GSEB Class 10 Maths Solutions Chapter 6 ત્રિકોણ Exercise 6.4

 

Question 1. Triangle ABC is similar to triangle DEF. Their areas are 64 cm\(^2\) and 121 cm\(^2\) respectively. If EF = 15.4 cm, then find BC.
Answer: Given that \( \triangle ABC \sim \triangle DEF \).
According to Theorem 6.6, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
So, we have:
\[ \frac{\text{Area}(ABC)}{\text{Area}(DEF)} = \left(\frac{BC}{EF}\right)^2 \]
Substitute the given values:
\[ \frac{64}{121} = \left(\frac{BC}{15.4}\right)^2 \]
Take the square root on both sides:
\[ \sqrt{\frac{64}{121}} = \frac{BC}{15.4} \]
\[ \frac{8}{11} = \frac{BC}{15.4} \]
Now, we will solve for BC:
\[ BC = \frac{8 \times 15.4}{11} \]
\[ BC = 8 \times 1.4 \]
\[ BC = 11.2 \text{ cm} \]
Thus, the length of side BC is 11.2 cm.
In simple words: જ્યારે બે ત્રિકોણ સમાન હોય, ત્યારે તેમના ક્ષેત્રફળનો ગુણોત્તર તેમની બાજુઓના ગુણોત્તરના વર્ગ જેટલો હોય છે. આપણને ક્ષેત્રફળ અને એક બાજુની લંબાઈ આપી છે, તેથી આપણે આ સૂત્રનો ઉપયોગ કરીને ખૂટતી બાજુ BCની લંબાઈ શોધી શકીએ છીએ.

Exam Tip: Remember Theorem 6.6 regarding the ratio of areas of similar triangles to the square of their corresponding sides. Always make sure to take the square root correctly when solving for unknown side lengths.

 

Question 2. In a trapezium ABCD, AB || CD. Its diagonals intersect each other at point O. If AB = 2CD, then find the ratio of the areas of \( \triangle AOB \) and \( \triangle COD \).

Answer: In trapezium ABCD, AB is parallel to CD, and diagonals AC and BD intersect at point O. Now, consider \( \triangle AOB \) and \( \triangle COD \): The alternate interior angles are equal: \( \angle OAB = \angle OCD \) (Alternate Interior Angles) \( \angle OBA = \angle ODC \) (Alternate Interior Angles) The vertically opposite angles are equal: \( \angle AOB = \angle COD \) (Vertically Opposite Angles) Therefore, by the AAA (Angle-Angle-Angle) similarity criterion, \( \triangle AOB \sim \triangle COD \). For similar triangles, the ratio of their areas is equivalent to the square of the ratio of their corresponding sides.
\[ \frac{\text{Area}(AOB)}{\text{Area}(COD)} = \left(\frac{AB}{CD}\right)^2 \]
Given that \( AB = 2CD \), substitute this into the equation:
\[ \frac{\text{Area}(AOB)}{\text{Area}(COD)} = \left(\frac{2CD}{CD}\right)^2 \]
\[ \frac{\text{Area}(AOB)}{\text{Area}(COD)} = (2)^2 \]
\[ \frac{\text{Area}(AOB)}{\text{Area}(COD)} = \frac{4}{1} \]
So, the ratio of the areas of \( \triangle AOB \) and \( \triangle COD \) is 4:1.
In simple words: સમલંબ ચતુષ્કોણમાં, કર્ણ છેદે ત્યારે બે ત્રિકોણ બને છે. જો સમાંતર બાજુઓમાં એક બાજુ બીજી બાજુ કરતાં બમણી હોય, તો આ બે ત્રિકોણ સમરૂપ હોય છે. તેમના ક્ષેત્રફળનો ગુણોત્તર તેમની અનુરૂપ બાજુઓના ગુણોત્તરના વર્ગ જેટલો હોય છે.

Exam Tip: For problems involving similar triangles in a trapezium, remember that the triangles formed by the diagonals intersecting are always similar. This allows you to use the area ratio theorem effectively.

 

Question 3. In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, prove that \( \frac{\text{Area}(ABC)}{\text{Area}(DBC)} = \frac{AO}{DO} \).

Answer: Let's draw perpendiculars AM to BC and DN to BC. Thus, the area of \( \triangle ABC = \frac{1}{2} \times BC \times AM \) and the area of \( \triangle DBC = \frac{1}{2} \times BC \times DN \).
\[ \frac{\text{Area}(ABC)}{\text{Area}(DBC)} = \frac{\frac{1}{2} \times BC \times AM}{\frac{1}{2} \times BC \times DN} \]
\[ \frac{\text{Area}(ABC)}{\text{Area}(DBC)} = \frac{AM}{DN} \quad \ldots(1) \]
Now, consider \( \triangle AMO \) and \( \triangle DNO \): We observe that: \( \angle AMO = \angle DNO \) (Each is 90 degrees, as AM and DN are perpendiculars) \( \angle AOM = \angle DON \) (Vertically Opposite Angles) Therefore, by the AA (Angle-Angle) similarity criterion, \( \triangle AMO \sim \triangle DNO \).
From the property of similar triangles, the ratio of corresponding sides is equal:
\[ \frac{AM}{DN} = \frac{AO}{DO} \quad \ldots(2) \]
From equations (1) and (2), we can conclude that:
\[ \frac{\text{Area}(ABC)}{\text{Area}(DBC)} = \frac{AO}{DO} \]
Hence, the proof is complete.
In simple words: એક જ પાયા પર બનેલા બે ત્રિકોણના ક્ષેત્રફળનો ગુણોત્તર, તેમની સંગત ઊંચાઈઓના ગુણોત્તર જેટલો હોય છે. પછી, આપણે બે નાના ત્રિકોણને સમરૂપ સાબિત કરીએ છીએ અને તેમની બાજુઓના ગુણોત્તરનો ઉપયોગ કરીને મૂળ ગુણોત્તર મેળવીએ છીએ.

Exam Tip: When proving ratios of areas of triangles, remember to draw perpendiculars to the common base. Also, keep an eye out for similar triangles that can help establish relationships between line segments.

 

Question 4. If the areas of two similar triangles are equal, then prove that they are congruent.
Answer: **Given:** \( \triangle ABC \sim \triangle PQR \) and \( \text{Area}(ABC) = \text{Area}(PQR) \).
**To Prove:** \( \triangle ABC \cong \triangle PQR \).
**Proof:** Since \( \triangle ABC \sim \triangle PQR \), the ratio of their areas is equal to the square of the ratio of their corresponding sides.
\[ \frac{\text{Area}(ABC)}{\text{Area}(PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{CA}{RP}\right)^2 \quad \ldots(1) \]
Given that \( \text{Area}(ABC) = \text{Area}(PQR) \).
So, we can write:
\[ \frac{\text{Area}(ABC)}{\text{Area}(PQR)} = 1 \quad \ldots(2) \]
From equations (1) and (2), we get:
\[ \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{CA}{RP}\right)^2 = 1 \]
Taking the square root, we have:
\[ \frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} = 1 \]
This implies:
\( AB = PQ \)
\( BC = QR \)
\( CA = RP \)
Therefore, by the SSS (Side-Side-Side) congruence criterion, \( \triangle ABC \cong \triangle PQR \).
Hence, if the areas of two similar triangles are equal, they are congruent.
In simple words: જો બે સમરૂપ ત્રિકોણના ક્ષેત્રફળ સરખા હોય, તો તેમની બધી બાજુઓ પણ સરખી હોય છે. જ્યારે બધી બાજુઓ સરખી હોય, ત્યારે ત્રિકોણ એકરૂપ કહેવાય છે, એટલે કે તેઓ એકબીજાની બરાબર હોય છે.

Exam Tip: This is a fundamental theorem. Clearly state the given conditions, what needs to be proven, and then use the properties of similar triangles and area ratios to establish side equality, leading to congruence by SSS.

 

Question 5. D, E, and F are the midpoints of sides AB, BC, and CA respectively of \( \triangle ABC \). Find the ratio of the areas of \( \triangle DEF \) and \( \triangle ABC \).

Answer: In \( \triangle ABC \), D, E, and F are the midpoints of sides AB, BC, and CA respectively. According to the Midpoint Theorem: The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side. Thus, we have: \( EF \parallel AB \) and \( EF = \frac{1}{2} AB \) \( DE \parallel AC \) and \( DE = \frac{1}{2} AC \) \( DF \parallel BC \) and \( DF = \frac{1}{2} BC \) Now, let's consider \( \triangle DEF \) and \( \triangle ABC \). From the properties above, we can see that all three sides of \( \triangle DEF \) are proportional to the corresponding sides of \( \triangle ABC \) with a ratio of 1:2. Therefore, by the SSS similarity criterion, \( \triangle DEF \sim \triangle ABC \). The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
\[ \frac{\text{Area}(DEF)}{\text{Area}(ABC)} = \left(\frac{EF}{AB}\right)^2 \quad \ldots(1) \]
We know that \( EF = \frac{1}{2} AB \). So, \( \frac{EF}{AB} = \frac{1}{2} \quad \ldots(2) \)
Substitute (2) into (1):
\[ \frac{\text{Area}(DEF)}{\text{Area}(ABC)} = \left(\frac{1}{2}\right)^2 \]
\[ \frac{\text{Area}(DEF)}{\text{Area}(ABC)} = \frac{1}{4} \]
Hence, the ratio of the areas of \( \triangle DEF \) to \( \triangle ABC \) is 1:4.
In simple words: જો તમે એક ત્રિકોણની બધી બાજુઓના મધ્યબિંદુઓને જોડો છો, તો તમને એક નવો નાનો ત્રિકોણ મળે છે. આ નાનો ત્રિકોણ મોટા ત્રિકોણ જેવો જ હોય છે, પણ તેની બાજુઓ અડધી હોય છે. તેથી, નાના ત્રિકોણનું ક્ષેત્રફળ મોટા ત્રિકોણના ક્ષેત્રફળના ચોથા ભાગનું હોય છે.

Exam Tip: The Midpoint Theorem is crucial for this type of problem. Remember that connecting midpoints creates a similar triangle whose sides are half the original, leading to an area ratio of 1:4.

 

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: **Given:** \( \triangle ABC \sim \triangle PQR \). AD and PM are medians of \( \triangle ABC \) and \( \triangle PQR \) respectively. **To Prove:** \( \frac{\text{Area}(ABC)}{\text{Area}(PQR)} = \left(\frac{AD}{PM}\right)^2 \). **Proof:** Since \( \triangle ABC \sim \triangle PQR \), their corresponding angles are equal, and the ratio of their corresponding sides is equal.
So, \( \angle B = \angle Q \) and \( \frac{AB}{PQ} = \frac{BC}{QR} \).
Since AD is the median of \( \triangle ABC \), D is the midpoint of BC. Thus, \( BD = \frac{1}{2} BC \).
Similarly, since PM is the median of \( \triangle PQR \), M is the midpoint of QR. Thus, \( QM = \frac{1}{2} QR \).
Substitute these into the side ratio:
\[ \frac{AB}{PQ} = \frac{BC}{QR} = \frac{2BD}{2QM} = \frac{BD}{QM} \]
Now, consider \( \triangle ABD \) and \( \triangle PQM \): We have: \( \angle B = \angle Q \) (from similarity of \( \triangle ABC \) and \( \triangle PQR \)) \( \frac{AB}{PQ} = \frac{BD}{QM} \) (as shown above)
Therefore, by the SAS (Side-Angle-Side) similarity criterion, \( \triangle ABD \sim \triangle PQM \).
From the similarity of \( \triangle ABD \) and \( \triangle PQM \), the ratio of their corresponding sides is equal:
\[ \frac{AB}{PQ} = \frac{AD}{PM} \quad \ldots(1) \]
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Since \( \triangle ABC \sim \triangle PQR \):
\[ \frac{\text{Area}(ABC)}{\text{Area}(PQR)} = \left(\frac{AB}{PQ}\right)^2 \quad \ldots(2) \]
Substitute (1) into (2):
\[ \frac{\text{Area}(ABC)}{\text{Area}(PQR)} = \left(\frac{AD}{PM}\right)^2 \]
Hence, the proof is complete.
In simple words: બે સમાન ત્રિકોણના ક્ષેત્રફળોનો ગુણોત્તર તેમની મધ્યગાઓના ગુણોત્તરના વર્ગ જેટલો હોય છે. આપણે પહેલા સાબિત કરીએ છીએ કે નાના ત્રિકોણ (બાજુ અને મધ્યગા દ્વારા બનેલા) પણ સમાન હોય છે, અને પછી મોટા ત્રિકોણના ક્ષેત્રફળના ગુણોત્તરને તેમની મધ્યગાઓના ગુણોત્તર સાથે જોડીએ છીએ.

Exam Tip: This proof relies on extending the similarity property to medians. Remember to use the SAS similarity criterion for the smaller triangles formed by the medians and then connect it back to the area ratio of the original triangles.

 

Question 7. Prove that the area of an equilateral triangle described on one side of a square is half the area of the equilateral triangle described on one of its diagonals.

Answer: Let ABCD be a square. Let \( \triangle PAB \) be an equilateral triangle constructed on side AB, and let \( \triangle QAC \) be an equilateral triangle constructed on diagonal AC. In \( \triangle ABC \), \( \angle B = 90^\circ \) and \( AB = BC \) (properties of a square). By the Pythagorean theorem, for \( \triangle ABC \):
\( AC^2 = AB^2 + BC^2 \)
Substitute \( BC = AB \):
\( AC^2 = AB^2 + AB^2 \)
\( AC^2 = 2AB^2 \)
This means \( \frac{AB^2}{AC^2} = \frac{1}{2} \).
So, \( \left(\frac{AB}{AC}\right)^2 = \frac{1}{2} \quad \ldots(1) \)
We know that all equilateral triangles are similar to each other. Therefore, \( \triangle PAB \sim \triangle QAC \). The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
\[ \frac{\text{Area}(PAB)}{\text{Area}(QAC)} = \left(\frac{AB}{AC}\right)^2 \]
Substitute the value from equation (1):
\[ \frac{\text{Area}(PAB)}{\text{Area}(QAC)} = \frac{1}{2} \]
Thus, \( \text{Area}(PAB) = \frac{1}{2} \text{Area}(QAC) \).
Hence, the area of an equilateral triangle described on one side of a square is half the area of the equilateral triangle described on its diagonal.
In simple words: આપણે એક ચોરસ લઈએ છીએ. જો આપણે તેની એક બાજુ પર એક સમબાજુ ત્રિકોણ બનાવીએ અને પછી તેના વિકર્ણ પર બીજો સમબાજુ ત્રિકોણ બનાવીએ, તો બાજુ પરના ત્રિકોણનું ક્ષેત્રફળ વિકર્ણ પરના ત્રિકોણના ક્ષેત્રફળથી અડધું હોય છે. આ પાયથાગોરસ પ્રમેય અને સમરૂપ ત્રિકોણના ગુણધર્મોનો ઉપયોગ કરીને સાબિત કરી શકાય છે.

Exam Tip: Remember that all equilateral triangles are similar. The key to this proof is using the Pythagorean theorem to establish the relationship between the side of the square and its diagonal, and then applying the area ratio theorem for similar triangles.

 

Question 8. D is the midpoint of BC, and there are two equilateral triangles ABC and BDE. The ratio of the areas of triangle ABC and BDE is .......... થાય.
(A) 2:1
(B) 1:2
(C) 4:1
(D) 1:4
Answer: (C) 4:1
In simple words: જો D BC નું મધ્યબિંદુ હોય, તો BC ની લંબાઈ BD કરતાં બમણી હોય છે. બે સમબાજુ ત્રિકોણ હંમેશા સમાન હોય છે. તેમના ક્ષેત્રફળનો ગુણોત્તર તેમની બાજુઓના ગુણોત્તરના વર્ગ જેટલો હોય છે. આ રીતે, ABC અને BDE ના ક્ષેત્રફળનો ગુણોત્તર 4:1 મળે છે.

Exam Tip: For problems involving equilateral triangles and midpoints, remember that all equilateral triangles are similar. Use the midpoint property to find the ratio of corresponding sides, then square this ratio to get the area ratio.

 

Question 9. The ratio of the sides of two similar triangles is 4:9. The ratio of the areas of these triangles is .......... થાય.
(A) 2:3
(B) 4:9
(C) 81:16
(D) 16:81
Answer: (D) 16:81
According to Theorem 6.6, for two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Given that the ratio of the sides is 4:9.
Ratio of areas = \( \left(\text{Ratio of sides}\right)^2 \)
Ratio of areas = \( \left(\frac{4}{9}\right)^2 \)
Ratio of areas = \( \frac{4^2}{9^2} \)
Ratio of areas = \( \frac{16}{81} \)
Thus, the ratio of the areas is 16:81.
In simple words: જો બે ત્રિકોણ સમાન હોય, તો તેમના ક્ષેત્રફળનો ગુણોત્તર તેમની બાજુઓના ગુણોત્તરના વર્ગ જેટલો હોય છે. તેથી, જો બાજુઓનો ગુણોત્તર 4:9 હોય, તો ક્ષેત્રફળનો ગુણોત્તર (4 નો વર્ગ):(9 નો વર્ગ) એટલે કે 16:81 થશે.

Exam Tip: This is a direct application of the theorem regarding the ratio of areas of similar triangles. Simply square the given ratio of sides to find the ratio of their areas.