CBSE Class 7 Mathematics Triangles and Its Properties Assignment Set 07

Question 1. The exterior angle \(\angle ACD\) of \(\Delta ABC\) is \(120^\circ\) and \(\angle ABC = 50^\circ\). Find
(i) \(\angle ACB\)
(ii) \(\angle BAC\)
Is (iii) \(\angle ACD > \angle A\) ?
(iv) \(\angle ACD = \angle A + \angle B\) ?
Answer:
(i) Since BCD is a straight line, the angles on it must add up to \(180^\circ\).
\(\angle ACB + \angle ACD = 180^\circ\)
\(\implies \angle ACB + 120^\circ = 180^\circ\)
\(\implies \angle ACB = 180^\circ - 120^\circ = 60^\circ\)

(ii) We know that all three angles inside a triangle add up to \(180^\circ\).
In \(\Delta ABC\):
\(\angle BAC + \angle ABC + \angle ACB = 180^\circ\)
\(\implies \angle BAC + 50^\circ + 60^\circ = 180^\circ\)
\(\implies \angle BAC + 110^\circ = 180^\circ\)
\(\implies \angle BAC = 180^\circ - 110^\circ = 70^\circ\)

(iii) Here, \(\angle ACD = 120^\circ\) and \(\angle A = \angle BAC = 70^\circ\).
Since \(120^\circ\) is larger than \(70^\circ\), the answer is Yes, \(\angle ACD > \angle A\).

(iv) Let us add \(\angle A\) and \(\angle B\):
\(\angle A + \angle B = 70^\circ + 50^\circ = 120^\circ\)
Since the exterior angle \(\angle ACD\) is also \(120^\circ\), the answer is Yes, \(\angle ACD = \angle A + \angle B\).
In simple words: The angles on a straight line make \(180^\circ\), so the inner corner is \(60^\circ\). All three inner corners of a triangle add up to \(180^\circ\), which makes the top angle \(70^\circ\). The outer angle is larger than the top angle, and it is equal to the sum of the two opposite inner angles.

Exam Tip: Remember that an exterior angle of a triangle is always equal to the sum of its two opposite interior angles. This rule helps you find angles quickly without showing too many steps.

Question 2. In the adjoining figure \(\angle BAC = 30^\circ\), \(\angle ABD = 32^\circ\), \(\angle CBD = 25^\circ\). Find the marked angles.
Answer:
First, let us look at \(\Delta ABD\) (the upper triangle).
We know that the three angles in a triangle always add up to \(180^\circ\).
So, in \(\Delta ABD\):
\(\angle DAB + \angle ABD + \angle ADB = 180^\circ\)
\(\implies 30^\circ + 32^\circ + a = 180^\circ\)
\(\implies 62^\circ + a = 180^\circ\)
\(\implies a = 180^\circ - 62^\circ = 118^\circ\)

Next, since \(ADC\) is a straight line, the angles \(a\) and \(b\) make a linear pair. This means they add up to \(180^\circ\).
\(a + b = 180^\circ\)
\(\implies 118^\circ + b = 180^\circ\)
\(\implies b = 180^\circ - 118^\circ = 62^\circ\)

Now, let us look at the lower triangle \(\Delta BCD\).
The three angles in this triangle must also add up to \(180^\circ\).
\(\angle CBD + \angle BDC + \angle BCD = 180^\circ\)
\(\implies 25^\circ + b + c = 180^\circ\)
\(\implies 25^\circ + 62^\circ + c = 180^\circ\)
\(\implies 87^\circ + c = 180^\circ\)
\(\implies c = 180^\circ - 87^\circ = 93^\circ\)

Finally, since \(BCE\) is a straight line, the angles \(c\) and \(d\) add up to \(180^\circ\).
\(c + d = 180^\circ\)
\(\implies 93^\circ + d = 180^\circ\)
\(\implies d = 180^\circ - 93^\circ = 87^\circ\)

Therefore, the marked angles are:
\(a = 118^\circ\)
\(b = 62^\circ\)
\(c = 93^\circ\)
\(d = 87^\circ\)
In simple words: Inside any triangle, the three corners always add up to \(180^\circ\). We use this rule for the top triangle to find \(a = 118^\circ\). Since straight lines make \(180^\circ\), we find \(b = 62^\circ\). Using the bottom triangle gives us \(c = 93^\circ\), and the bottom straight line then gives us \(d = 87^\circ\).

Exam Tip: Label the steps clearly and mention the property you are using, such as the Angle Sum Property of a triangle or the Linear Pair Axiom, to score full marks.

Question 3. In the adjoining figure, D is a point on the side BC of \(\Delta ABC\). Prove that \(AB + AC + BC > 2AD\)
Answer:
We know that in any triangle, the sum of the lengths of any two sides is always greater than the length of the third side.

Let us apply this rule to two smaller triangles in the figure:

1. In \(\Delta ABD\):
\(AB + BD > AD\) --- (Equation 1)

2. In \(\Delta ACD\):
\(AC + CD > AD\) --- (Equation 2)

Now, let us add Equation 1 and Equation 2 together:
\((AB + BD) + (AC + CD) > AD + AD\)
\(\implies AB + AC + (BD + CD) > 2AD\)

Since \(D\) is a point on the line segment \(BC\), we can write:
\(BD + CD = BC\)

Substituting this back into our inequality, we get:
\(AB + AC + BC > 2AD\)

Hence proved.
In simple words: Any two sides of a triangle added together are always longer than the third side. By using this rule on both sides of the inner line \(AD\), we show that the outer sides of the triangle are together more than twice the length of \(AD\).

Exam Tip: Clearly write down the property "the sum of any two sides of a triangle is greater than the third side" before writing the inequalities. This is key to securing full marks in proof questions.

Question 4. O is a point in the interior of \(\Delta ABC\). Show that \(2(OA + OB + OC) > (AB + BC + CA)\) [Hint: Join OA, OB, OC and apply result]
Answer:
Let us join the point \(O\) inside the triangle to all three corners \(A\), \(B\), and \(C\). This creates three smaller triangles: \(\Delta OAB\), \(\Delta OBC\), and \(\Delta OCA\).

We will use the property that the sum of any two sides of a triangle is greater than its third side.

1. In \(\Delta OAB\):
\(OA + OB > AB\) --- (Equation 1)

2. In \(\Delta OBC\):
\(OB + OC > BC\) --- (Equation 2)

3. In \(\Delta OCA\):
\(OC + OA > CA\) --- (Equation 3)

Now, let us add all three equations together:
\((OA + OB) + (OB + OC) + (OC + OA) > AB + BC + CA\)
\(\implies 2OA + 2OB + 2OC > AB + BC + CA\)
\(\implies 2(OA + OB + OC) > (AB + BC + CA)\)

Hence proved.
In simple words: Connecting the middle point to each corner makes three smaller triangles. Since any two sides of a triangle are longer than the third side, adding their sides together shows that twice the inner lines is longer than the three outer edges.

Exam Tip: Don't forget to state that you are adding the corresponding sides of the three inequalities. Grouping terms with a common factor of 2 makes your proof neat and easy for the examiner to follow.

Question 5. The length of two sides of a \(\Delta\) are 6cm and 7cm. Between what two measure should the length of the third side fall?
Answer:
Let the two given sides of the triangle be \(a = 6\text{ cm}\) and \(b = 7\text{ cm}\). Let the third side be \(c\).

According to the properties of a triangle:
1. The third side must be shorter than the sum of the other two sides:
\(c < a + b\)
\(\implies c < 6 + 7 = 13\text{ cm}\)

2. The third side must be longer than the difference between the other two sides:
\(c > b - a\)
\(\implies c > 7 - 6 = 1\text{ cm}\)

Therefore, the length of the third side must be **greater than 1 cm and less than 13 cm**.
In simple words: The third side of a triangle must always be shorter than the other two sides added together, which is \(13\text{ cm}\). It must also be longer than the difference between those two sides, which is \(1\text{ cm}\).

Exam Tip: Always state both the upper limit (the sum of the two sides) and the lower limit (the positive difference of the two sides) to give a complete range.

Question 6. Is it possible to construct a \(\Delta\) with sides 4.5cm, 5.5cm and 6.5cm, if yes how?
Answer:
To check if a triangle can be made, the sum of any two sides must be greater than the third side.

Let the sides be \(a = 4.5\text{ cm}\), \(b = 5.5\text{ cm}\), and \(c = 6.5\text{ cm}\).
Let us check all three pairs:
- \(4.5 + 5.5 = 10\text{ cm}\), which is greater than \(6.5\text{ cm}\).
- \(4.5 + 6.5 = 11\text{ cm}\), which is greater than \(5.5\text{ cm}\).
- \(5.5 + 6.5 = 12\text{ cm}\), which is greater than \(4.5\text{ cm}\).

Since the sum of any two sides is always greater than the third side, **yes, it is possible** to construct this triangle.

**How to construct it:**
1. Draw a straight line segment \(BC = 6.5\text{ cm}\) using a ruler.
2. Set your compass to a width of \(4.5\text{ cm}\). Place the pointer on \(B\) and draw an arc above the line.
3. Set your compass to a width of \(5.5\text{ cm}\). Place the pointer on \(C\) and draw another arc that crosses the first one. Label this meeting point as \(A\).
4. Draw straight lines to connect point \(A\) to \(B\) and point \(A\) to \(C\).
In simple words: Yes, we can make this triangle because any two sides added together are longer than the third side. We draw it by first making the longest side with a ruler, and then using a compass to find where the other two sides meet.

Exam Tip: When proving a triangle is possible, always show all three addition checks. Showing only one is a common mistake that can cost you marks.

Question 7. Length of two diagonals of a rhombus are 24cm and 10cm. Find its perimeter.
Answer:
Let \(ABCD\) be a rhombus whose diagonals \(AC\) and \(BD\) intersect each other at point \(O\).
Given:
- \(AC = 24\text{ cm}\)`
- \(BD = 10\text{ cm}\)`

We know that the diagonals of a rhombus bisect each other at right angles (\(90^\circ\)).
Therefore:
- \(OA = \frac{AC}{2} = \frac{24}{2} = 12\text{ cm}\)
- \(OB = \frac{BD}{2} = \frac{10}{2} = 5\text{ cm}\)
- \(\angle AOB = 90^\circ\)

In the right-angled triangle \(\Delta AOB\), we can use Pythagoras' theorem:
\(AB^2 = OA^2 + OB^2\)
\(\implies AB^2 = 12^2 + 5^2\)
\(\implies AB^2 = 144 + 25\)
\(\implies AB^2 = 169\)
\(\implies AB = \sqrt{169} = 13\text{ cm}\)

Since all four sides of a rhombus are equal in length, the perimeter is:
\(\text{Perimeter} = 4 \times \text{side}\)
\(\implies \text{Perimeter} = 4 \times 13 = 52\text{ cm}\)
In simple words: The diagonals of a rhombus cut each other in half at a right angle. This creates a small right triangle inside with sides of \(12\text{ cm}\) and \(5\text{ cm}\). Using Pythagoras' rule, we find the outer side is \(13\text{ cm}\). Since all four sides are equal, the total distance around is \(52\text{ cm}\).

Exam Tip: Remember to state the property that diagonals of a rhombus bisect each other at right angles. This is a very common exam question, and mentioning this property clearly is essential for getting full marks.

Question 8. Find the perimeter of a square if the length of one diagonal is \(5\sqrt{2}\text{ cm}\).
Answer:
Let the side of the square be \(s\).
We know that the formula for the diagonal of a square is:
\(\text{Diagonal} = \text{side} \times \sqrt{2}\)

Given that the diagonal is \(5\sqrt{2}\text{ cm}\):
\(s\sqrt{2} = 5\sqrt{2}\)

Dividing both sides by \(\sqrt{2}\), we get:
\(s = 5\text{ cm}\)

So, the side of the square is \(5\text{ cm}\).

Now, we can find the perimeter of the square:
\(\text{Perimeter} = 4 \times \text{side}\)
\(\implies \text{Perimeter} = 4 \times 5 = 20\text{ cm}\)
In simple words: The diagonal of a square is always the side multiplied by \(\sqrt{2}\). Since the diagonal is \(5\sqrt{2}\text{ cm}\), each side must be \(5\text{ cm}\). Since a square has four equal sides, the perimeter is \(4 \times 5 = 20\text{ cm}\).

Exam Tip: You can also find the side by using Pythagoras' theorem on the right-angled triangle formed by two sides and the diagonal: \(s^2 + s^2 = (5\sqrt{2})^2\). Both methods are correct and score full marks.

Question 9. The length of diagonal of a rectangle is 17cm and one of its side is 15cm. Find its perimeter.
Answer:
Let the rectangle have a length of \(l = 15\text{ cm}\), a width of \(w\), and a diagonal of \(d = 17\text{ cm}\).

In a rectangle, the diagonal and two adjacent sides form a right-angled triangle. By using Pythagoras' theorem, we can write:
\(d^2 = l^2 + w^2\)
\(\implies 17^2 = 15^2 + w^2\)
\(\implies 289 = 225 + w^2\)
\(\implies w^2 = 289 - 225\)
\(\implies w^2 = 64\)
\(\implies w = \sqrt{64} = 8\text{ cm}\)

So, the width of the rectangle is \(8\text{ cm}\).

Now, we can find the perimeter of the rectangle:
\(\text{Perimeter} = 2(l + w)\)
\(\implies \text{Perimeter} = 2(15 + 8)\)
\(\implies \text{Perimeter} = 2 \times 23 = 46\text{ cm}\)
In simple words: The diagonal and the sides of a rectangle make a right-angled triangle. Using Pythagoras' rule, we find the other side is \(8\text{ cm}\). Adding the two sides together and doubling the result gives us a perimeter of \(46\text{ cm}\).

Exam Tip: Be careful not to confuse the formula for area (\(l \times w\)) with the formula for perimeter (\(2(l + w)\)). Read the question carefully to make sure you calculate the perimeter.

Question 10. The upper part (CD) of a tree (BD) is broken over by the wind and the top touches the ground at A which is 9 m away from the base of tree. The remaining part (BC) of the tree is 12m. Find the total height of the tree.
Answer:
Let \(BC\) be the standing part of the tree, which is \(12\text{ m}\) tall.
Let \(CD\) be the broken part of the tree. When it breaks, its top \(D\) falls to the ground at \(A\).
So, the length of the broken part \(CD\) is equal to the length of \(AC\).

Since the tree stands straight, it makes a right angle with the ground at \(B\). Thus, \(\Delta ABC\) is a right-angled triangle.
Given:
- \(AB = 9\text{ m}\) (distance from the base to the fallen top)
- \(BC = 12\text{ m}\) (height of the standing part)

By using Pythagoras' theorem in \(\Delta ABC\):
\(AC^2 = AB^2 + BC^2\)
\(\implies AC^2 = 9^2 + 12^2\)
\(\implies AC^2 = 81 + 144\)
\(\implies AC^2 = 225\)
\(\implies AC = \sqrt{225} = 15\text{ m}\)

Since the broken part \(CD\) is equal to \(AC\), we have:
\(CD = 15\text{ m}\)

To find the total height of the tree before it broke, we add the standing part and the broken part together:
\(\text{Total height of the tree} = BC + CD\)
\(\implies \text{Total height of the tree} = 12\text{ m} + 15\text{ m} = 27\text{ m}\)
In simple words: The standing part of the tree and the ground make a right angle. This gives us a right triangle with sides of \(12\text{ m}\) and \(9\text{ m}\). Using Pythagoras' rule, we find the leaning part is \(15\text{ m}\) long. Adding this to the standing height of \(12\text{ m}\) gives the original height of \(27\text{ m}\).

Exam Tip: A very common mistake is finding only the hypotenuse (\(15\text{ m}\)) and stopping there. Always read the question carefully and add the broken part to the remaining standing height to find the total original height.