OP Malhotra Class 9 Maths Solutions Chapter 10 Pythagoras Theorem Exercise 10 (A)

S Chand Class 9 ICSE Maths Solutions Chapter 10 Pythagoras Theorem Ex 10(A)

 

Question 1. In a right angled triangle ABC, c is the length of the hypotenuse, and a and b are other two sides.
(a) If \( a = 6 \) and \( b = 8 \), then find \( c \).
(b) If \( c = 25 \) and \( a = 24 \), then find \( b \).
(c) If \( c = 13 \) and \( b = 5 \), then find \( a \).
(d) If \( a = 10 \) and \( c = 21 \), then find \( b \).
(e) If \( a = 9 \) and \( b = 9 \), then find \( c \).
Answer:
(a) Given \( a = 6 \) and \( b = 8 \).
We use the Pythagoras theorem: \( c^2 = a^2 + b^2 \).
Substitute the given values:
\( c^2 = (6)^2 + (8)^2 \)
\( c^2 = 36 + 64 \)
\( c^2 = 100 \)
To find \( c \), take the square root of both sides:
\( c = \sqrt{100} \)
\( c = 10 \)

(b) Given \( c = 25 \) and \( a = 24 \).
Using the Pythagoras theorem: \( c^2 = a^2 + b^2 \).
Substitute the given values:
\( (25)^2 = (24)^2 + b^2 \)
\( 625 = 576 + b^2 \)
To find \( b^2 \), subtract 576 from 625:
\( b^2 = 625 - 576 \)
\( b^2 = 49 \)
To find \( b \), take the square root of both sides:
\( b = \sqrt{49} \)
\( b = 7 \)

(c) Given \( c = 13 \) and \( b = 5 \).
Using the Pythagoras theorem: \( c^2 = a^2 + b^2 \).
Substitute the given values:
\( (13)^2 = a^2 + (5)^2 \)
\( 169 = a^2 + 25 \)
To find \( a^2 \), subtract 25 from 169:
\( a^2 = 169 - 25 \)
\( a^2 = 144 \)
To find \( a \), take the square root of both sides:
\( a = \sqrt{144} \)
\( a = 12 \)

(d) Given \( a = 10 \) and \( c = 21 \).
Using the Pythagoras theorem: \( c^2 = a^2 + b^2 \).
Substitute the given values:
\( (21)^2 = (10)^2 + b^2 \)
\( 441 = 100 + b^2 \)
To find \( b^2 \), subtract 100 from 441:
\( b^2 = 441 - 100 \)
\( b^2 = 341 \)
To find \( b \), take the square root of both sides:
\( b = \sqrt{341} \)

(e) Given \( a = 9 \) and \( b = 9 \).
Using the Pythagoras theorem: \( c^2 = a^2 + b^2 \).
Substitute the given values:
\( c^2 = (9)^2 + (9)^2 \)
\( c^2 = 81 + 81 \)
\( c^2 = 162 \)
To find \( c \), take the square root of both sides:
\( c = \sqrt{162} \)
We can simplify the square root:
\( c = \sqrt{81 \times 2} \)
\( c = 9\sqrt{2} \)
In simple words: The Pythagorean theorem helps us find the length of a missing side in a right-angled triangle. We just put the known side lengths into the formula \( a^2 + b^2 = c^2 \) and solve for the unknown. Remember \( c \) is always the longest side, called the hypotenuse.

🎯 Exam Tip: Always remember to state the Pythagorean theorem clearly before substituting values. Make sure to perform calculations carefully and simplify square roots if possible to get the final answer.

 

Question 2. A rectangular field is 30 m by 40 m. What distance is saved by walking diagonally across the field.
Answer: For a rectangular field of 30 m by 40 m, let the length be \( l = 40 \) m and the width be \( w = 30 \) m.
If a person walks along the sides, the distance covered would be \( l + w = 40 + 30 = 70 \) m.
If the person walks diagonally, the distance will be the length of the diagonal. In a rectangle, the diagonal forms the hypotenuse of a right-angled triangle with the length and width as the other two sides.
Using the Pythagoras theorem, the diagonal \( d \):
\( d^2 = l^2 + w^2 \)
\( d^2 = (40)^2 + (30)^2 \)
\( d^2 = 1600 + 900 \)
\( d^2 = 2500 \)
\( d = \sqrt{2500} \)
\( d = 50 \) m
The distance saved is the difference between walking along the sides and walking diagonally:
Distance saved \( = (l + w) - d \)
Distance saved \( = (40 + 30) - 50 \)
Distance saved \( = 70 - 50 \)
Distance saved \( = 20 \) m
So, walking diagonally saves 20 meters.
In simple words: If you walk around the two sides of a rectangular field, it's a longer path. But if you walk straight across from one corner to the opposite corner (diagonally), it's shorter. The difference in these two paths is the distance you save. We use Pythagoras theorem to find the diagonal length.

🎯 Exam Tip: For problems involving distances saved by taking a diagonal path, remember that the diagonal always forms the hypotenuse of a right-angled triangle within the rectangular shape. Clearly show both path calculations and the subtraction step.

 

Question 3. A man travels 7 km due north, then goes 3 km due east and then 3 km due south. How far is he from his starting point?
Answer: Let the starting point be O.
First, the man travels 7 km due North from O to A. So, \( OA = 7 \) km.
Next, he travels 3 km due East from A to B. So, \( AB = 3 \) km.
Then, he travels 3 km due South from B to C. So, \( BC = 3 \) km.
To find the distance from the starting point O to the final point C, we can draw a perpendicular from C to OA, let's call the intersection point D.
Now, we have a rectangle formed by points A, B, C, D if we consider point D such that ABCD is a rectangle. (This implicitly makes CD parallel to AB and AD parallel to BC).
Since \( AB = 3 \) km, then \( CD = 3 \) km.
Since \( BC = 3 \) km, then \( AD = 3 \) km.
Now, consider triangle ODC.
The base OD will be \( OA - AD \).
\( OD = 7 \text{ km} - 3 \text{ km} = 4 \text{ km} \).
The height DC is \( 3 \) km.
Triangle ODC is a right-angled triangle at D.
Using the Pythagoras theorem to find OC (the distance from the starting point):
\( OC^2 = OD^2 + DC^2 \)
\( OC^2 = (4)^2 + (3)^2 \)
\( OC^2 = 16 + 9 \)
\( OC^2 = 25 \)
\( OC = \sqrt{25} \)
\( OC = 5 \) km
So, the man is 5 km from his starting point.
In simple words: Imagine the man's journey as steps on a map. He goes up, then right, then down. To find how far he is from where he started, we can draw a straight line from start to finish. This creates a right-angled triangle, and we use the Pythagoras theorem to find the length of that straight line.

🎯 Exam Tip: When solving movement problems, always draw a clear diagram to visualize the path. Break down complex paths into simpler right-angled triangles to apply the Pythagoras theorem effectively. This problem uses vector addition concepts simplified by geometry.

 

Question 4. The diagonals of a rhombus are 12 cm and 9 cm long. Calculate the length of one side of the rhombus.
Answer: In a rhombus, the diagonals bisect each other at right angles.
Let the diagonals be \( d_1 = 12 \) cm and \( d_2 = 9 \) cm.
When they bisect, they divide into half.
So, half of \( d_1 \) is \( \frac{12}{2} = 6 \) cm.
And half of \( d_2 \) is \( \frac{9}{2} = 4.5 \) cm.
Let the intersection point of the diagonals be O. Then we have four right-angled triangles formed inside the rhombus.
Consider one of these right-angled triangles, for example, \( \triangle AOB \). The sides \( AO \) and \( OB \) are the half-diagonals, and \( AB \) is the side of the rhombus (which is the hypotenuse for this triangle).
Using the Pythagoras theorem:
\( AB^2 = AO^2 + OB^2 \)
\( AB^2 = (6)^2 + (4.5)^2 \)
\( AB^2 = 36 + 20.25 \)
\( AB^2 = 56.25 \)
To find \( AB \), take the square root of both sides:
\( AB = \sqrt{56.25} \)
\( AB = 7.5 \) cm
Since all sides of a rhombus are equal, the length of one side of the rhombus is 7.5 cm.
In simple words: A rhombus has four equal sides and its diagonals cross in the middle at a perfect square corner (90 degrees). If you cut the diagonals in half, you get two shorter lines that form a right-angled triangle with one side of the rhombus. We use Pythagoras to find that side.

🎯 Exam Tip: Remember the properties of a rhombus: diagonals bisect each other at right angles. This creates four congruent right-angled triangles. Identify the half-diagonals as the perpendicular sides and the rhombus side as the hypotenuse in one of these triangles.

 

Question 5.
(a) In a right-angled triangle ABC it is given that the hypotenuse, AC = 2.5 cm, and the side AB = 1.5 cm. Calculate the side BC.
(b) AD is drawn perpendicular to BC, base of an equilateral triangle ABC. Given BC = 10 cm, find the length of AD, correct to 1 place of decimal.
Answer:
(a) Given a right-angled triangle ABC, with \( \angle B = 90^\circ \).
The hypotenuse \( AC = 2.5 \) cm.
The side \( AB = 1.5 \) cm.
We need to find the length of side BC.
Using the Pythagoras theorem:
\( AC^2 = AB^2 + BC^2 \)
Substitute the given values:
\( (2.5)^2 = (1.5)^2 + BC^2 \)
\( 6.25 = 2.25 + BC^2 \)
To find \( BC^2 \), subtract 2.25 from 6.25:
\( BC^2 = 6.25 - 2.25 \)
\( BC^2 = 4.00 \)
To find \( BC \), take the square root of both sides:
\( BC = \sqrt{4.00} \)
\( BC = 2 \) cm

(b) Given an equilateral triangle ABC, with side length \( BC = 10 \) cm.
In an equilateral triangle, all sides are equal, so \( AB = BC = AC = 10 \) cm.
AD is drawn perpendicular to BC. In an equilateral triangle, the altitude (perpendicular from a vertex to the opposite side) also bisects the base.
So, D is the midpoint of BC.
Therefore, \( BD = DC = \frac{BC}{2} = \frac{10}{2} = 5 \) cm.
Now, consider the right-angled triangle ABD (with \( \angle D = 90^\circ \)).
We know \( AB = 10 \) cm (hypotenuse) and \( BD = 5 \) cm.
Using the Pythagoras theorem:
\( AB^2 = AD^2 + BD^2 \)
Substitute the values:
\( (10)^2 = AD^2 + (5)^2 \)
\( 100 = AD^2 + 25 \)
To find \( AD^2 \), subtract 25 from 100:
\( AD^2 = 100 - 25 \)
\( AD^2 = 75 \)
To find \( AD \), take the square root of both sides:
\( AD = \sqrt{75} \)
We can simplify \( \sqrt{75} \) as \( \sqrt{25 \times 3} = 5\sqrt{3} \).
Using the approximate value of \( \sqrt{3} \approx 1.732 \):
\( AD = 5 \times 1.732 \)
\( AD = 8.66 \) cm
Rounding to one decimal place, \( AD = 8.7 \) cm.
In simple words: For part (a), we use Pythagoras theorem directly to find a missing side in a right triangle. For part (b), an equilateral triangle has all sides equal, and when you draw a line straight down from the top corner to the middle of the bottom side, it creates a right-angled triangle. We can then use Pythagoras to find the length of this height.

🎯 Exam Tip: Always identify the right angle in the triangle before applying the Pythagorean theorem. For equilateral triangles, remember that the altitude bisects the base, creating two congruent right-angled triangles.

 

Question 6. ABC is right-angled triangle. Angle ABC = 90°, AC = 25 cm and AB = 24 cm. Calculate the area of a \( \triangle ABC \).
Answer: Given a right-angled triangle ABC, with \( \angle ABC = 90^\circ \).
The hypotenuse \( AC = 25 \) cm.
The side \( AB = 24 \) cm.
To calculate the area of \( \triangle ABC \), we need the lengths of its base and height. For a right-angled triangle, the two sides forming the right angle can be considered as the base and height.
First, find the length of side BC using the Pythagoras theorem:
\( AC^2 = AB^2 + BC^2 \)
Substitute the given values:
\( (25)^2 = (24)^2 + BC^2 \)
\( 625 = 576 + BC^2 \)
To find \( BC^2 \), subtract 576 from 625:
\( BC^2 = 625 - 576 \)
\( BC^2 = 49 \)
To find \( BC \), take the square root of both sides:
\( BC = \sqrt{49} \)
\( BC = 7 \) cm
Now that we have both perpendicular sides, we can calculate the area:
Area of \( \triangle ABC = \frac{1}{2} \times \text{Base} \times \text{Height} \)
Area \( = \frac{1}{2} \times BC \times AB \)
Area \( = \frac{1}{2} \times 7 \times 24 \)
Area \( = 7 \times 12 \)
Area \( = 84 \) cm\( ^2 \)
In simple words: To find the area of a right-angled triangle, you need the lengths of the two sides that meet at the right angle (base and height). If you only know one of these sides and the longest side (hypotenuse), first use the Pythagoras theorem to find the missing base or height. Then, use the simple formula: half times base times height.

🎯 Exam Tip: Always remember the formula for the area of a triangle (\( \frac{1}{2} \times \text{base} \times \text{height} \)). In a right-angled triangle, the two sides forming the right angle are the base and height. Use the Pythagorean theorem to find any missing side first.

 

Question 7. A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from ground.
Answer: Let the vertical wall be represented by AB, the ground by BC, and the ladder by AC.
The ladder AC has a length of 13 m.
The foot of the ladder (C) is 5 m away from the foot of the wall (B). So, \( BC = 5 \) m.
The wall is vertical, so the angle at B is \( 90^\circ \), forming a right-angled triangle ABC.
We need to find the distance of the other end of the ladder from the ground, which is the height AB.
Using the Pythagoras theorem:
\( AC^2 = AB^2 + BC^2 \)
Substitute the given values:
\( (13)^2 = AB^2 + (5)^2 \)
\( 169 = AB^2 + 25 \)
To find \( AB^2 \), subtract 25 from 169:
\( AB^2 = 169 - 25 \)
\( AB^2 = 144 \)
To find \( AB \), take the square root of both sides:
\( AB = \sqrt{144} \)
\( AB = 12 \) m
So, the height of the wall where the ladder touches is 12 m from the ground.
In simple words: Imagine a ladder leaning against a wall. The ladder, the wall, and the ground form a triangle with a square corner where the wall meets the ground. We know the ladder's length and how far its base is from the wall. Using the Pythagoras rule, we can figure out how high up the wall the ladder reaches.

🎯 Exam Tip: Problems involving ladders, walls, and the ground almost always form a right-angled triangle. Clearly label the hypotenuse (the ladder) and the two perpendicular sides (wall height and distance from wall) before applying the Pythagoras theorem.

 

Question 8. Use the given information to make a neat diagram of the figure having given properties and write down the name of the figure.
In triangle ABC, \( AB^2 = BC^2 + AC^2 \) and \( AC = 2 BC \).
Answer: Given the properties for triangle ABC:
1. \( AB^2 = BC^2 + AC^2 \)
2. \( AC = 2 BC \)
From the first property, \( AB^2 = BC^2 + AC^2 \), according to the converse of the Pythagoras theorem, this means that \( \triangle ABC \) is a right-angled triangle, and the angle opposite the longest side (AB) is \( 90^\circ \). So, \( \angle C = 90^\circ \).
Now, let's draw the diagram based on these properties.
(i) Draw a line segment BC of some length.
(ii) At point C, draw a ray CX perpendicular to BC (making a \( 90^\circ \) angle).
(iii) Along the ray CX, cut off a segment CA such that its length is twice the length of BC (i.e., \( AC = 2 \times BC \)).
(iv) Join points A and B.
The figure obtained is a right-angled triangle ABC with the right angle at C, and side AC is twice the length of side BC. The name of the figure is a **Right-angled triangle**.
In simple words: We are told two things about a triangle. First, that one side squared is equal to the sum of the squares of the other two sides. This tells us it's a triangle with a right angle. Second, one side is twice as long as another. We draw it by making a square corner and then making sure one side is double the length of another. It's just a special type of right triangle.

🎯 Exam Tip: When given properties of a triangle, use the converse of the Pythagoras theorem to determine if it is a right-angled triangle. Then, use the other conditions to draw the figure accurately and name it correctly. A construction type of question means you need to show how to build the figure.

 

Question 9. In the figure, ABCD represents a quadrilateral in which AD = 13 cm, DC = 12 cm, BC = 3 cm, and \( \angle ABD = \angle BCD = 90^\circ \). Calculate the length of AB.
Answer: Given a quadrilateral ABCD with the following information:
\( AD = 13 \) cm
\( DC = 12 \) cm
\( BC = 3 \) cm
\( \angle ABD = 90^\circ \)
\( \angle BCD = 90^\circ \)
We need to find the length of AB.
First, consider the right-angled triangle BCD (since \( \angle BCD = 90^\circ \)).
We know \( DC = 12 \) cm and \( BC = 3 \) cm.
Using the Pythagoras theorem to find BD (the hypotenuse of \( \triangle BCD \)):
\( BD^2 = CD^2 + BC^2 \)
\( BD^2 = (12)^2 + (3)^2 \)
\( BD^2 = 144 + 9 \)
\( BD^2 = 153 \)
Next, consider the right-angled triangle ABD (since \( \angle ABD = 90^\circ \)).
We know \( AD = 13 \) cm (hypotenuse) and we just found \( BD^2 = 153 \).
Using the Pythagoras theorem:
\( AD^2 = AB^2 + BD^2 \)
Substitute the known values:
\( (13)^2 = AB^2 + 153 \)
\( 169 = AB^2 + 153 \)
To find \( AB^2 \), subtract 153 from 169:
\( AB^2 = 169 - 153 \)
\( AB^2 = 16 \)
To find \( AB \), take the square root of both sides:
\( AB = \sqrt{16} \)
\( AB = 4 \) cm
So, the length of AB is 4 cm.
In simple words: This shape is made of two right-angled triangles joined together. To find the missing side of the first triangle, we first use the Pythagoras theorem on the second triangle to find their shared side. Once we have that shared side, we can use Pythagoras again on the first triangle to find the length we need.

🎯 Exam Tip: When dealing with quadrilaterals that contain right angles, often you can split them into multiple right-angled triangles. Solve for a shared side in one triangle first, then use that information to solve for the unknown in the other triangle.

 

Question 10. Which of the triangles whose sides are given below are right angled?
(i) 4 cm, 5 cm, 6 cm
(ii) 1.2 cm, 3.7 cm, 3.5 cm
(iii) 4 cm, 9.6 cm, 10.4 cm
(iv) 2.2 cm, 3.3 cm, 4.4 cm
Answer: For a triangle to be right-angled, its sides must satisfy the converse of the Pythagoras theorem. This means the square of the longest side (hypotenuse) must be equal to the sum of the squares of the other two sides. Let the sides be \( a, b, c \) where \( c \) is the longest side. Then \( c^2 = a^2 + b^2 \).

(i) Given sides: 4 cm, 5 cm, 6 cm.
The longest side is 6 cm. So, \( c = 6 \). The other sides are \( a = 4 \) and \( b = 5 \).
Calculate \( c^2 \): \( (6)^2 = 36 \)
Calculate \( a^2 + b^2 \): \( (4)^2 + (5)^2 = 16 + 25 = 41 \)
Since \( 36 \neq 41 \), this is not a right-angled triangle.

(ii) Given sides: 1.2 cm, 3.7 cm, 3.5 cm.
The longest side is 3.7 cm. So, \( c = 3.7 \). The other sides are \( a = 1.2 \) and \( b = 3.5 \).
Calculate \( c^2 \): \( (3.7)^2 = 13.69 \)
Calculate \( a^2 + b^2 \): \( (1.2)^2 + (3.5)^2 = 1.44 + 12.25 = 13.69 \)
Since \( 13.69 = 13.69 \), this is a right-angled triangle.

(iii) Given sides: 4 cm, 9.6 cm, 10.4 cm.
The longest side is 10.4 cm. So, \( c = 10.4 \). The other sides are \( a = 4 \) and \( b = 9.6 \).
Calculate \( c^2 \): \( (10.4)^2 = 108.16 \)
Calculate \( a^2 + b^2 \): \( (4)^2 + (9.6)^2 = 16.00 + 92.16 = 108.16 \)
Since \( 108.16 = 108.16 \), this is a right-angled triangle.

(iv) Given sides: 2.2 cm, 3.3 cm, 4.4 cm.
The longest side is 4.4 cm. So, \( c = 4.4 \). The other sides are \( a = 2.2 \) and \( b = 3.3 \).
Calculate \( c^2 \): \( (4.4)^2 = 19.36 \)
Calculate \( a^2 + b^2 \): \( (2.2)^2 + (3.3)^2 = 4.84 + 10.89 = 15.73 \)
Since \( 19.36 \neq 15.73 \), this is not a right-angled triangle.
Therefore, triangles (ii) and (iii) are right-angled triangles.
In simple words: To check if a triangle has a right angle, we use a simple rule: if you square the longest side, it must be exactly equal to the sum of the squares of the two shorter sides. If the numbers match, it's a right-angled triangle; if they don't, it isn't.

🎯 Exam Tip: Always identify the longest side first, as this will be the hypotenuse if the triangle is right-angled. Then, carefully calculate the square of the longest side and the sum of the squares of the other two sides. A common error is misidentifying the longest side.

 

Question 11. In the figure, find the distance of D from A, unit of length is 1 cm, (AB = 2, BC = 4, CD = 2).
Answer: Given the figure with lengths AB = 2 units, BC = 4 units, CD = 2 units, and \( \angle ABC = \angle BCD = 90^\circ \). We need to find the distance AD.
Let's join AD and draw a perpendicular from A to the line containing BC, and another perpendicular from D to the line containing BC. Or, draw a line through D parallel to BC, meeting the extension of AB at a point E.
Let's extend AB to E such that ADE is a right-angled triangle. (This means drawing a line from D parallel to BC, let it meet the extension of AB at E).
Since \( \angle ABC = \angle BCD = 90^\circ \), this means AB is parallel to CD in a way that suggests a trapezoid or a combined figure.
Let's draw a line from D parallel to AB, meeting BC at a point F. Then ABFD forms a rectangle. So \( BF = AB = 2 \) and \( DF = AD \) in the diagram.
Alternatively, consider points A, B, C, D in a coordinate plane. Let B be at the origin \( (0,0) \).
Then A is at \( (0,2) \) (since AB = 2 along y-axis).
C is at \( (4,0) \) (since BC = 4 along x-axis).
Since \( \angle BCD = 90^\circ \) and CD = 2, D would be at \( (4,2) \).
Now we need to find the distance between A \( (0,2) \) and D \( (4,2) \).
This isn't using the provided solution's method.

Let's follow the provided solution's approach, which implies construction of a point E and forming triangles.
From the figure and solution: Let AD intersect BC at E.
Consider \( \triangle ABE \) and \( \triangle DCE \).
Given \( AB = CD = 2 \) units.
\( \angle AEB = \angle DEC \) (vertically opposite angles).
\( \angle ABE = \angle DCE = 90^\circ \).
By Angle-Side-Angle (ASA) axiom, \( \triangle ABE \cong \triangle DCE \).
Since the triangles are congruent, their corresponding parts are equal (c.p.c.t.).
So, \( AE = ED \).
Also, \( BE = EC \).
Since \( BC = 4 \) and \( BE + EC = BC \), and \( BE = EC \), then \( BE = EC = \frac{4}{2} = 2 \) units.
Now, consider the right-angled triangle ABE (with \( \angle B = 90^\circ \)).
We know \( AB = 2 \) and \( BE = 2 \).
Using the Pythagoras theorem to find AE (hypotenuse):
\( AE^2 = AB^2 + BE^2 \)
\( AE^2 = (2)^2 + (2)^2 \)
\( AE^2 = 4 + 4 \)
\( AE^2 = 8 \)
\( AE = \sqrt{8} \)
\( AE = \sqrt{4 \times 2} \)
\( AE = 2\sqrt{2} \) units
We need to find the distance AD. Since \( AD = AE + ED \) and \( AE = ED \), then \( AD = 2 \times AE \).
\( AD = 2 \times 2\sqrt{2} \)
\( AD = 4\sqrt{2} \) units
Using the approximate value of \( \sqrt{2} \approx 1.414 \):
\( AD = 4 \times 1.414 \)
\( AD = 5.656 \) cm
Rounding to two decimal places, \( AD \approx 5.66 \) cm.
In simple words: We have a shape that looks like a Z. By drawing an extra line and finding how two small triangles are perfectly identical (congruent), we can figure out the lengths of their sides. Then, using the Pythagoras theorem on one of these right-angled triangles helps us find a part of the total distance. Finally, we add these parts to get the full distance.

🎯 Exam Tip: For complex figures, look for congruent triangles or ways to break the figure into right-angled triangles. Often, drawing an auxiliary line can help create these simpler shapes. Remember that congruent triangles have equal corresponding parts.

 

Question 12. The sides of a right-angled triangle containing the right angle are 5x and \( (3x – 1) \) cm. If the area of the triangle be 60 cm\( ^2 \), calculate the lengths of the sides of the triangle.
Answer: Let the right-angled triangle be ABC, with \( \angle B = 90^\circ \).
The sides containing the right angle are \( AB = 5x \) cm and \( BC = (3x - 1) \) cm.
The area of the triangle is given as 60 cm\( ^2 \).
The formula for the area of a right-angled triangle is: \( \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \)
So, \( 60 = \frac{1}{2} \times (5x) \times (3x - 1) \)
Multiply both sides by 2:
\( 120 = 5x(3x - 1) \)
Distribute \( 5x \) on the right side:
\( 120 = 15x^2 - 5x \)
Rearrange the equation to form a quadratic equation:
\( 15x^2 - 5x - 120 = 0 \)
Divide the entire equation by 5 to simplify:
\( 3x^2 - x - 24 = 0 \)
Now, solve this quadratic equation. We can use factorization by splitting the middle term. We need two numbers that multiply to \( 3 \times (-24) = -72 \) and add up to -1. These numbers are -9 and 8.
\( 3x^2 - 9x + 8x - 24 = 0 \)
Factor by grouping:
\( 3x(x - 3) + 8(x - 3) = 0 \)
\( (x - 3)(3x + 8) = 0 \)
This gives two possible values for \( x \):
Case 1: \( x - 3 = 0 \implies x = 3 \)
Case 2: \( 3x + 8 = 0 \implies 3x = -8 \implies x = -\frac{8}{3} \)
Since length cannot be negative, we discard \( x = -\frac{8}{3} \).
So, \( x = 3 \).
Now, calculate the lengths of the sides:
\( AB = 5x = 5 \times 3 = 15 \) cm
\( BC = 3x - 1 = (3 \times 3) - 1 = 9 - 1 = 8 \) cm
Now, we need to find the length of the hypotenuse AC.
Using the Pythagoras theorem:
\( AC^2 = AB^2 + BC^2 \)
\( AC^2 = (15)^2 + (8)^2 \)
\( AC^2 = 225 + 64 \)
\( AC^2 = 289 \)
\( AC = \sqrt{289} \)
\( AC = 17 \) cm
Thus, the lengths of the sides of the triangle are 15 cm, 8 cm, and 17 cm.
In simple words: We are given the area of a right-angled triangle and expressions for its two shorter sides. First, we use the area formula to create an equation and solve for the unknown value 'x'. Since length can't be negative, we pick the positive value for 'x'. Then, we plug 'x' back into the expressions to find the actual lengths of the two shorter sides. Finally, we use the Pythagoras theorem to find the length of the longest side (hypotenuse).

🎯 Exam Tip: When given an algebraic expression for side lengths and the area, set up the area formula as a quadratic equation. Solving the quadratic equation will give possible values for the variable, but always discard negative solutions for lengths or distances. Remember to calculate all three side lengths at the end.

 

Question 13. If the figure, it is given that AB = BC = 25 m. If AE = 7 m, and CD = 24 m, find the length of DE and also show that triangle ABE and triangle BDC are congruent.
Answer: Given the figure with \( AB = BC = 25 \) m, \( AE = 7 \) m, and \( CD = 24 \) m.
We need to find the length of DE and prove \( \triangle ABE \cong \triangle BDC \).

Part 1: Find the length of DE.
First, consider the right-angled triangle BDC (with right angle at D).
We know the hypotenuse \( BC = 25 \) m and side \( CD = 24 \) m.
Using the Pythagoras theorem to find BD:
\( BC^2 = BD^2 + CD^2 \)
\( (25)^2 = BD^2 + (24)^2 \)
\( 625 = BD^2 + 576 \)
\( BD^2 = 625 - 576 \)
\( BD^2 = 49 \)
\( BD = \sqrt{49} \)
\( BD = 7 \) m
Next, consider the right-angled triangle ABE (with right angle at E).
We know the hypotenuse \( AB = 25 \) m and side \( AE = 7 \) m.
Using the Pythagoras theorem to find BE:
\( AB^2 = AE^2 + BE^2 \)
\( (25)^2 = (7)^2 + BE^2 \)
\( 625 = 49 + BE^2 \)
\( BE^2 = 625 - 49 \)
\( BE^2 = 576 \)
\( BE = \sqrt{576} \)
\( BE = 24 \) m
Now, we can find the length of DE. From the figure, DE is the sum of DB and BE.
\( DE = DB + BE \)
\( DE = 7 \text{ m} + 24 \text{ m} \)
\( DE = 31 \) m

Part 2: Show that \( \triangle ABE \cong \triangle BDC \).
We have the following side lengths:
\( AB = 25 \) m (Given)
\( BC = 25 \) m (Given)
\( AE = 7 \) m (Given)
\( BD = 7 \) m (Calculated from \( \triangle BDC \))
\( BE = 24 \) m (Calculated from \( \triangle ABE \))
\( CD = 24 \) m (Given)
Compare the sides of \( \triangle ABE \) and \( \triangle BDC \):
1. \( AB = BC \) (Each is 25 m, Given)
2. \( AE = BD \) (Each is 7 m, Given and Calculated)
3. \( BE = CD \) (Each is 24 m, Calculated and Given)
Since all three corresponding sides of \( \triangle ABE \) are equal to the corresponding sides of \( \triangle BDC \), by the Side-Side-Side (SSS) congruence axiom, the triangles are congruent.
Therefore, \( \triangle ABE \cong \triangle BDC \). This is proven.
In simple words: First, we use the Pythagoras theorem in the two right-angled triangles to find their missing side lengths. Then, to show the triangles are identical, we check if all three sides of one triangle are equal to the three matching sides of the other triangle. If they are, the triangles are congruent, meaning they are exact copies of each other.

🎯 Exam Tip: For problems requiring both length calculation and congruence proof, perform the calculations first. The calculated lengths are crucial for establishing the Side-Side-Side (SSS) congruence condition. Clearly state the congruence axiom used.

 

Question 14. A ladder rests against a vertical wall at a height of 12 m from the ground with its foot at a distance of 9 m from the wall on the ground. If the foot of the ladder is shifted 3 m away from the wall, how much lower will the ladder slide down?

Answer: Let AC be the wall and BC be the ground. The ladder is AB. Given: Original height of the wall \( AC = 12 \text{ m} \) Original distance of the ladder's foot from the wall \( BC = 9 \text{ m} \) In right-angled triangle ABC (at C): \( AB^2 = AC^2 + BC^2 \) (By Pythagoras Theorem) \( \implies AB^2 = (12)^2 + (9)^2 \) \( \implies AB^2 = 144 + 81 \) \( \implies AB^2 = 225 \) \( \implies AB = \sqrt{225} \) \( \implies AB = 15 \text{ m} \) So, the length of the ladder is \( 15 \text{ m} \). Now, the foot of the ladder is shifted \( 3 \text{ m} \) away from the wall. New distance of the foot from the wall \( EC = BC + 3 = 9 + 3 = 12 \text{ m} \). The ladder length remains the same, so \( DE = 15 \text{ m} \). Let DC be the new height on the wall. In right-angled triangle DEC (at C): \( DE^2 = DC^2 + EC^2 \) (By Pythagoras Theorem) \( \implies (15)^2 = DC^2 + (12)^2 \) \( \implies 225 = DC^2 + 144 \) \( \implies DC^2 = 225 - 144 \) \( \implies DC^2 = 81 \) \( \implies DC = \sqrt{81} \) \( \implies DC = 9 \text{ m} \) The ladder slides down by \( AD = AC - DC = 12 - 9 = 3 \text{ m} \). This means the top of the ladder moved down the wall.In simple words: First, we find the ladder's full length using the old height and base. Then, we use this same ladder length with the new, longer base distance to find the new height on the wall. The difference between the old height and the new height tells us how far the ladder slid down.

🎯 Exam Tip: Always draw a clear diagram for such problems to visualize the initial and final positions. Clearly label the known and unknown lengths to avoid confusion when applying the Pythagoras theorem.

 

Question 15. AD is an altitude of a ∆ABC and AD is 12 cm, BD = 9 cm and DC = 16 cm long respectively. Prove that the angle BAC is a right angle.

Answer: Given: In \( \triangle ABC \), AD is the altitude to BC, meaning \( AD \perp BC \). \( AD = 12 \text{ cm} \), \( BD = 9 \text{ cm} \), \( DC = 16 \text{ cm} \). We need to prove that \( \angle BAC \) is a right angle (\( 90^\circ \)). Proof: 1. In right-angled triangle ABD (at D): \( AB^2 = AD^2 + BD^2 \) (By Pythagoras Theorem) \( \implies AB^2 = (12)^2 + (9)^2 \) \( \implies AB^2 = 144 + 81 \) \( \implies AB^2 = 225 \) \( \implies AB = \sqrt{225} \) \( \implies AB = 15 \text{ cm} \) 2. In right-angled triangle ACD (at D): \( AC^2 = AD^2 + DC^2 \) (By Pythagoras Theorem) \( \implies AC^2 = (12)^2 + (16)^2 \) \( \implies AC^2 = 144 + 256 \) \( \implies AC^2 = 400 \) \( \implies AC = \sqrt{400} \) \( \implies AC = 20 \text{ cm} \) 3. Now, calculate the length of BC: \( BC = BD + DC = 9 + 16 = 25 \text{ cm} \) 4. Check if \( \triangle ABC \) is a right-angled triangle using the converse of Pythagoras Theorem: \( BC^2 = (25)^2 = 625 \) \( AB^2 + AC^2 = (15)^2 + (20)^2 \) \( \implies AB^2 + AC^2 = 225 + 400 \) \( \implies AB^2 + AC^2 = 625 \) Since \( AB^2 + AC^2 = BC^2 \), by the converse of Pythagoras Theorem, \( \triangle ABC \) is a right-angled triangle with the right angle at A. Therefore, \( \angle BAC = 90^\circ \). The sum of the squares of the two shorter sides equals the square of the longest side. Hence, proved.In simple words: We find the lengths of the sides AB and AC using the altitude AD and Pythagoras. Then, we add the parts BD and DC to get BC. Finally, we check if the square of BC is equal to the sum of the squares of AB and AC. If it is, then angle BAC is 90 degrees, proving it's a right triangle.

🎯 Exam Tip: For proving right angles using side lengths, the converse of Pythagoras theorem is crucial. Always clearly show the calculations for all three squares and their sum/comparison.

 

Question 16. The shortest distance AP from a point A to a straight line QR is 12 cm, and Q, R are 15 cm and 20 cm distance from A on opposite sides of AP. Prove that QAR is a right angle.

Answer: Given: AP is the shortest distance from point A to line QR, which means \( AP \perp QR \). \( AP = 12 \text{ cm} \). \( AQ = 15 \text{ cm} \) (distance of Q from A). \( AR = 20 \text{ cm} \) (distance of R from A). We need to prove that \( \angle QAR \) is a right angle (\( 90^\circ \)). Proof: 1. In right-angled triangle APQ (at P): \( AQ^2 = AP^2 + QP^2 \) (By Pythagoras Theorem) \( \implies (15)^2 = (12)^2 + QP^2 \) \( \implies 225 = 144 + QP^2 \) \( \implies QP^2 = 225 - 144 \) \( \implies QP^2 = 81 \) \( \implies QP = \sqrt{81} \) \( \implies QP = 9 \text{ cm} \) 2. In right-angled triangle APR (at P): \( AR^2 = AP^2 + PR^2 \) (By Pythagoras Theorem) \( \implies (20)^2 = (12)^2 + PR^2 \) \( \implies 400 = 144 + PR^2 \) \( \implies PR^2 = 400 - 144 \) \( \implies PR^2 = 256 \) \( \implies PR = \sqrt{256} \) \( \implies PR = 16 \text{ cm} \) 3. Now, calculate the length of QR: Since Q and R are on opposite sides of AP, \( QR = QP + PR \) \( \implies QR = 9 + 16 \) \( \implies QR = 25 \text{ cm} \) 4. Check if \( \triangle QAR \) is a right-angled triangle using the converse of Pythagoras Theorem: \( QR^2 = (25)^2 = 625 \) \( AQ^2 + AR^2 = (15)^2 + (20)^2 \) \( \implies AQ^2 + AR^2 = 225 + 400 \) \( \implies AQ^2 + AR^2 = 625 \) Since \( AQ^2 + AR^2 = QR^2 \), by the converse of Pythagoras Theorem, \( \triangle QAR \) is a right-angled triangle with the right angle at A. This confirms the angle formed by the two distances is \( 90^\circ \). Hence, proved.In simple words: We find the lengths of the two segments QP and PR on the line QR by using Pythagoras theorem in the smaller right triangles. Then we add them to get the total length of QR. Finally, we check if the squares of the sides AQ and AR add up to the square of QR. If they do, then angle QAR is 90 degrees.

🎯 Exam Tip: When a point is connected to a line, the "shortest distance" always implies a perpendicular. Break the main triangle into two right-angled triangles using this perpendicular to apply Pythagoras theorem effectively.

 

Question 17. In the figure, PQR is a triangle in which PT is an altitude. Given PQ = 25 cm, PR = 17 cm, PT = 15 cm, and QR = x cm. Calculate x.

Answer: Given: In \( \triangle PQR \), PT is an altitude, which means \( PT \perp QR \). \( PQ = 25 \text{ cm} \), \( PR = 17 \text{ cm} \), \( PT = 15 \text{ cm} \), \( QR = x \text{ cm} \). We need to calculate the value of x. 1. In right-angled triangle PTQ (at T): \( PQ^2 = PT^2 + QT^2 \) (By Pythagoras Theorem) \( \implies (25)^2 = (15)^2 + QT^2 \) \( \implies 625 = 225 + QT^2 \) \( \implies QT^2 = 625 - 225 \) \( \implies QT^2 = 400 \) \( \implies QT = \sqrt{400} \) \( \implies QT = 20 \text{ cm} \) 2. In right-angled triangle PTR (at T): \( PR^2 = PT^2 + TR^2 \) (By Pythagoras Theorem) \( \implies (17)^2 = (15)^2 + TR^2 \) \( \implies 289 = 225 + TR^2 \) \( \implies TR^2 = 289 - 225 \) \( \implies TR^2 = 64 \) \( \implies TR = \sqrt{64} \) \( \implies TR = 8 \text{ cm} \) 3. Now, calculate the total length of QR (x): Since T lies on QR, \( QR = QT + TR \) \( \implies x = 20 + 8 \) \( \implies x = 28 \text{ cm} \) The value of x is \( 28 \text{ cm} \). The altitude PT divides the base into two segments, which we can find individually.In simple words: First, we use the known side PQ and the altitude PT in the left right-angled triangle to find the segment QT. Next, we use the side PR and altitude PT in the right right-angled triangle to find the segment TR. Finally, we add QT and TR together to get the full length of QR, which is x.

🎯 Exam Tip: For problems involving altitudes, break the main triangle into two right-angled triangles. Apply Pythagoras theorem to each smaller triangle to find the unknown segments before combining them.

 

Question 18. In the figure, AB = 8 cm, BC = 6 cm, AC = 3 cm, and the angle ADC = 90°. Calculate CD.

Answer: Given: \( AB = 8 \text{ cm} \), \( BC = 6 \text{ cm} \), \( AC = 3 \text{ cm} \), and \( \angle ADC = 90^\circ \). We need to calculate the length of CD. Let \( CD = x \text{ cm} \). From the figure and given information, D is a point on the line containing BC, and C is between B and D. So, \( BD = BC + CD = (6 + x) \text{ cm} \). 1. In right-angled triangle ABD (at D): \( AB^2 = BD^2 + AD^2 \) (By Pythagoras Theorem) \( \implies (8)^2 = (6 + x)^2 + AD^2 \) \( \implies 64 = (6 + x)^2 + AD^2 \) \( \implies AD^2 = 64 - (6 + x)^2 \) ... (Equation i) 2. In right-angled triangle ACD (at D): \( AC^2 = CD^2 + AD^2 \) (By Pythagoras Theorem) \( \implies (3)^2 = x^2 + AD^2 \) \( \implies 9 = x^2 + AD^2 \) \( \implies AD^2 = 9 - x^2 \) ... (Equation ii) 3. Equating (i) and (ii) (since both represent \( AD^2 \)): \( 64 - (6 + x)^2 = 9 - x^2 \) \( \implies 64 - (36 + 12x + x^2) = 9 - x^2 \) \( \implies 64 - 36 - 12x - x^2 = 9 - x^2 \) \( \implies 28 - 12x = 9 \) \( \implies -12x = 9 - 28 \) \( \implies -12x = -19 \) \( \implies x = \frac{-19}{-12} \) \( \implies x = \frac{19}{12} \) \( \implies x = 1\frac{7}{12} \text{ cm} \) So, \( CD = 1\frac{7}{12} \text{ cm} \). By using the altitude and the Pythagoras theorem in two triangles, we can set up an equation to find the unknown length.In simple words: We call the unknown length CD as 'x'. Then we use Pythagoras theorem twice: once for the larger triangle ABD and once for the smaller triangle ACD, both involving the height AD. By making AD squared the subject in both equations and setting them equal, we create an algebraic equation to solve for x.

🎯 Exam Tip: When a shared altitude is involved, express its square in terms of the other sides from two different right-angled triangles. Equating these expressions allows you to form a solvable equation for the unknown length.

 

Question 19. In the figure, the angle BAC is a right angle and AD is perpendicular to BC; AB = 4 cm, AC = 3 cm and BD = x. Calculate x.

Answer: Given: In \( \triangle ABC \), \( \angle BAC = 90^\circ \). AD is perpendicular to BC (\( AD \perp BC \)). \( AB = 4 \text{ cm} \), \( AC = 3 \text{ cm} \), \( BD = x \text{ cm} \). We need to calculate the value of x. 1. In right-angled triangle ABC (at A): \( BC^2 = AB^2 + AC^2 \) (By Pythagoras Theorem) \( \implies BC^2 = (4)^2 + (3)^2 \) \( \implies BC^2 = 16 + 9 \) \( \implies BC^2 = 25 \) \( \implies BC = \sqrt{25} \) \( \implies BC = 5 \text{ cm} \) 2. Now, find DC in terms of x: \( DC = BC - BD = (5 - x) \text{ cm} \) 3. In right-angled triangle ABD (at D): \( AD^2 = AB^2 - BD^2 \) (By Pythagoras Theorem) \( \implies AD^2 = (4)^2 - x^2 \) \( \implies AD^2 = 16 - x^2 \) ... (Equation i) 4. In right-angled triangle ACD (at D): \( AD^2 = AC^2 - DC^2 \) (By Pythagoras Theorem) \( \implies AD^2 = (3)^2 - (5 - x)^2 \) \( \implies AD^2 = 9 - (25 - 10x + x^2) \) ... (Equation ii) 5. Equating (i) and (ii) (since both represent \( AD^2 \)): \( 16 - x^2 = 9 - (25 - 10x + x^2) \) \( \implies 16 - x^2 = 9 - 25 + 10x - x^2 \) \( \implies 16 = -16 + 10x \) \( \implies 16 + 16 = 10x \) \( \implies 32 = 10x \) \( \implies x = \frac{32}{10} \) \( \implies x = 3.2 \text{ cm} \) So, \( BD = 3.2 \text{ cm} \). This problem shows how to find segment lengths within a right triangle when an altitude is drawn.In simple words: First, we find the full length of the base BC using Pythagoras in the main triangle ABC. Then, we use Pythagoras in the two smaller triangles (ABD and ACD) to write an equation for AD squared. By setting these two expressions for AD squared equal, we can solve for x.

🎯 Exam Tip: When an altitude divides a right triangle into two smaller right triangles, use the Pythagoras theorem in all three triangles. This often leads to equations that can be solved for unknown lengths.

 

Question 20. In the figure, the angle BAD and ADC are right angles and AX is parallel to BC. If AB = BC = 5 cm, and DC = 8 cm. Calculate the area of ABCX.

Answer: Given: In the figure, \( \angle BAD = 90^\circ \) and \( \angle ADC = 90^\circ \). This means AB is parallel to DC, and AD is perpendicular to both AB and DC. Thus, ABCD is a right-angled trapezium. Also given: \( AB = BC = 5 \text{ cm} \) and \( DC = 8 \text{ cm} \). A line AX is drawn parallel to BC (\( AX \parallel BC \)). Since \( AB \parallel XC \) (both perpendicular to AD), this forms a parallelogram ABXC. Since \( AB = BC = 5 \text{ cm} \) and \( AX \parallel BC \), ABXC is a rhombus (a parallelogram with all sides equal). From the properties of a rhombus: \( AX = BC = 5 \text{ cm} \) \( XC = AB = 5 \text{ cm} \) Now, calculate the length of DX: \( DX = DC - XC = 8 - 5 = 3 \text{ cm} \) Consider the right-angled triangle ADX (at D, because \( \angle ADC = 90^\circ \)): \( AX^2 = AD^2 + DX^2 \) (By Pythagoras Theorem) \( \implies (5)^2 = AD^2 + (3)^2 \) \( \implies 25 = AD^2 + 9 \) \( \implies AD^2 = 25 - 9 \) \( \implies AD^2 = 16 \) \( \implies AD = \sqrt{16} \) \( \implies AD = 4 \text{ cm} \) This length AD is the height of the rhombus ABXC if we consider AX or BC as the base. Now, calculate the area of ABCX. Since ABXC is a rhombus, its area can be calculated as Base \( \times \) Height. Using BC as the base and AD as the corresponding height: Area of ABCX = \( BC \times AD \) \( \implies \) Area = \( 5 \text{ cm} \times 4 \text{ cm} \) \( \implies \) Area = \( 20 \text{ cm}^2 \) The area of the quadrilateral ABCX is \( 20 \text{ cm}^2 \). By carefully identifying the geometric shapes formed, we can apply the correct formulas.In simple words: First, we recognize that ABCD is a special shape (a right-angled trapezium). Then, we see that AX being parallel to BC makes ABXC a parallelogram, and because AB equals BC, it's actually a rhombus. We find the height AD of this rhombus using Pythagoras theorem in triangle ADX. Finally, we multiply the base (BC) by the height (AD) to get the area of the rhombus.

🎯 Exam Tip: Always be precise in identifying geometric shapes based on given properties and constructions. A rhombus has equal sides, and its area can be found by multiplying its base by its perpendicular height.