GSEB Class 7 Maths Solutions Chapter 6 The Triangles and Its Properties InText Questions

Try These (Page 113)

Question 1. Write the six elements (i.e. the 3 sides and the 3 angles) of \( \triangle ABC \).
Answer: Six elements of \( \triangle ABC \) are: \( \angle A \), \( \angle B \), \( \angle C \), \( \overline{AB} \), \( \overline{BC} \) and \( \overline{CA} \).
In simple words: A triangle has six basic parts: three corners where lines meet (called angles) and three straight lines that make up its edges (called sides).

Exam Tip: Remember that a triangle always has exactly three angles and three sides. Each side is opposite to an angle, and each angle is formed by two sides.

Question 2. Write the:
(i) Side opposite to the vertex Q of \( \triangle PQR \)
(ii) Angle opposite to the side LM of \( \triangle LMN \)
(iii) Vertex opposite to the side RT of \( \triangle RST \)
Answer:
(i) Side opposite to the vertex Q of \( \triangle PQR \) is \( \overline{PR} \).
(ii) Angle opposite to the side LM of \( \triangle LMN \) is \( \angle N \).
(iii) Vertex opposite to the side RT of \( \triangle RST \) is S.
In simple words: To find the opposite side from a corner, just look across the triangle. To find the opposite angle from a side, look across the triangle at the corner that isn't touching that side.

Exam Tip: Always visualize the triangle. The side opposite a vertex does not touch that vertex. The angle opposite a side does not have that side as one of its arms.

Question 3. Look at figures and classify each of the triangles according to its –
(a) Sides
(b) Angles
Answer:
(i) (a) In the first figure, \( \overline{AC} = \overline{BC} = 8 \) cm. So, \( \triangle ABC \) is an isosceles triangle.
(b) Since all angles of \( \triangle ABC \) are less than 90°. It is an acute triangle.
(ii) (a) In the second figure, \( PQ \neq QR \neq RP \). Thus, \( \triangle PQR \) is a scalene triangle.
(b) Since \( \angle R = 90^\circ \). Therefore, \( \triangle PQR \) is a right triangle.
(iii) (a) In the third figure, in \( \triangle LMN \), \( LN = MN = 7 \) cm. So, \( \triangle LMN \) is an isosceles triangle.
(b) In \( \triangle LMN \), \( \angle N > 90^\circ \). Hence, \( \triangle LMN \) is an obtuse triangle.
(iv) (a) In the fourth figure, in \( \triangle RST \), \( RS = ST = TR = 5.2 \) cm. So, \( \triangle RST \) is an equilateral triangle.
(b) All the angles of \( \triangle RST \) are acute. Thus, it is an acute triangle.
(v) (a) In the fifth figure, in \( \triangle ABC \), \( \overline{AB} = \overline{BC} = 3 \) cm. So, \( \triangle ABC \) is an isosceles triangle.
(b) In \( \triangle ABC \), \( \angle B > 90^\circ \). Therefore, it is an obtuse triangle.
(vi) (a) In the sixth figure, in \( \triangle PQR \), \( \overline{PQ} = \overline{QR} = 6 \) cm. So, \( \triangle PQR \) is an isosceles triangle.
(b) In \( \triangle PQR \), \( \angle Q = 90^\circ \). Therefore, it is a right triangle.
In simple words: Triangles are named by their sides (equal sides or all different) and their angles (all small, one big, or one square). Look at the lengths of the sides and the size of the angles to tell what kind it is.

Exam Tip: Classifying triangles requires checking both side lengths (equal sides for isosceles, all equal for equilateral, all different for scalene) and angle measures (all acute for acute, one right angle for right, one obtuse angle for obtuse).

Think, Discuss and Write (Page 114)

Question 1. How many medians can a triangle have?
Answer: Since a triangle has three sides, and by joining the mid-point of each side to the opposite corner, we can draw three line segments. Thus, a triangle can have three medians.
In simple words: A triangle has three medians, one from each corner to the middle of the opposite side.

Exam Tip: A median always connects a vertex to the midpoint of the opposite side. There are always three such lines in any triangle.

Question 2. Does a median lie wholly in the interior of the triangle? (If you think that this is not true, draw a figure to show such a case.)
Answer: Yes, a median lies entirely inside the triangle.
In simple words: Every median of a triangle is always found completely within the triangle's boundaries.

Exam Tip: Understand the definition of a median: it connects a vertex to the midpoint of the opposite side, which by definition keeps it within the triangle's shape.

Think, Discuss and Write (Page 115)

Question 1. How many altitudes can a triangle have?
Answer: An altitude can be drawn from each vertex of a triangle. So, a triangle can have three altitudes.
In simple words: A triangle has three altitudes, one from each corner drawn straight down to the opposite side (or its extended line).

Exam Tip: An altitude is a perpendicular line segment from a vertex to the opposite side (or its extension). Like medians, there are always three altitudes in a triangle.

Question 2. Draw rough sketches of altitudes from A to \( \overline{BC} \) for the following triangles.
Answer: We have \( \overline{AL} \) as an altitude in each case as shown below:

Exam Tip: Be careful with obtuse triangles – the altitude from an acute vertex will fall outside the triangle, requiring the base to be extended.

Question 3. Will an altitude always lie in the interior of a triangle? If you think that this need not be true, draw a rough sketch to show such a case.
Answer: No, the altitude does not always lie in the interior of a triangle. For example, in the adjoining figure, the altitude AD is not within \( \triangle ABC \).

In simple words: No, a triangle's height isn't always inside it. For triangles with a wide angle (obtuse), the height from one of the sharper corners can fall outside.

Exam Tip: This is a common point of confusion. Remember that for obtuse triangles, two of the three altitudes will lie outside the triangle, on the extensions of the sides.

Question 4. Can you think of a triangle in which two altitudes of the triangle are two of its sides?
Answer: Yes, a right triangle has two of its sides as its altitudes. In a right triangle ABC, right-angled at B, \( \overline{AB} \) and \( \overline{BC} \) are its altitudes.

In simple words: Yes, in a triangle with a square corner (a right triangle), the two sides that form that square corner are actually its heights (altitudes).

Exam Tip: This property is unique to right triangles. The legs of a right triangle serve as altitudes from the acute vertices to the opposite legs.

Question 5. Can the altitude and median be same for a triangle?
Answer: Yes, a median and altitude can be the same line segment. In the adjoining figure, AL is an altitude as well as a median of \( \triangle ABC \).

In simple words: Yes, a height (altitude) and a middle-line (median) can be the same in a triangle, but only if that triangle is special, like an isosceles triangle from its unique vertex.

Exam Tip: An altitude and a median coincide only in isosceles or equilateral triangles, drawn from the vertex angle to the base.

Think, Discuss and Write (Page 117)

Question 1. Exterior angles can be formed for a triangle in many ways. Three of them are shown here. There are three more ways of getting exterior angles. Try to produce those rough sketches.
Answer: Three other exterior angles can be formed by extending the other sides at each vertex, as illustrated below:


In simple words: Besides the ways already shown, you can make an outside angle by extending any side of the triangle, creating a new angle next to one of the inside angles.

Exam Tip: An exterior angle is formed by one side of a triangle and the extension of an adjacent side. There are always two exterior angles at each vertex, and these two are vertically opposite, thus equal.

Question 2. Are the exterior angles formed at each vertex of a triangle equal?
Answer: No.
In simple words: The outside angles at each corner of a triangle are not always the same size.

Exam Tip: Exterior angles at different vertices are generally not equal, unless the triangle has specific symmetries (e.g., an equilateral triangle where all exterior angles are 120°).

Question 3. What can you say about the sum of an exterior angle of a triangle and its adjacent interior angle?
Answer: The sum of an exterior angle and its adjacent interior angle forms a linear pair.
\( \implies \) [Exterior angle] + [Interior adjacent angle] = 180°.
In simple words: An outside angle and the inside angle next to it always add up to 180 degrees because they form a straight line.

Exam Tip: Always remember that an exterior angle and its adjacent interior angle are supplementary, meaning their sum is 180 degrees. This is a fundamental property of angles on a straight line.

Think, Discuss and Write (Page 118)

Question 1. What can you say about each of the interior opposite angles, when the exterior angle is –
(i) a right angle?
(ii) an obtuse angle?
(iii) an acute angle?
Answer:
(i) Each of the interior opposite angles is an acute angle.
(ii) At least one of the interior opposite angles must be an acute angle.
(iii) Each of the interior opposite angles is an acute angle.
In simple words: (i) If the outside angle is a square angle, then both inside opposite angles must be sharp angles. (ii) If the outside angle is a wide angle, at least one of the inside opposite angles must be a sharp angle. (iii) If the outside angle is a sharp angle, then both inside opposite angles must be sharp angles too.

Exam Tip: The exterior angle theorem states that an exterior angle is equal to the sum of its two interior opposite angles. Use this theorem to deduce properties of the interior opposite angles based on the type of exterior angle.

Question 2. Can the exterior angle of a triangle be a straight angle?
Answer: No.
In simple words: An outside angle of a triangle cannot be a perfectly flat, straight line angle.

Exam Tip: A straight angle (180°) for an exterior angle would imply the adjacent interior angle is 0°, which means the "triangle" would collapse into a line, which is not a true triangle.

Try These (Page 118)

Question 1. An exterior angle of a triangle is of measure 70° and one of its interior opposite angles is of measure 25°. Find the measure of the other interior opposite angle.
Answer:

Exterior angle = 70°
Interior opposite angles are 25° and \(x\).
\( \implies x + 25^\circ = 70^\circ \) (Using the exterior angle property of a triangle)
or \( x = 70^\circ - 25^\circ = 45^\circ \)
\( \implies \) The required interior opposite angle = 45°.
In simple words: The outside angle is 70 degrees. One of the inside angles far away is 25 degrees. To find the other inside angle far away, subtract 25 from 70, which gives 45 degrees.

Exam Tip: Always state the exterior angle property clearly: "An exterior angle of a triangle is equal to the sum of its two interior opposite angles." This will help you remember the formula and score full marks.

Question 2. The two interior opposite angles of an exterior angle of triangle are 60° and 80°. Find the measure of the exterior angle.
Answer: Interior angles are 60° and 80°.
\( \implies \) [Exterior angle] = [Sum of the interior opposite angles]
\( \implies \) [Exterior angle] = 60° + 80° = 140°
In simple words: If two inside angles far from an outside angle are 60 and 80 degrees, then the outside angle is simply their total, which is 140 degrees.

Exam Tip: Clearly write down the exterior angle property before applying it. This shows your understanding of the theorem.

Question 3. Is something wrong in this diagram? Comment.
Answer: We know that an exterior angle of a triangle is equal to the sum of interior opposite angles.

Here, the exterior angle is 50°, and the interior opposite angles are 50° and 50°.
\( \implies \) This triangle cannot be formed. (Because \( 50^\circ \neq 50^\circ + 50^\circ \))
In simple words: Yes, there is a mistake in the picture. An outside angle should be equal to the total of the two inside angles that are not next to it. Here, 50 is not equal to 50 plus 50, so this triangle cannot exist as drawn.

Exam Tip: Always double-check diagrams against fundamental geometric theorems. If a diagram violates a basic rule like the exterior angle property, it indicates an error in the diagram itself.

Try These (Page 122)

Question 1. Two angles of a triangle are 30° and 80°. Find the third angle.
Answer: Let the third angle be \(x\).
\( \implies \) Using the angle sum property of a triangle, we have
\( 30^\circ + 80^\circ + x = 180^\circ \)
or \( 110^\circ + x = 180^\circ \)
or \( x = 180^\circ - 110^\circ = 70^\circ \)
\( \implies \) The measure of the required third angle is 70°.
In simple words: The three angles in any triangle always add up to 180 degrees. If two angles are 30 and 80 degrees, their sum is 110. Subtract 110 from 180 to find the last angle, which is 70 degrees.

Exam Tip: Always clearly state "angle sum property of a triangle" when using it, as this is a key geometric theorem and shows your understanding.

Question 2. One of the angles of a triangle is 80° and the other two angles are equal. Find the measure of each of the equal angles.
Answer: Let each of the equal angles be \(x\).
Using the angle sum property of a triangle, we have
\( x + x + 80^\circ = 180^\circ \)
or \( 2x + 80^\circ = 180^\circ \)
or \( 2x = 180^\circ - 80^\circ = 100^\circ \)
or \( \frac{2x}{2} = \frac{100^\circ}{2} \)
or \( x = 50^\circ \)
\( \implies \) The required measure of each of the equal angles is 50°.
In simple words: If one angle is 80 degrees and the other two are the same, first take 80 from 180. That leaves 100 degrees for the two equal angles. Then, divide 100 by 2, which means each of those equal angles is 50 degrees.

Exam Tip: When dealing with isosceles triangles (which have two equal angles), represent the equal angles with the same variable (e.g., \(x\)) to simplify calculations using the angle sum property.

Question 3. The three angles of a triangle are in the ratio 1:2:1. Find all the angles of the triangle. Classify the triangle in two different ways.
Answer: Let the angles of the triangle be \(x\), \(2x\), \(x\).
\( \implies \) Using the angle sum property, we have
\( x + 2x + x = 180^\circ \)
or \( 4x = 180^\circ \)
or \( \frac{4x}{4} = \frac{180^\circ}{4} \)
or \( x = 45^\circ \)
\( \implies 2x = 2 \times 45^\circ = 90^\circ \)
Thus, the three angles of the triangle are 45°, 90°, 45°.
(i) Its two angles are equal.
\( \implies \) It is an isosceles triangle.
(ii) Its one angle is 90°.
\( \implies \) It is a right-angled triangle.
In simple words: The angles are like parts: 1 part, 2 parts, 1 part. Total 4 parts. Since all angles add up to 180, each part is 45 degrees. So the angles are 45, 90, and 45 degrees. This means it's an isosceles triangle (two angles are same) and a right-angled triangle (one angle is 90 degrees).

Exam Tip: When angles are given in a ratio, represent them as \(x\), \(2x\), \(3x\), etc., then use the angle sum property to find the value of \(x\) and calculate each angle. Then, use both angle and side properties to classify the triangle.

Think, Discuss and Write (Page 122)

Question 1. Can you have a triangle with two right angles?
Answer: No, because the sum of two right angles is 180°. When adding the measure of the third angle, the total of three angles will be more than 180°, which is not true for a triangle.
In simple words: No, because two 90-degree angles already add up to 180 degrees. If you add a third angle, the total would be more than 180, which is impossible for a triangle.

Exam Tip: Always remember that the sum of angles in any triangle is exactly 180°. This fundamental rule helps eliminate impossible triangle configurations.

Question 2. Can you have a triangle with two obtuse angles?
Answer: No, because an obtuse angle has its measure more than 90°. Therefore, the sum of two obtuse angles is greater than 180°, which is not possible in a triangle.
In simple words: No, because each wide angle is bigger than 90 degrees. So, two wide angles would already add up to more than 180 degrees, which cannot happen in a triangle.

Exam Tip: Similar to two right angles, two obtuse angles (each greater than 90°) would also sum to more than 180°, violating the triangle angle sum property.

Question 3. Can you have a triangle with two acute angles?
Answer: Yes, because the sum of two acute angles can be less than 180°. Therefore, a triangle can have two acute angles.
In simple words: Yes, a triangle can have two sharp angles, as their total could be less than 180 degrees, leaving room for a third angle.

Exam Tip: All triangles must have at least two acute angles. The third angle can be acute, right, or obtuse.

Question 4. Can you have a triangle with all the three angles greater than 60°?
Answer: No, because the sum of three angles (each of them being greater than 60°) is greater than 180°, which is not possible in a triangle.
In simple words: No, if all three angles were bigger than 60 degrees, their total would be more than 180 degrees, which is not allowed for a triangle.

Exam Tip: If all angles are greater than 60°, their sum will exceed 180°. The only triangle where all angles are equal is an equilateral triangle, where each angle is exactly 60°.

Question 5. Can you have a triangle with all the three angles equal to 60°?
Answer: Yes, because the sum of three angles, each being equal to 60°, is 180°, which is true for a triangle.
In simple words: Yes, if all three angles are 60 degrees, they add up to exactly 180 degrees, which is perfect for a triangle. This kind of triangle is called an equilateral triangle.

Exam Tip: An equilateral triangle has three equal angles, each measuring 60°. This is a valid and common type of triangle.

Question 6. Can you have a triangle with all the three angles less than 60°?
Answer: No, because the sum of the three angles, each being less than 60°, is less than 180°, which is not possible in a triangle.
In simple words: No, if all three angles were smaller than 60 degrees, their total would be less than 180 degrees, which cannot happen in a triangle.

Exam Tip: If all angles are less than 60°, their sum will be less than 180°, which means the figure cannot close to form a triangle.

Try These (Page 123)

Question 1. Find angle x in each figure:
Answer:
(i) In the figure, two sides of the triangle are equal.
\( \implies \) The base angles opposite to equal sides are equal.
So, \( x = 40^\circ \).
Thus, the required value of \( x \) is 40°.

(ii) In the figure, two sides of the triangle are equal.
\( \implies \) The base angles opposite to equal sides are equal.
The other base angle is 45°.
Now, the sum of three angles = \( x + 45^\circ + 45^\circ \)
But the sum of three angles of a triangle is 180°.
\( \implies x + 90^\circ = 180^\circ \)
or \( x = 180^\circ - 90^\circ = 90^\circ \).
Thus, the required value of \( x \) is 90°.

(iii) In the figure, two sides of the triangle are equal.
\( \implies \) The base angles opposite to equal sides are equal, i.e., \( x = 50^\circ \).
Thus, the required value of \( x \) is 50°.

(iv) In the figure, two sides of the triangle are equal.
\( \implies \) The base angles opposite to the equal sides are equal.
The other base angle is \( x \).
Now, sum of three angles of a triangle = 180°
or \( x + x + 100^\circ = 180^\circ \)
or \( 2x + 100^\circ = 180^\circ \)
or \( 2x = 180^\circ - 100^\circ = 80^\circ \)
or \( \frac{2x}{2} = \frac{80^\circ}{2} \)
Thus, the required value of \( x \) is 40°.

(v) In the figure, two sides of the triangle are equal and it is a right triangle.
\( \implies \) Its base angles opposite to the equal sides are equal.
The other base angle = \( x \).
Sum of the three angles of a triangle is 180°.
\( \implies x + x + 90^\circ = 180^\circ \)
or \( 2x + 90^\circ = 180^\circ \)
or \( 2x = 180^\circ - 90^\circ = 90^\circ \)
or \( \frac{2x}{2} = \frac{90^\circ}{2} \)
or \( x = 45^\circ \).
Thus, the required value of \( x \) is 45°.

(vi) In the figure, the two sides of the triangle are equal.
\( \implies \) The base angles opposite to the equal sides are equal.
Since one of the base angle is \( x \).
The other base angle is also \( x \).
Sum of the three angles of a triangle is 180°.
\( \implies x + x + 40^\circ = 180^\circ \)
or \( 2x + 40^\circ = 180^\circ \)
or \( 2x = 180^\circ - 40^\circ = 140^\circ \)
or \( \frac{2x}{2} = \frac{140^\circ}{2} \)
or \( x = 70^\circ \).
Thus, the required value of \( x \) is 70°.

(vii) In the figure, two sides of the triangle are equal.
\( \implies \) The base angles opposite to equal sides are equal.
One of the base angle = \( x \).
The other base angle = \( x \).
Now, \( x \) and 120° form a linear pair = 180°.
\( \implies x + 120^\circ = 180^\circ \)
or \( x = 180^\circ - 120^\circ = 60^\circ \)
Thus the value of \( x = 60^\circ \).

(viii) In the figure, two sides of the triangle are equal.
\( \implies \) The base angles opposite to the equal sides are equal.
Since, one of the base angles = \( x \).
The other base angle = \( x \).
Since, exterior angle is equal to sum of the interior opposite angles.
\( \implies x + x = 110^\circ \)
or \( 2x = 110^\circ \)
or \( x = \frac{110^\circ}{2} = 55^\circ \).

(ix) Two sides of the triangle are equal.
\( \implies \) The base angles opposite to the equal sides are equal.
Since, one of the base angles = \( x \).
The other base angle = \( x \).
Also, the vertically opposite angles 30° and \( x \) are equal.
\( \implies x = 30^\circ \).

In simple words: For each shape, we use the rules of triangles. If two sides are equal, the angles opposite them are also equal. Also, angles on a straight line add up to 180 degrees, and opposite angles when two lines cross are equal. We use these rules to find the unknown angle 'x'.

Exam Tip: For problems involving angles in triangles, recall key properties: angle sum property (180°), isosceles triangle property (angles opposite equal sides are equal), linear pair property (angles on a straight line sum to 180°), and vertically opposite angles (equal).

Question 2. Find x and y in each figure:
Answer:
(i) In the figure, two sides of the triangle are equal.
\( \implies \) The base angles opposite to the equal sides are equal.
So the base angles are \( y \), and angle = \( y \).
Now, \( y \) and 120° form a linear pair,
\( \implies y + 120^\circ = 180^\circ \)
or \( y = 180^\circ - 120^\circ = 60^\circ \).
Now, sum of the three angles = 180°
\( \implies x + y + y = 180^\circ \)
or \( x + 60^\circ + 60^\circ = 180^\circ \)
or \( x + 120^\circ = 180^\circ \)
or \( x = 180^\circ - 120^\circ = 60^\circ \).
Thus, \( x = 60^\circ \) and \( y = 60^\circ \).

(ii) In the given figure, two sides of the triangle are equal.
\( \implies \) The base angles opposite to equal sides are equal.
Since, one of the base angles is \( x \),
The other base angle = \( x \).
Also, the triangle is a right-angled triangle.
\( \implies \) Third angle of the triangle = 90°.
Now, sum of the three angles of the triangle = 180°.
\( \implies x + x + 90^\circ = 180^\circ \)
or \( 2x + 90^\circ = 180^\circ \)
or \( 2x = 180^\circ - 90^\circ = 90^\circ \)
or \( \frac{2x}{2} = \frac{90^\circ}{2} \)
or \( x = 45^\circ \).
Now, \( x \) and \( y \) form a linear pair,
\( \implies x + y = 180^\circ \)
or \( y = 180^\circ - 45^\circ = 135^\circ \).
Thus, \( x = 45^\circ \) and \( y = 135^\circ \).

(iii) In the figure, two sides of the triangle are equal.
\( \implies \) Base angles are \( x \) and \( x \).
The third angle of the triangle = The vertically opposite angle of 92° = 92°.
Now, sum of the three angles of the triangle = 180°.
\( \implies x + x + 92^\circ = 180^\circ \)
or \( 2x + 92^\circ = 180^\circ \)
or \( 2x = 180^\circ - 92^\circ = 88^\circ \)
or \( \frac{2x}{2} = \frac{88^\circ}{2} \)
or \( x = 44^\circ \).
Now, \( x \) and \( y \) form a linear pair,
\( \implies x + y = 180^\circ \)
or \( y = 180^\circ - 44^\circ = 136^\circ \).
Thus, \( x = 44^\circ \) and \( y = 136^\circ \).

In simple words: For each diagram, we first use the fact that angles opposite equal sides are equal. Then we use that angles on a straight line add to 180 degrees, and the total of angles inside a triangle is 180 degrees. Also, if lines cross, opposite angles are equal. We combine these rules to figure out the values of 'x' and 'y'.

Exam Tip: For problems with multiple unknown angles (like x and y), identify all geometric relationships: linear pairs, vertically opposite angles, and properties of isosceles/right triangles. Solve step-by-step using these relations.

Try These (Page 129)

Question 1. Find the unknown length x in the following figures.

Answer:
(i) In the given right triangle, the longest side (hypotenuse) is \( x \).
Using the Pythagoras theorem, we have:
\( 3^2 + 4^2 = x^2 \)
\( 9 + 16 = x^2 \)
\( 25 = x^2 \)
\( x = \sqrt{25} \)
\( \implies x = 5 \)
The needed value for \( x \) is 5.
(ii) In the given right-angled triangle:
By using the Pythagoras rule, we get:
\( x^2 = 6^2 + 8^2 \)
\( x^2 = 36 + 64 \)
\( x^2 = 100 \)
\( x^2 = 10^2 \)
\( \implies x = 10 \)
The value of \( x \) is 10.
(iii) In the given right-angled triangle:
By using the Pythagoras theorem, we have:
\( x^2 = 8^2 + 15^2 \)
\( x^2 = 64 + 225 \)
\( x^2 = 289 \)
\( x^2 = 17^2 \)
\( \implies x = 17 \) cm
The required value for \( x \) is 17 cm.
(iv) By applying the Pythagoras property in the given right triangle, we find:
\( x^2 = 7^2 + 24^2 \)
\( x^2 = 49 + 576 \)
\( x^2 = 625 \)
\( x^2 = 25^2 \)
\( \implies x = 25 \)
The value of \( x \) is 25.
(v) The given figure can be shown with labels as follows:
In the right triangle I, by using the Pythagoras rule, we find:
\( y^2 + 12^2 = 37^2 \)
\( y^2 + 144 = 1369 \)
\( y^2 = 1369 - 144 \)
\( y^2 = 1225 \)
\( y^2 = 35^2 \)
\( \implies y = 35 \)
In the right triangle II, by using the Pythagoras rule, we find:
\( (x - y)^2 + 12^2 = 37^2 \)
\( (x - 35)^2 + 144 = 1369 \)
\( (x - 35)^2 = 1369 - 144 \)
\( (x - 35)^2 = 1225 \)
\( (x - 35)^2 = (35)^2 \)
\( x - 35 = 35 \)
\( x = 35 + 35 \)
\( x = 70 \)
The value of \( x \) is 70.
(vi) By using the Pythagoras property in the right triangle ABC, we get:
\( 12^2 + 5^2 = x^2 \)
\( 144 + 25 = x^2 \)
\( 169 = x^2 \)
\( (13)^2 = x^2 \)
\( \implies x = 13 \)
The value of \( x \) is 13.
In simple words: For each right-angled triangle, we use the Pythagoras theorem where the square of the hypotenuse equals the sum of the squares of the other two sides to find the unknown length.

Exam Tip: Remember the Pythagorean triplet formula \( a^2 + b^2 = c^2 \) for right-angled triangles; this helps to quickly calculate unknown side lengths.

Think, Discuss and Write (Page 131)

Question 1. Which is the longest side in the triangle PQR, right-angled at P?
Answer: The right angle is at vertex P. The sides that form this right angle are PQ and PR. The hypotenuse is the side opposite the right angle, which is QR. Therefore, the longest side is QR.
In simple words: In a right-angled triangle, the longest side is always the one opposite the 90-degree angle, which is called the hypotenuse. For triangle PQR with a right angle at P, QR is the hypotenuse.

Exam Tip: Always identify the right angle first; the side opposite it is the hypotenuse and will always be the longest side of the right triangle.

Question 2. Which is the longest side in the triangle ABC, right-angled at B?
Answer: The right angle is at vertex B. The sides that make up the right angle are AB and BC. The hypotenuse is the side opposite the right angle, which is AC. Hence, the longest side is AC.
In simple words: When triangle ABC has a right angle at B, the longest side is AC because it's the side across from the right angle.

Exam Tip: Visualizing the triangle helps; the hypotenuse 'stretches' across the largest angle (90 degrees), making it the longest side.

Question 3. Which is the longest side of a right triangle?
Answer: In any right triangle, the longest side is its hypotenuse.
In simple words: The hypotenuse is always the longest side in a right-angled triangle.

Exam Tip: Define the hypotenuse as the side opposite the right angle to score full marks.

Question 4. 'The diagonal of a rectangle produce by itself the same area as produced by its length and breadth " This is Baudhayan Theorem. Compare it with the Pythagoras property.
Answer: In the given figure, ABCD is a rectangle, and BD is one of its diagonals.
According to the Baudhayan Theorem, we have:
\( (\text{Diagonal})^2 = (\text{Length})^2 + (\text{Breadth})^2 \)
This means, \( (DB)^2 = (CD)^2 + (BC)^2 \) ... (1)
Now, triangle BCD is a right triangle, and DB serves as its hypotenuse.
Using the Pythagoras property, we find:
\( (DB)^2 = \) Sum of the squares of the legs
Or \( (DB)^2 = (DC)^2 + (CB)^2 \)
Or \( (DB)^2 = (CD)^2 + (BC)^2 \) ... (2)
From equations (1) and (2), we can observe that the Baudhayan Theorem and the Pythagoras property are the same.
In simple words: The Baudhayan Theorem says a rectangle's diagonal squared is its length squared plus its breadth squared. This is exactly what the Pythagoras theorem states for a right-angled triangle, where the diagonal acts as the hypotenuse. So, they both explain the same idea in geometry.

Exam Tip: Clearly state both theorems' formulas and show how they lead to the same mathematical expression to prove their equivalence.