GSEB Class 7 Maths Solutions Chapter 6 The Triangles and Its Properties Exercise 6.5

Question 1. PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
Answer: In the correct triangle PQR, by applying the Pythagoras rule, we find:
\( \text{QR}^2 = \text{PR}^2 + \text{PQ}^2 \)
\( \implies \text{QR}^2 = 24^2 + 10^2 \)
\( \implies \text{QR}^2 = 576 + 100 \)
\( \implies \text{QR}^2 = 676 \)
\( \implies \text{QR}^2 = 26^2 \)
\( \implies \text{QR} = 26 \)
Therefore, the side QR measures 26 cm.

Exam Tip: Remember to correctly identify the hypotenuse (the side opposite the right angle) before applying the Pythagoras theorem. It's always the longest side.

Question 2. ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.
Answer: In this right triangle, applying the Pythagoras rule, we find:
\( \text{AC}^2 + \text{BC}^2 = \text{AB}^2 \)
\( \implies 7^2 + x^2 = 25^2 \)
\( \implies 49 + x^2 = 625 \)
\( \implies x^2 = 625 - 49 \)
\( \implies x^2 = 576 \)
\( \implies x^2 = 24^2 \)
\( \implies x = 24 \)
Therefore, side BC measures 24 cm.

Exam Tip: Always draw a clear diagram to visualize the triangle and label the given sides and the unknown side 'x'. This helps prevent errors.

Question 3. A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance 'a'. Find the distance of the foot of the ladder from the wall.
Answer: The ladder's base is 'a' meters from the wall. By using the Pythagoras theorem, we get:
\( a^2 + 12^2 = 15^2 \)
\( \implies a^2 + 144 = 225 \)
\( \implies a^2 = 225 - 144 \)
\( \implies a^2 = 81 \)
\( \implies a^2 = 9^2 \)
\( \implies a = 9 \, \text{m} \)
So, the necessary distance of the ladder's base from the wall is 9 m.

Exam Tip: Questions involving ladders leaning against walls often form a right-angled triangle. Identify the hypotenuse (the ladder itself) and the two legs (wall height and distance from wall).

Question 4. Which of the following can be the sides of a right triangle? In the case of right-angled triangles, identify the right angles.
Answer:
(i) 2.5 cm, 6.5 cm, 6 cm
The greatest side measures 6.5 cm. Then, \( (2.5)^2 + (6)^2 = 6.25 + 36 = 42.25 \), which is equal to \( (6.5)^2 \). Therefore, these given lengths can form the sides of a right triangle. Clearly, the right angle is found between the sides measuring 2.5 cm and 6 cm.
(ii) 2 cm, 2 cm, 5 cm
The greatest side is 5 cm. So, \( 2^2 + 2^2 = 4 + 4 = 8 \). However, 8 is not equal to \( 5^2 \). Therefore, these given lengths cannot form the sides of a right triangle.
(iii) 1.5 cm, 2 cm, 2.5 cm
The greatest side measures 2.5 cm. Now, \( (1.5)^2 + (2)^2 = 2.25 + 4 = 6.25 \). Also, \( (2.5)^2 = 6.25 \). So, \( (1.5)^2 + (2)^2 = (2.5)^2 \). Therefore, these given lengths can form the sides of a right triangle. Clearly, the right angle is located between the sides measuring 1.5 cm and 2 cm.

Exam Tip: To check if sides form a right triangle, use the converse of the Pythagoras theorem: if the square of the longest side equals the sum of the squares of the other two sides, it's a right triangle.

Question 5. A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
Answer: Imagine the tree BC snaps at point C, so that the broken part CD becomes equal to CA. Now, triangle ABC forms a right-angled shape. By applying the Pythagoras theorem, we find:
\( \text{AB}^2 + \text{BC}^2 = \text{AC}^2 \)
\( \implies 12^2 + 5^2 = \text{AC}^2 \)
\( \implies 144 + 25 = \text{AC}^2 \)
\( \implies \text{AC}^2 = 169 \)
\( \implies \text{AC}^2 = 13^2 \)
\( \implies \text{AC} = 13 \, \text{m} \)
Next, the total height of the tree is BD. This equals BC plus CD, which is also BC plus AC since AC and CD are the same length. So, \( 5 \, \text{m} + 13 \, \text{m} = 18 \, \text{m} \). Therefore, the tree's original height is 18 m.

Exam Tip: When a tree breaks and its top touches the ground, the broken part of the tree becomes the hypotenuse of the right triangle formed.

Question 6. Angles Q and R of a \( \triangle \)PQR are 25° and 65°. Write which of the following is true:
(i) \( \text{PQ}^2 + \text{QR}^2 = \text{RP}^2 \)
(ii) \( \text{PQ}^2 + \text{RP}^2 = \text{QR}^2 \)
(iii) \( \text{RP}^2 + \text{QR}^2 = \text{PQ}^2 \)
Answer: In triangle PQR, we know that the sum of angles is 180°:
\( \angle P + \angle Q + \angle R = 180^\circ \)
\( \implies \angle P + 25^\circ + 65^\circ = 180^\circ \)
\( \implies \angle P + 90^\circ = 180^\circ \)
\( \implies \angle P = 180^\circ - 90^\circ \)
\( \implies \angle P = 90^\circ \)
So, triangle PQR is a right-angled triangle, with its right angle at P. Now, the hypotenuse is the side opposite to angle P, which is QR. Using the Pythagoras theorem, we get:
\( \text{QR}^2 = \text{PQ}^2 + \text{RP}^2 \)
Thus, the relation (ii), meaning \( \text{PQ}^2 + \text{RP}^2 = \text{QR}^2 \), holds true.

Exam Tip: For problems involving triangles and angles, always begin by using the angle sum property of a triangle (\( \angle A + \angle B + \angle C = 180^\circ \)). This helps determine if it's a right-angled triangle.

Question 7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Answer: Let the breadth be x cm. In right triangle BAD, we use the Pythagorean theorem:
\( \text{BA}^2 + \text{AD}^2 = \text{BD}^2 \)
\( \implies 40^2 + x^2 = 41^2 \)
\( \implies x^2 = 41^2 - 40^2 \)
\( \implies x^2 = 1681 - 1600 \)
\( \implies x^2 = 81 \)
\( \implies x^2 = 9^2 \)
\( \implies x = 9 \)
Therefore, the breadth is 9 cm.
Now, the perimeter is calculated as 2(length + breadth). Substituting the values, we get \( 2(40 \, \text{cm} + 9 \, \text{cm}) = 2(49 \, \text{cm}) = 98 \, \text{cm} \). Thus, the perimeter of the rectangle comes out to be 98 cm.

Exam Tip: A diagonal in a rectangle divides it into two right-angled triangles. Use the given diagonal and one side as two sides of a right triangle to find the third side using Pythagoras theorem.

Question 8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Answer: Let the given figure be a rhombus where AC and BD are its diagonals, with AC = 30 cm and BD = 16 cm. Since the diagonals of a rhombus cut each other into two equal parts at right angles (they bisect at point O), we know that \( \angle AOB = \angle BOC = \angle COD = \angle DOA = 90^\circ \). Also, \( \text{OA} = \text{OC} = \frac{1}{2} \text{AC} = \frac{1}{2} \times 30 = 15 \, \text{cm} \), and \( \text{OB} = \text{OD} = \frac{1}{2} \text{BD} = \frac{1}{2} \times 16 = 8 \, \text{cm} \).
Now, in the right-angled triangle AOB, we have:
\( \text{AB}^2 = \text{AO}^2 + \text{BO}^2 \)
\( \implies \text{AB}^2 = 15^2 + 8^2 \)
\( \implies \text{AB}^2 = 225 + 64 \)
\( \implies \text{AB}^2 = 289 \)
\( \implies \text{AB}^2 = 17^2 \)
\( \implies \text{AB} = 17 \, \text{cm} \)
Similarly, in right triangle BOC, BC = 17 cm; in right triangle COD, CD = 17 cm; and in right triangle AOD, AD = 17 cm.
Next, the perimeter of the rhombus is the sum of all its sides: \( \text{AB} + \text{BC} + \text{CD} + \text{AD} \). This results in \( 17 \, \text{cm} + 17 \, \text{cm} + 17 \, \text{cm} + 17 \, \text{cm} = 68 \, \text{cm} \).

Exam Tip: Remember that the diagonals of a rhombus bisect each other at right angles. This creates four congruent right-angled triangles within the rhombus, allowing you to use the Pythagoras theorem.