CBSE Class 7 Mathematics Triangle And Its Properties Worksheet Set 03

Chapter 6: Triangles - Properties

Question 1. Fine the lengths of the hypotenuses (x) of the triangles whose legs are given
i) 3 cm, 4 cm
ii) 5 cm, 12 cm
iii) 12 cm, 16 cm
iv) 9 cm, 12 cm
iv) 7 cm, 24 cm
v) 8 cm, 15 cm
Answer:
We find the hypotenuse (\( x \)) using the Pythagoras theorem, which states that \( x^2 = a^2 + b^2 \), where \( a \) and \( b \) are the two shorter legs of the right-angled triangle.

i) Legs: 3 cm, 4 cm
\( \implies x = \sqrt{3^2 + 4^2} \)
\( \implies x = \sqrt{9 + 16} \)
\( \implies x = \sqrt{25} = 5\text{ cm} \)

ii) Legs: 5 cm, 12 cm
\( \implies x = \sqrt{5^2 + 12^2} \)
\( \implies x = \sqrt{25 + 144} \)
\( \implies x = \sqrt{169} = 13\text{ cm} \)

iii) Legs: 12 cm, 16 cm
\( \implies x = \sqrt{12^2 + 16^2} \)
\( \implies x = \sqrt{144 + 256} \)
\( \implies x = \sqrt{400} = 20\text{ cm} \)

iv) Legs: 9 cm, 12 cm
\( \implies x = \sqrt{9^2 + 12^2} \)
\( \implies x = \sqrt{81 + 144} \)
\( \implies x = \sqrt{225} = 15\text{ cm} \)

iv) Legs: 7 cm, 24 cm
\( \implies x = \sqrt{7^2 + 24^2} \)
\( \implies x = \sqrt{49 + 576} \)
\( \implies x = \sqrt{625} = 25\text{ cm} \)

v) Legs: 8 cm, 15 cm
\( \implies x = \sqrt{8^2 + 15^2} \)
\( \implies x = \sqrt{64 + 225} \)
\( \implies x = \sqrt{289} = 17\text{ cm} \)
In simple words: To find the longest side of a right-angled triangle, square the two shorter sides, add them up, and then find the square root of that total.

Exam Tip: Memorizing standard Pythagorean triples like (3, 4, 5) and (5, 12, 13) can help you solve these problems much faster during tests.

Question 2. Find the length of the diagonal of a rectangle using Pythagoras theorem, given l = 8 cm, b = 6 cm
Answer:
A rectangle's length (\( l \)), breadth (\( b \)), and diagonal (\( d \)) form a right-angled triangle where the diagonal acts as the hypotenuse.
Using the Pythagoras theorem:
\( \implies d^2 = l^2 + b^2 \)
\( \implies d^2 = 8^2 + 6^2 \)
\( \implies d^2 = 64 + 36 \)
\( \implies d^2 = 100 \)
\( \implies d = \sqrt{100} = 10\text{ cm} \)
The length of the diagonal is 10 cm.
In simple words: A diagonal cuts a rectangle into two right-angled triangles. We square the length and width, add them to get 100, and find the square root to get 10 cm.

Exam Tip: Draw a small labeled diagram of the rectangle to clearly show how the diagonal acts as the hypotenuse.

Question 3. An electric pole is 9 m high. A steel wire tied to the top of the pole is affixed at a point on the ground at a distance of 12 m from the foot of the pole. Find the length of the wire.
Answer:
The pole, the ground, and the wire form a right-angled triangle.
The height of the pole is one leg (\( a = 9\text{ m} \)) and the distance along the ground is the other leg (\( b = 12\text{ m} \)).
Let the length of the wire be the hypotenuse (\( w \)).
Using the Pythagoras theorem:
\( \implies w^2 = a^2 + b^2 \)
\( \implies w^2 = 9^2 + 12^2 \)
\( \implies w^2 = 81 + 144 \)
\( \implies w^2 = 225 \)
\( \implies w = \sqrt{225} = 15\text{ m} \)
The length of the steel wire is 15 meters.
In simple words: Think of the pole and the ground as the two sides of a square corner. Squaring both lengths and adding them gives 225. The square root of 225 is 15, which is the length of the wire.

Exam Tip: Be sure to keep track of the units used in the question. Here, the units are in meters (m), not centimeters (cm).

Question 4. ABCD is a quadrilateral . Ac is a diagonal
∠ B = 90°, ∠ DAC = 90°
If AB = 4 cm, BC = 3 cm
Find DC.
Answer:
First, let us find the length of diagonal AC using right-angled triangle ABC (right-angled at B).
Using the Pythagoras theorem:
\( \implies AC^2 = AB^2 + BC^2 \)
\( \implies AC^2 = 4^2 + 3^2 \)
\( \implies AC^2 = 16 + 9 \)
\( \implies AC^2 = 25 \)
\( \implies AC = \sqrt{25} = 5\text{ cm} \)

Next, let us find the length of DC using right-angled triangle DAC (right-angled at A, as \( \angle DAC = 90^\circ \)).
The hypotenuse in this triangle is DC, and the given side AD is 12 cm.
Using the Pythagoras theorem:
\( \implies DC^2 = AD^2 + AC^2 \)
\( \implies DC^2 = 12^2 + 5^2 \)
\( \implies DC^2 = 144 + 25 \)
\( \implies DC^2 = 169 \)
\( \implies DC = \sqrt{169} = 13\text{ cm} \)
The length of DC is 13 cm.
In simple words: First, we use the bottom triangle ABC to find the hidden line AC, which is 5 cm. Then, we use this 5 cm and the top line of 12 cm in the second triangle to find the final side DC, which is 13 cm.

Exam Tip: For composite figures with multiple triangles, solve them step-by-step. Use the calculated side of the first triangle as a known side for the next triangle.