JEE Mathematics Properties of Triangles MCQs Set 01

Answer: (d) \( 16R^2 \)

Explanation:
1. \( r=4R\sin\frac A2\sin\frac B2\sin\frac C2 \) and \( r_1=4R\sin\frac A2\cos\frac B2\cos\frac C2 \) (similarly r₂, r₃). A known result is \( r^2+r_1^2+r_2^2+r_3^2=16R^2-(a^2+b^2+c^2) \).
2. Adding \( a^2+b^2+c^2 \) gives \( 16R^2 \).

Teacher's Note:
1. Recall the standard identity connecting the squares of the inradius, exradii and sides.
2. Check with an equilateral triangle of side a: R = a/√3, r = a/(2√3), r₁ = r₂ = r₃ = (√3/2)a.

Answer: (a) 1

Explanation:
1. In an AP, B = 60° and A + C = 120°.
2. \( 4\sin A\sin C=2[\cos(A-C)-\cos(A+C)]=2\cos(A-C)+1 \), so \( 3-4\sin A\sin C=2-2\cos(A-C)=4\sin^2\frac{A-C}{2} \).
3. \( \frac{\sqrt{\cdot}}{|A-C|}=\frac{2|\sin\frac{A-C}2|}{|A-C|}\to1 \).

Teacher's Note:
1. Use the AP condition to get A + C = 120°.
2. Convert the product of sines to cosines, then use sin t/t → 1.

Answer: (a) 5

Explanation:
1. \( \frac Rr=\frac1{\cos(\pi/n)} \) for a regular n-gon.
2. \( \cos\frac\pi n=\frac1{\sqrt5-1}=\frac{\sqrt5+1}{4}=\cos36^\circ \), so \( \frac\pi n=\frac\pi5 \) and n = 5.

Teacher's Note:
1. For a regular polygon, r = R cos(π/n).
2. cos 36° = (√5 + 1)/4.

Answer: (d) \( \frac78 \)

Explanation:
1. \( 2a^2+4b^2+c^2-4ab-2ac=(a-2b)^2+(a-c)^2=0 \), so a = 2b and a = c.
2. Take b = 1: a = c = 2. \( \cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{4+4-1}{8}=\frac78 \).

Teacher's Note:
1. Group the terms into two perfect squares.
2. A sum of two squares is zero only if each is zero.

Answer: (b) \( \frac13(a^2+b^2+c^2) \)

Explanation:
1. Each median is \( m_a^2=\frac{2b^2+2c^2-a^2}{4} \) and \( GA=\frac23m_a \).
2. \( GA^2+GB^2+GC^2=\frac49\cdot\frac{3(a^2+b^2+c^2)}{4}=\frac13(a^2+b^2+c^2) \).

Teacher's Note:
1. The centroid divides each median in the ratio 2 : 1.
2. Use Apollonius' theorem for the median lengths.

Answer: (c) \( 3c^2a \)

Explanation:
1. Since b = c, \( a=2b\sin10^\circ \). Let \( s=\frac ab=2\sin10^\circ \).
2. From \( \sin30^\circ=3\sin10^\circ-4\sin^310^\circ=\frac12 \): \( 8\sin^310^\circ=6\sin10^\circ-1 \), i.e. \( s^3=3s-1 \).
3. Multiply by \( b^3 \): \( a^3=3ab^2-b^3\Rightarrow a^3+b^3=3ab^2=3c^2a \) (as b = c).

Teacher's Note:
1. Find the side ratio a/b from the isosceles triangle, then use sin 3θ.
2. Replace b² by c² to match the listed form.

Answer: (c) 0 < k < 4

Explanation:
1. \( (b+c)^2-a^2=b^2+c^2-a^2+2bc=2bc\cos A+2bc=2bc(1+\cos A) \).
2. So \( k=2(1+\cos A) \). Since 0 < A < π, −1 < cos A < 1, so 0 < k < 4.

Teacher's Note:
1. Use the cosine rule to replace b² + c² − a².
2. Range of cos A for a triangle is (−1, 1).

Answer: (c) \( \frac14 \)

Explanation:
1. \( a^2-(b-c)^2=a^2-b^2-c^2+2bc=2bc-2bc\cos A=2bc(1-\cos A) \).
2. Also \( \Delta=\frac12bc\sin A \). So \( \frac12\sin A=2(1-\cos A) \), i.e. \( \frac{\sin A}{1-\cos A}=\cot\frac A2=4 \).
3. Therefore \( \tan\frac A2=\frac14 \).

Teacher's Note:
1. Express a² − (b − c)² using the cosine rule.
2. sin A/(1 − cos A) = cot(A/2).

Answer: (c) right angled

Explanation:
1. Using \( a=2R\sin A \) etc.: \( 4R^2(\sin^2A+\sin^2B+\sin^2C)=8R^2 \), so \( \sin^2A+\sin^2B+\sin^2C=2 \).
2. This gives \( \cos^2A+\cos^2B+\cos^2C=1 \), and \( \cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1 \) in any triangle, so \( \cos A\cos B\cos C=0 \).
3. One angle is 90°: the triangle is right angled.

Teacher's Note:
1. Replace each side by 2R sin of the opposite angle.
2. Use the identity cos²A + cos²B + cos²C + 2cosA cosB cosC = 1.

Answer: (a) 9

Explanation:
1. \( \frac{\sin^2A+\sin A+1}{\sin A}=\sin A+1+\frac1{\sin A} \).
2. Since 0 < sin A ≤ 1 with sin A < 1 unless A = 90°, \( \sin A+\frac1{\sin A}\ge2 \) with equality only if sin A = 1.
3. The three angles cannot all be right angles, so the sum \( >3\cdot(2+1)=9 \) strictly.

Teacher's Note:
1. Rewrite each term as sin A + 1 + 1/sin A and apply AM-GM.
2. Equality would need all three sines equal to 1, impossible in a triangle.

Answer: (c) \( \frac\pi2 \)

Explanation:
1. Multiply by abc: \( 2bc\cos A+ca\cos B+2ab\cos C=a^2+b^2 \).
2. Using the cosine rule: \( (b^2+c^2-a^2)+\frac{c^2+a^2-b^2}{2}+(a^2+b^2-c^2)=a^2+b^2 \).
3. Simplify: \( 2b^2+\frac{c^2+a^2-b^2}2=a^2+b^2\Rightarrow b^2+c^2=a^2 \), so A = π/2.

Teacher's Note:
1. Clear the denominators and replace each product with a cosine-rule expression.
2. b² + c² = a² means a right angle at A.

Answer: (a) \( 3r \)

Explanation:
1. \( \frac1{r_1}+\frac1{r_2}+\frac1{r_3}=\frac{s-a}\Delta+\frac{s-b}\Delta+\frac{s-c}\Delta=\frac s\Delta=\frac1r \).
2. The harmonic mean is \( \frac{3}{\frac1{r_1}+\frac1{r_2}+\frac1{r_3}}=3r \).

Teacher's Note:
1. Use r_i = Δ/(s − a) etc.
2. The sum of the three (s − a) values is s.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( r_1+r_2+r_3=4R+r \). Since R ≥ 2r (Euler's inequality), \( r\le\frac R2 \), so \( r_1+r_2+r_3\le\frac{9R}2 \) with equality for an equilateral triangle.
2. Both statements are true and R is exactly what bounds the sum.

Teacher's Note:
1. Recall r₁ + r₂ + r₃ = 4R + r.
2. Euler's inequality R ≥ 2r gives r ≤ R/2.