CBSE Class 7 Mathematics Triangles and Its Properties Assignment Set 08

VII - Mathematics Assignment No. 04 - Triangle and its Properties

Question 1. In a right angled \(\Delta\), one of the acute angles is half of the other. Find the third angle.
Answer: Let the acute angles be \(x\) and \(\frac{x}{2}\).
In a right-angled triangle, one angle is always \(90^\circ\).
The sum of all angles in a triangle is \(180^\circ\).
\(x + \frac{x}{2} + 90^\circ = 180^\circ\)
\(\frac{3x}{2} = 90^\circ\)
\(3x = 180^\circ\)
\(x = 60^\circ\)
Therefore, the acute angles are \(60^\circ\) and \(30^\circ\).
The third angle is the right angle itself, which is \(90^\circ\).
In simple words: Since the triangle is right-angled, one angle is \(90^\circ\). The other two angles add up to \(90^\circ\), and using their given ratio, we find the angles are \(30^\circ\) and \(60^\circ\). The third angle is \(90^\circ\).

Exam Tip: Remember that in any right-angled triangle, the sum of the two acute angles is always \(90^\circ\).

Question 2. Angles of a \(\Delta\) are in the ratio 5:6:7. Find the greatest and the smallest.
Answer: Let the angles be \(5x\), \(6x\), and \(7x\).
Using the angle sum property of a triangle:
\(5x + 6x + 7x = 180^\circ\)
\(18x = 180^\circ\)
\(x = 10^\circ\)
Smallest angle = \(5x = 5 \times 10^\circ = 50^\circ\)
Greatest angle = \(7x = 7 \times 10^\circ = 70^\circ\)
In simple words: Divide the total angle sum of \(180^\circ\) according to the given ratio parts to find the individual angles.

Exam Tip: When angles are given in a ratio, always introduce a common variable \(x\) and equate their sum to \(180^\circ\).

Question 3. In \(\Delta ABC\), BO and CO are the angle bisectors of \(\angle B\) and \(\angle C\) also \(AB = AC\), \(\angle BAC = 80^\circ\). Find:
(i) \(\angle ABC\) and \(\angle ACB\)
(ii) \(\angle BOC\)
(iii) \(\angle OCB\)
Answer:
(i) Given \(AB = AC\), so \(\Delta ABC\) is an isosceles triangle.
Therefore, \(\angle ABC = \angle ACB\).
Sum of angles in \(\Delta ABC = 180^\circ\)
\(\angle BAC + \angle ABC + \angle ACB = 180^\circ\)
\(80^\circ + 2(\angle ABC) = 180^\circ\)
\(2(\angle ABC) = 100^\circ\)
\(\angle ABC = 50^\circ\) and \(\angle ACB = 50^\circ\).

(ii) Since BO and CO are angle bisectors of \(\angle B\) and \(\angle C\):
\(\angle OBC = \frac{1}{2} \angle ABC = \frac{50^\circ}{2} = 25^\circ\)
\(\angle OCB = \frac{1}{2} \angle ACB = \frac{50^\circ}{2} = 25^\circ\)
In \(\Delta OBC\), using angle sum property:
\(\angle BOC + \angle OBC + \angle OCB = 180^\circ\)
\(\angle BOC + 25^\circ + 25^\circ = 180^\circ\)
\(\angle BOC = 180^\circ - 50^\circ = 130^\circ\)

(iii) From part (ii), \(\angle OCB = 25^\circ\).
In simple words: First find the base angles of the main isosceles triangle. Then use the angle bisectors to find the angles inside the smaller triangle \(OBC\).

Exam Tip: Angle bisectors divide an angle into two equal halves. In an isosceles triangle, angles opposite to equal sides are equal.

Question 4. From the adjoining figure in which \(AB = AC\), \(AD \perp BC\), \(BC = 4\) cm and \(\angle ABC = 40^\circ\). Find:
(i) \(\angle ACB\)
(ii) \(\angle BAC\)
(iii) \(BD\)
Answer:
(i) Since \(AB = AC\), \(\Delta ABC\) is an isosceles triangle, so base angles are equal.
\(\angle ACB = \angle ABC = 40^\circ\).

(ii) Using the angle sum property in \(\Delta ABC\):
\(\angle BAC + \angle ABC + \angle ACB = 180^\circ\)
\(\angle BAC + 40^\circ + 40^\circ = 180^\circ\)
\(\angle BAC = 180^\circ - 80^\circ = 100^\circ\)

(iii) In an isosceles triangle, the altitude drawn from the vertex to the base also bisects the base.
\(BD = \frac{BC}{2} = \frac{4\text{ cm}}{2} = 2\text{ cm}\).
In simple words: The altitude in an isosceles triangle splits the base into two equal halves and also acts as an angle bisector.

Exam Tip: In an isosceles triangle, the altitude to the base is perpendicular bisector of the base.

Question 5. In this figure, ABC is an isosceles \(\Delta\) in which \(AB = AC\). Ext \(\angle ABE = 150^\circ\) and \(AD \perp BC\). Find (i) \(\angle BAD\) (ii) \(\angle BAC\)
Answer:
(i) The exterior angle at B is given as \(\angle ABE = 150^\circ\).
Since ABC is a straight line, the interior angle \(\angle ABC = 180^\circ - 150^\circ = 30^\circ\).
In right-angled triangle \(ABD\) (\(\angle ADB = 90^\circ\)):
\(\angle BAD + \angle ABC + \angle ADB = 180^\circ\)
\(\angle BAD + 30^\circ + 90^\circ = 180^\circ\)
\(\angle BAD = 180^\circ - 120^\circ = 60^\circ\).

(ii) Since \(AD\) is the altitude of an isosceles triangle \(ABC\), it also acts as the angle bisector of \(\angle BAC\).
Therefore, \(\angle BAC = 2 \times \angle BAD = 2 \times 60^\circ = 120^\circ\).
In simple words: Use the linear pair to find the interior angle, then use the right-angled triangle \(ABD\) to find \(\angle BAD\), and double it for the vertex angle.

Exam Tip: An exterior angle and its adjacent interior angle always form a linear pair (sum to \(180^\circ\)).

Question 6. The length of two sides of a right angled \(\Delta\) are 3 cm and 4 cm respectively. Find the length of its hypotenuse.
Answer:
Using Pythagoras Theorem:
\((\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Perpendicular})^2\)
\((\text{Hypotenuse})^2 = 3^2 + 4^2\)
\((\text{Hypotenuse})^2 = 9 + 16 = 25\)
\(\text{Hypotenuse} = \sqrt{25} = 5\text{ cm}\).
In simple words: Apply Pythagoras theorem by squaring the two given sides, adding them, and taking the square root.

Exam Tip: Always identify that the two given sides are the legs (perpendicular sides) since the hypotenuse is always the longest side.

Question 7. The square of hypotenuse of a right angled \(\Delta\) is 80 cm and one side is half the other. Find the length of two sides.
Answer:
Let the two sides (legs) be \(x\) and \(2x\).
By Pythagoras Theorem:
\((\text{Base})^2 + (\text{Perpendicular})^2 = (\text{Hypotenuse})^2\)
\(x^2 + (2x)^2 = 80\)
\(x^2 + 4x^2 = 80\)
\(5x^2 = 80\)
\(x^2 = \frac{80}{5} = 16\)
\(x = 4\text{ cm}\)
The other side = \(2x = 2 \times 4 = 8\text{ cm}\).
Therefore, the lengths of the two sides are 4 cm and 8 cm.
In simple words: Assume the sides as \(x\) and \(2x\), substitute them into the Pythagoras theorem along with the given squared hypotenuse, and solve for \(x\).

Exam Tip: Pay careful attention when the problem gives the square of the hypotenuse directly instead of the length.

Question 8. Two poles of heights 23 m and 15 m stand parallel to each other at a distance of 15 m. Find the distance between their tops.
Answer:
Let the height of the first pole be \(h_1 = 23\text{ m}\) and the second pole be \(h_2 = 15\text{ m}\).
Distance between their bases = \(15\text{ m}\).
Difference in heights = \(23\text{ m} - 15\text{ m} = 8\text{ m}\).
The distance between their tops forms the hypotenuse of a right-angled triangle where the other two sides are the distance between the poles (15 m) and the difference in their heights (8 m).
Using Pythagoras Theorem:
\((\text{Distance between tops})^2 = 15^2 + 8^2\)
\((\text{Distance between tops})^2 = 225 + 64 = 289\)
\(\text{Distance between tops} = \sqrt{289} = 17\text{ m}\).
In simple words: Form a right-angled triangle by drawing a horizontal line from the top of the shorter pole to the taller pole. The base is 15 m and the height is the difference in pole heights (8 m). Then apply Pythagoras theorem.

Exam Tip: Real-life problems involving poles and shadows can almost always be modeled using right-angled triangles and the Pythagoras theorem.