GSEB Class 10 Maths Solutions Chapter 6 Triangle Exercise 6.4

Question 1. Let \( \triangle ABC \sim \triangle DEF \) and their areas be, respectively 64 cm\(^2\) and 121 cm\(^2\). If EF = 15.4 cm, find BC.
Answer: We have the following information:
Area of \( \triangle ABC = 64 \) cm\(^2\)
Area of \( \triangle DEF = 121 \) cm\(^2\)
Length of EF = 15.4 cm
Since \( \triangle ABC \sim \triangle DEF \), the ratio of their areas is equal to the square of the ratio of their corresponding sides:
\( \frac {ar \triangle ABC}{ar \triangle DEF} = \frac {BC^2}{EF^2} \)
Substitute the given values into the formula:
\( \frac {64}{121} = \frac {BC^2}{(15.4)^2} \)
To find BC, take the square root of both sides:
\( \sqrt{\frac {64}{121}} = \sqrt{\frac {BC^2}{(15.4)^2}} \)
\( \frac {8}{11} = \frac {BC}{15.4} \)
Now, solve for BC:
\( BC = \frac {8}{11} \times 15.4 \)
\( BC = 8 \times 1.4 \)
\( BC = 11.2 \) cm
In simple words: When two triangles are similar, the ratio of their areas is the square of the ratio of their sides. By using this rule and the given numbers, we can calculate the missing side length.

Exam Tip: Remember the theorem for areas of similar triangles. It's crucial to correctly identify corresponding sides when setting up the ratio.

Question 2. Diagonals of a trapezium ABCD with \( AB \parallel DC \) intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD. (CBSE 2012)
Answer: Let's consider a trapezium ABCD where \( AB \parallel DC \). The diagonals AC and BD meet at point O.
Now, let's look at \( \triangle AOB \) and \( \triangle COD \):
\( \angle AOB = \angle COD \) (These are vertically opposite angles, so they are always equal.)
\( \angle OAB = \angle OCD \) (These are alternate interior angles because \( AB \parallel DC \) and AC is a transversal.)
Therefore, \( \triangle AOB \sim \triangle COD \) (The triangles are similar by the AA (Angle-Angle) similarity criterion.)
When two triangles are similar, the ratio of their areas is equivalent to the square of the ratio of their corresponding sides.
So, \( \frac {ar \triangle AOB}{ar \triangle COD} = \frac {AB^2}{CD^2} \)
We are given that \( AB = 2 CD \). Let's substitute this into the ratio:
\( \frac {ar \triangle AOB}{ar \triangle COD} = \frac {(2 CD)^2}{CD^2} \)
\( \frac {ar \triangle AOB}{ar \triangle COD} = \frac {4 CD^2}{CD^2} \)
\( \frac {ar \triangle AOB}{ar \triangle COD} = 4 \)
Thus, the ratio of the areas of \( \triangle AOB \) and \( \triangle COD \) is \( 4 : 1 \).
In simple words: In a trapezium, the triangles formed by the intersecting diagonals are similar. If one parallel side is double the other, the area of the larger triangle will be four times the area of the smaller one.

Exam Tip: For problems involving trapeziums and diagonals, always remember that the triangles formed by the intersecting diagonals are similar. This allows you to use the property of areas of similar triangles.

Question 3. In Figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \( \frac {ar \triangle ABC}{ar \triangle DBC} = \frac {AO}{DO} \) (CBSE 2005, 2012)
Answer: We have two triangles, \( \triangle ABC \) and \( \triangle DBC \), that share the same base BC. The line segment AD intersects BC at point O.
To prove the relationship between their areas and the segments AO and DO, we will draw two altitudes:
Let's draw \( AE \perp BC \) (AE is the altitude from A to BC).
Let's draw \( DF \perp BC \) (DF is the altitude from D to BC).
Now, consider \( \triangle AOE \) and \( \triangle DOF \):
\( \angle AOE = \angle DOF \) (These angles are vertically opposite, so they are equal.)
\( \angle AEO = \angle DFO = 90^\circ \) (These angles are right angles because AE and DF are perpendiculars to BC.)
Therefore, \( \triangle AOE \sim \triangle DOF \) (The triangles are similar by the AA (Angle-Angle) similarity criterion.)
Since these triangles are similar, their corresponding sides are proportional:
\( \frac {AE}{DF} = \frac {AO}{DO} \) ...(1)
Next, let's consider the areas of the main triangles:
The area of \( \triangle ABC = \frac {1}{2} \times \text{base} \times \text{height} = \frac {1}{2} \times BC \times AE \)
The area of \( \triangle DBC = \frac {1}{2} \times \text{base} \times \text{height} = \frac {1}{2} \times BC \times DF \)
Let's find the ratio of their areas:
\( \frac {ar \triangle ABC}{ar \triangle DBC} = \frac {\frac {1}{2} \times BC \times AE}{\frac {1}{2} \times BC \times DF} \)
We can cancel out \( \frac {1}{2} \times BC \) from both the numerator and denominator:
\( \frac {ar \triangle ABC}{ar \triangle DBC} = \frac {AE}{DF} \) ...(2)
By comparing equation (1) and equation (2), we can conclude:
\( \frac {ar \triangle ABC}{ar \triangle DBC} = \frac {AO}{DO} \)
This proves the desired relationship.
In simple words: When two triangles share the same base, the ratio of their areas is the same as the ratio of their heights. If the heights come from similar smaller triangles, then the ratio of heights can be replaced by the ratio of parts of the line connecting their vertices.

Exam Tip: When dealing with ratios of areas for triangles on the same base, remember that the ratio of their areas is equal to the ratio of their corresponding altitudes. Then, look for similar triangles to relate these altitudes to other line segments.

Question 4. If the area of two similar triangles are equal, prove that they are congruent. (CBSE 2012)
Answer: Let's assume we have two similar triangles, \( \triangle ABC \) and \( \triangle DEF \). We are given that their areas are equal, meaning \( ar \triangle ABC = ar \triangle DEF \). Our goal is to prove that these triangles are congruent, i.e., \( \triangle ABC \cong \triangle DEF \).
Here's the proof:
We know that \( \triangle ABC \sim \triangle DEF \) (This is given).
A property of similar triangles is that the ratio of their areas is equal to the square of the ratio of their corresponding sides. So, we can write:
\( \frac {ar \triangle ABC}{ar \triangle DEF} = \frac {AB^2}{DE^2} = \frac {BC^2}{EF^2} = \frac {AC^2}{DF^2} \)
Since we are given that \( ar \triangle ABC = ar \triangle DEF \), their ratio is 1:
\( \frac {ar \triangle ABC}{ar \triangle DEF} = 1 \)
Substituting this into the area ratio equation:
\( 1 = \frac {AB^2}{DE^2} = \frac {BC^2}{EF^2} = \frac {AC^2}{DF^2} \)
This implies that:
\( AB^2 = DE^2 \implies AB = DE \)
\( BC^2 = EF^2 \implies BC = EF \)
\( AC^2 = DF^2 \implies AC = DF \)
So, all three corresponding sides of the two triangles are equal.
Additionally, since \( \triangle ABC \sim \triangle DEF \), their corresponding angles are also equal:
\( \angle A = \angle D \)
\( \angle B = \angle E \)
\( \angle C = \angle F \)
Since we have proved that two corresponding angles and the included side (e.g., \( \angle B = \angle E \), \( BC = EF \), \( \angle C = \angle F \)) are equal, we can conclude that \( \triangle ABC \cong \triangle DEF \) by the ASA (Angle-Side-Angle) congruence criterion.
Therefore, if the areas of two similar triangles are equal, the triangles must be congruent.
In simple words: If two triangles look the same shape (similar) and also cover the same amount of space (equal area), then they must be exactly the same size and shape (congruent).

Exam Tip: This is a fundamental theorem. Always start by stating the property of areas of similar triangles, then use the given condition of equal areas to show that corresponding sides must be equal. Finally, use a congruence criterion (like SSS or ASA) to prove congruence.

Question 5. D, E and F are respectively the midpoints of sides AB, BC and CA of \( \triangle ABC \). Find the ratio of the area of \( \triangle DEF \) and \( \triangle ABC \).
Answer: Let's consider \( \triangle ABC \). We are given that D, E, and F are the midpoints of sides AB, BC, and CA, respectively. When we connect these midpoints, we form a new triangle, \( \triangle DEF \).
By the Midpoint Theorem, which states that the line segment connecting the midpoints of any two sides of a triangle is parallel to the third side and is half its length:
1. The line segment DF connects the midpoints of AB and AC. So, \( DF \parallel BC \) and \( DF = \frac {1}{2} BC \).
2. The line segment DE connects the midpoints of AB and BC. So, \( DE \parallel AC \) and \( DE = \frac {1}{2} AC \).
3. The line segment EF connects the midpoints of BC and CA. So, \( EF \parallel AB \) and \( EF = \frac {1}{2} AB \).
From these relationships, we can see that the sides of \( \triangle DEF \) are proportional to the corresponding sides of \( \triangle ABC \) with a ratio of \( \frac {1}{2} \). That is:
\( \frac {DF}{BC} = \frac {DE}{AC} = \frac {EF}{AB} = \frac {1}{2} \)
Since all corresponding sides are proportional, \( \triangle DEF \sim \triangle ABC \) (by the SSS (Side-Side-Side) similarity criterion).
Now, we know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides:
\( \frac {ar \triangle DEF}{ar \triangle ABC} = \left( \frac {DF}{BC} \right)^2 \)
Substitute the ratio \( \frac {DF}{BC} = \frac {1}{2} \) into the equation:
\( \frac {ar \triangle DEF}{ar \triangle ABC} = \left( \frac {1}{2} \right)^2 \)
\( \frac {ar \triangle DEF}{ar \triangle ABC} = \frac {1}{4} \)
Therefore, the ratio of the area of \( \triangle DEF \) to the area of \( \triangle ABC \) is \( 1 : 4 \).
In simple words: When you connect the middle points of all three sides of a triangle, you create a smaller triangle inside. This smaller triangle is always similar to the original one, and its area is exactly one-fourth of the original triangle's area.

Exam Tip: The Midpoint Theorem is essential here. Understanding that the inner triangle formed by midpoints is similar to the original triangle with a side ratio of 1:2 directly leads to an area ratio of 1:4. Memorize this common result.

Question 6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Answer: Let's consider two similar triangles, \( \triangle ABC \) and \( \triangle DEF \). Let AM and DN be the medians corresponding to sides BC and EF, respectively. We aim to prove that \( \frac {ar \triangle ABC}{ar \triangle DEF} = \frac {AM^2}{DN^2} \).
Here's the proof:
1. We are given that \( \triangle ABC \sim \triangle DEF \).
This means their corresponding sides are proportional:
\( \frac {AB}{DE} = \frac {BC}{EF} = \frac {AC}{DF} \) ...(1)
Also, their corresponding angles are equal, specifically \( \angle B = \angle E \).
2. Since AM and DN are medians, they bisect the sides BC and EF respectively.
So, \( BM = \frac {1}{2} BC \) and \( EN = \frac {1}{2} EF \).
3. From equation (1), we have \( \frac {BC}{EF} = \frac {AB}{DE} \). Substituting the median relations:
\( \frac {2BM}{2EN} = \frac {AB}{DE} \implies \frac {BM}{EN} = \frac {AB}{DE} \) ...(2)
4. Now, let's consider \( \triangle ABM \) and \( \triangle DEN \):
We have \( \frac {AB}{DE} = \frac {BM}{EN} \) (from equation (2)).
We also have \( \angle B = \angle E \) (corresponding angles of similar triangles \( \triangle ABC \) and \( \triangle DEF \)).
Therefore, \( \triangle ABM \sim \triangle DEN \) (by the SAS (Side-Angle-Side) similarity criterion).
5. Since \( \triangle ABM \sim \triangle DEN \), their corresponding sides are proportional:
\( \frac {AB}{DE} = \frac {AM}{DN} \) ...(3)
6. We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides:
\( \frac {ar \triangle ABC}{ar \triangle DEF} = \left( \frac {AB}{DE} \right)^2 \) ...(4)
7. From equation (3), we can replace \( \frac {AB}{DE} \) with \( \frac {AM}{DN} \) in equation (4):
\( \frac {ar \triangle ABC}{ar \triangle DEF} = \left( \frac {AM}{DN} \right)^2 = \frac {AM^2}{DN^2} \)
This proves that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
In simple words: When two triangles are similar, the way their areas compare is the same as how the squares of their medians compare. If one median is twice as long as the other, the area of that triangle will be four times larger.

Exam Tip: The key steps in this proof are establishing the similarity of the smaller triangles formed by the medians (\( \triangle ABM \sim \triangle DEN \)) and then relating the ratio of their sides (which includes the medians) to the ratio of the main triangles' areas.

Question 7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals. (CBSE 2012)
Answer: Let's consider a square ABCD. We will describe an equilateral triangle on one of its sides and another equilateral triangle on one of its diagonals.
Let the side of the square be 'a' units. So, \( BC = a \).
An equilateral triangle, say \( \triangle BQC \), is described on side BC. All sides of \( \triangle BQC \) are equal to 'a'.
Now, let's find the length of the diagonal of the square. Using the Pythagorean theorem for \( \triangle ABC \):
\( AC^2 = AB^2 + BC^2 \)
\( AC^2 = a^2 + a^2 \)
\( AC^2 = 2a^2 \)
\( AC = \sqrt {2a^2} = a\sqrt {2} \)
An equilateral triangle, say \( \triangle APC \), is described on the diagonal AC. All sides of \( \triangle APC \) are equal to \( a\sqrt {2} \).
All equilateral triangles are similar to each other because each angle in an equilateral triangle is \( 60^\circ \) (AAA similarity).
Therefore, \( \triangle APC \sim \triangle BQC \).
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. So:
\( \frac {ar \triangle APC}{ar \triangle BQC} = \left( \frac {AC}{BC} \right)^2 \)
Substitute the side lengths we found:
\( \frac {ar \triangle APC}{ar \triangle BQC} = \left( \frac {a\sqrt {2}}{a} \right)^2 \)
\( \frac {ar \triangle APC}{ar \triangle BQC} = (\sqrt {2})^2 \)
\( \frac {ar \triangle APC}{ar \triangle BQC} = 2 \)
From this, we can conclude:
\( ar \triangle APC = 2 \times ar \triangle BQC \)
Or, rearranging the equation:
\( ar \triangle BQC = \frac {1}{2} ar \triangle APC \)
This proves that the area of the equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
In simple words: Imagine a square. If you draw a tiny equilateral triangle on one side, and a bigger one on its diagonal, the area of the small one will be exactly half the area of the big one.

Exam Tip: This problem combines properties of squares (diagonal length) with properties of equilateral and similar triangles. Clearly defining the side lengths based on the square's side 'a' is a crucial first step.

Question 8. In an equilateral triangle ABC, D is the midpoint of BC. If \( \triangle ABC \) and \( \triangle BDE \) are two equilateral triangles, find the ratio of the areas of \( \triangle ABC \) and \( \triangle BDE \).
(a) 2:1
(b) 1:2
(c) 4:1
(d) 1:4
Answer: (c) 4:1
In simple words: Since D is the middle point of BC, the side AB is twice as long as BD. Because both triangles are equilateral (and thus similar), the ratio of their areas is the square of the ratio of their sides, which gives 4:1.

Exam Tip: Remember that all equilateral triangles are similar. The ratio of their areas is always the square of the ratio of their corresponding sides. Identify the relationship between the side lengths using the midpoint property.

Question 9. Sides of two similar triangles are in the ratio 4: 9. The area of these triangles are in the ratio.
(a) 2:3
(b) 4:9
(c) 81:16
(d) 16:81
Answer: (d) 16:81
In simple words: When triangles are similar, if you want to know how their areas compare, you need to square the ratio of their sides. Here, 4 squared is 16, and 9 squared is 81.

Exam Tip: This is a direct application of the theorem stating that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Be careful not to just use the side ratio directly; always square it.