. Find its height.
Answer:
In an isosceles triangle, an altitude drawn from the vertex angle to the base divides the base into two equal parts and also forms two right-angled triangles.
Here, the equal sides (hypotenuses) are 13 cm each, and the base is 24 cm.
When the altitude AD is drawn, it divides the base BC (24 cm) into two 12 cm segments (BD = DC = 12 cm).
Now consider the right-angled triangle ABD.
We have the hypotenuse AB = 13 cm and one leg BD = 12 cm.
Using the Pythagoras theorem, \( \mathrm{AB}^2 = \mathrm{AD}^2 + \mathrm{BD}^2 \).
\( 13^2 = \mathrm{AD}^2 + 12^2 \)
\( 169 = \mathrm{AD}^2 + 144 \)
\( \mathrm{AD}^2 = 169 - 144 \)
\( \mathrm{AD}^2 = 25 \)
\( \mathrm{AD} = \sqrt{25} \)
\( \mathrm{AD} = 5 \) cm.
The height of the triangle is 5 cm.
In simple words: The altitude splits the base into two 12 cm parts. Using Pythagoras on one half-triangle: \( \text{height}^2 = 13^2 - 12^2 = 169 - 144 = 25 \). So, the height is \( \sqrt{25} = 5 \) cm.

🎯 Exam Tip: Remember that an altitude in an isosceles triangle bisects the base and creates two congruent right-angled triangles, which simplifies finding the height using Pythagoras.

Question 6. Find the distance between the helicopter and the ship.
Answer:
We can imagine a right-angled triangle where:
The height of the helicopter (AP) is one leg = 80 m.
The horizontal distance of the ship from the point directly below the helicopter (PS) is the other leg = 150 m.
The distance between the helicopter and the ship (AS) is the hypotenuse.
Using the Pythagoras theorem, \( \mathrm{AS}^2 = \mathrm{AP}^2 + \mathrm{PS}^2 \).
\( \mathrm{AS}^2 = 80^2 + 150^2 \)
\( \mathrm{AS}^2 = 6400 + 22500 \)
\( \mathrm{AS}^2 = 28900 \)
\( \mathrm{AS} = \sqrt{28900} \)
\( \mathrm{AS} = 170 \) m.
The distance between the helicopter and the ship is 170 m. This is an example of applying geometry to real-world scenarios.
In simple words: The helicopter's height, the horizontal distance, and the distance to the ship form a right triangle. Using Pythagoras: \( \text{distance}^2 = 80^2 + 150^2 = 28900 \). So the distance is \( \sqrt{28900} = 170 \) m.

🎯 Exam Tip: Always visualize the problem and identify the right-angled triangle. Label the known sides and the unknown side before applying the theorem.

Question 7. In triangle ABC, line l, is a perpendicular bisector of BC. If BC = 12cm, SM = 8cm, find CS.
Answer:
Line \( l_1 \) is a perpendicular bisector of BC, meaning it cuts BC exactly in half at point M and forms a 90-degree angle with BC.
Given BC = 12 cm, so BM = MC = \( \frac{12}{2} = 6 \) cm.
We are also given SM = 8 cm.
Now consider the right-angled triangle SMC.
The legs are SM = 8 cm and MC = 6 cm. CS is the hypotenuse.
Using the Pythagoras theorem, \( \mathrm{CS}^2 = \mathrm{SM}^2 + \mathrm{MC}^2 \).
\( \mathrm{CS}^2 = 8^2 + 6^2 \)
\( \mathrm{CS}^2 = 64 + 36 \)
\( \mathrm{CS}^2 = 100 \)
\( \mathrm{CS} = \sqrt{100} \)
\( \mathrm{CS} = 10 \) cm.
Thus, the length of CS is 10 cm. This demonstrates how perpendicular bisectors create useful right triangles.
In simple words: Since \( l_1 \) bisects BC, MC = 6 cm. In the right triangle SMC, \( \mathrm{CS}^2 = \mathrm{SM}^2 + \mathrm{MC}^2 = 8^2 + 6^2 = 100 \). So, CS is 10 cm.

🎯 Exam Tip: When a perpendicular bisector is involved, remember it creates a right angle and divides the segment into two equal halves, which are key facts for applying Pythagoras.

Question 8. Identify the centroid of \( \Delta \) PQR.
Answer:
In triangle PQR, PT is a median from vertex P (as T is the midpoint of QR).
QS is a median from vertex Q (as S is the midpoint of PR).
When two medians intersect, their intersection point is the centroid. Here, medians QT and PS meet at point W. Therefore, W is the centroid of \( \Delta \) PQR. The centroid is the point where all medians of a triangle meet.
In simple words: Medians are lines from a vertex to the midpoint of the opposite side. The centroid is where these medians cross. In this triangle, W is the point where medians QT and PS meet, making it the centroid.

🎯 Exam Tip: The centroid is the point of intersection of the medians. Knowing the definition helps in identifying it from a diagram.

Question 9. Name the orthocentre of \( \Delta \) PQR.
Answer:
The orthocenter of a triangle is the point where all three altitudes intersect. An altitude is a line segment from a vertex perpendicular to the opposite side.
In a right-angled triangle, two of the altitudes are the legs of the triangle itself. For example, in a right-angled triangle PQR, if the right angle is at vertex P, then PQ is an altitude to PR, and PR is an altitude to PQ.
These two altitudes (the legs) intersect at the right-angle vertex P. The third altitude from R would also pass through P. Therefore, the orthocenter of a right-angled triangle is always the vertex containing the 90-degree angle.
In simple words: The orthocenter is where all the altitudes meet. In a right-angled triangle, the two sides that form the right angle are also altitudes, so they meet at the right-angle vertex. Therefore, P is the orthocenter.

🎯 Exam Tip: For a right-angled triangle, the orthocenter is the vertex at which the right angle is formed.

Question 10. In the given figure, A is the midpoint of YZ and G is the centroid of the triangle XYZ. If the length of GA is 3 cm, find XA.
Answer:
Given A is the midpoint of YZ, which means XA is a median.
G is the centroid of \( \Delta \) XYZ.
The centroid divides each median in a 2:1 ratio from the vertex.
So, \( \frac{\mathrm{XG}}{\mathrm{GA}} = \frac{2}{1} \).
Given GA = 3 cm.
\( \frac{\mathrm{XG}}{3} = \frac{2}{1} \)
\( \implies \) XG = \( 2 \times 3 \)
\( \implies \) XG = 6 cm.
The total length of the median XA is XG + GA.
XA = \( 6 + 3 \)
\( \implies \) XA = 9 cm.
In simple words: The centroid G divides the median XA into XG and GA in a 2:1 ratio. Since GA is 3 cm, XG must be twice that, which is 6 cm. So, the full length of the median XA is \( 6 + 3 = 9 \) cm.

🎯 Exam Tip: Clearly remember the 2:1 ratio for the centroid and which part is twice as long as the other. XA is the whole median, XG is the part from the vertex, and GA is the part from the midpoint.

Question 11. If I is the incentre of \( \Delta \) XYZ, \( \angle \mathrm{IYZ} = 30^\circ \) and \( \angle \mathrm{IZY} = 40^\circ \), find \( \angle \mathrm{YXZ} \).
Answer:
The incenter (I) is the point where the angle bisectors of a triangle meet. This means that line segment YI bisects angle Y (∠XYZ), and ZI bisects angle Z (∠XZY).
Since YI bisects \( \angle \mathrm{XYZ} \):
\( \angle \mathrm{IYZ} = 30^\circ \implies \angle \mathrm{XYI} = 30^\circ \).
So, \( \angle \mathrm{XYZ} = \angle \mathrm{XYI} + \angle \mathrm{IYZ} = 30^\circ + 30^\circ = 60^\circ \).
Since ZI bisects \( \angle \mathrm{XZY} \):
\( \angle \mathrm{IZY} = 40^\circ \implies \angle \mathrm{XZI} = 40^\circ \).
So, \( \angle \mathrm{XZY} = \angle \mathrm{XZI} + \angle \mathrm{IZY} = 40^\circ + 40^\circ = 80^\circ \).
Now, using the angle sum property of a triangle (sum of all angles is \( 180^\circ \)) for \( \Delta \) XYZ:
\( \angle \mathrm{YXZ} + \angle \mathrm{XYZ} + \angle \mathrm{XZY} = 180^\circ \)
\( \angle \mathrm{YXZ} + 60^\circ + 80^\circ = 180^\circ \)
\( \angle \mathrm{YXZ} + 140^\circ = 180^\circ \)
\( \angle \mathrm{YXZ} = 180^\circ - 140^\circ \)
\( \implies \angle \mathrm{YXZ} = 40^\circ \).
In simple words: The incenter I means YI and ZI are angle bisectors. So \( \angle Y \) is \( 30^\circ + 30^\circ = 60^\circ \), and \( \angle Z \) is \( 40^\circ + 40^\circ = 80^\circ \). Since all angles in a triangle add to \( 180^\circ \), \( \angle \mathrm{YXZ} = 180^\circ - (60^\circ + 80^\circ) = 180^\circ - 140^\circ = 40^\circ \).

🎯 Exam Tip: Always remember that the incenter is formed by angle bisectors, meaning it divides each vertex angle into two equal parts.

Objective Type Questions

Question 12. If \( \Delta \) GUT is isosceles and right angled, then \( \angle \mathrm{TUG} \) is
(A) 30°
(B) 40°
(C) 45°
(D) 55°
Answer: (C) 45°
In simple words: In a right-angled isosceles triangle, one angle is \( 90^\circ \), and the other two angles must be equal. Since the sum of angles in a triangle is \( 180^\circ \), each of the other two equal angles is \( (180^\circ - 90^\circ) / 2 = 45^\circ \).

🎯 Exam Tip: Remember that in any isosceles right-angled triangle, the two non-right angles are always \( 45^\circ \).

Question 13. The hypotenuse of a right angled triangle of sides 12cm and 16cm is
(A) 28 cm
(B) 20 cm
(C) 24 cm
(D) 21 cm
Answer: (B) 20 cm
In simple words: For a right-angled triangle with legs 12 cm and 16 cm, we find the hypotenuse 'c' using Pythagoras theorem: \( \mathrm{c}^2 = 12^2 + 16^2 = 144 + 256 = 400 \). So, \( \mathrm{c} = \sqrt{400} = 20 \) cm.

🎯 Exam Tip: Always apply the Pythagoras theorem (\( \mathrm{a}^2 + \mathrm{b}^2 = \mathrm{c}^2 \)) carefully, especially during squaring and finding the square root.

Question 14. The area of a rectangle of length 21cm and diagonal 29 cm is
(A) 609 cm\( ^2 \)
(B) 580 cm\( ^2 \)
(C) 420 cm\( ^2 \)
(D) 210 cm\( ^2 \)
Answer: (C) 420 cm\( ^2 \)
In simple words: The length (21 cm), width, and diagonal (29 cm) form a right triangle. Using Pythagoras, \( \text{width}^2 = 29^2 - 21^2 = 841 - 441 = 400 \), so the width is 20 cm. The area is length \( \times \) width = \( 21 \times 20 = 420 \) cm\( ^2 \).

🎯 Exam Tip: Remember that the diagonal of a rectangle creates two right-angled triangles with the sides, allowing you to use the Pythagoras theorem to find an unknown side.

Question 15. The sides of a right angled triangle are in the ratio 5:12:13 and its perimeter is 120 units then, the sides are
(A) 25, 36, 59
(B) 10, 24, 26
(C) 36, 39, 45
(D) 20, 48, 52
Answer: (D) 20,48,52
In simple words: Let the sides be \( 5a, 12a, 13a \). The perimeter is \( 5a + 12a + 13a = 30a \). Since the perimeter is 120, \( 30a = 120 \), so \( a = 4 \). The sides are then \( 5 \times 4 = 20 \), \( 12 \times 4 = 48 \), and \( 13 \times 4 = 52 \) units.

🎯 Exam Tip: When given a ratio of sides and a perimeter, introduce a common multiplier (like 'a') to represent the actual side lengths, then solve for the multiplier using the perimeter.