JEE Main Maths: Properties of Triangles questions with solutions

Q1. In triangle ABC, one angle is 60 degrees, the area is 10*sqrt(3) sq cm, and the perimeter is 20 cm. If a > b > c, where a, b, c are the sides opposite to angles A, B, C respectively, which statements are correct? (A) The inradius of the triangle equals sqrt(3). (B) The length of the longest side is 7. (C) The circumradius of the triangle equals 7/sqrt(3). (D) (incomplete)

1. (A) Inradius = sqrt(3)
2. (B) Longest side = 7
3. (C) Circumradius = 7/sqrt(3)
4. (D) (not available)

Answer: (A) Inradius = sqrt(3)

With angle C = 60 deg: area = (1/2)ab sin60 = 10*sqrt(3) gives ab = 40. Cosine rule gives c = 7. Then a+b = 13 and ab = 40, so a = 8, b = 5. Inradius r = Area/s = 10*sqrt(3)/10 = sqrt(3). Longest side = a = 8, not 7. Circumradius R = c/(2sinC) = 7/sqrt(3). So A and C are correct; B is false.

Q2. Let a, b, c be the sides of a triangle with semi-perimeter S. Given that b*c/(b+c) + c*a/(c+a) + a*b/(a+b) = S, what type of triangle must it be?

1. Equilateral
2. Scalene
3. Isosceles
4. Right isosceles

Answer: Equilateral

Semi-perimeter S = (a+b+c)/2. Each term bc/(b+c) <= (b+c)/4 by AM-HM. Summing: LHS <= (a+b+c)/2 = S, with equality iff a = b = c. So the given condition is satisfied if and only if the triangle is equilateral.

Q3. In a right-angled triangle the radius of the inscribed circle is 9 and the radius of the circumscribed circle is 37.5. Find the perimeter of the triangle.

1. 168
2. 150
3. 120
4. 180

Answer: 168

In a right triangle the circumradius is half the hypotenuse, and the inradius relates the two legs and hypotenuse by r = (a + b - c)/2. Combining gives the perimeter directly.

Q4. Can the three altitudes of a triangle be in the ratio 2: 5: 6? State whether such a triangle can exist and why.

1. Yes, because any three positive numbers can serve as the altitudes of a triangle.
2. No, because the corresponding side lengths would violate the triangle inequality.
3. Yes, because the altitudes are always directly proportional to the sides.
4. No, because the altitudes of any triangle must all be equal.

Answer: No, because the corresponding side lengths would violate the triangle inequality.

Since area = (1/2)*base*height is the same for each side, the sides are inversely proportional to the altitudes. If altitudes are 2: 5: 6, the sides are proportional to 1/2: 1/5: 1/6 = 15: 6: 5 (multiplying by 30). Check triangle inequality on sides 15, 6, 5: 6 + 5 = 11 < 15, so the inequality fails. Therefore no such triangle exists.

Q5. In triangle ABC, AD is the bisector of angle A (D on BC). Which statement correctly gives the ratio of the areas of triangle ABD to triangle ACD, and why?

1. Area(ABD)/Area(ACD) = AB/AC, because both triangles have the same height from A and BD/DC = AB/AC
2. Area(ABD)/Area(ACD) = AC/AB, because the bisector splits BC inversely to the adjacent sides
3. Area(ABD)/Area(ACD) = (AB/AC)², by similarity of the two sub-triangles
4. Area(ABD)/Area(ACD) = 1, because an angle bisector always halves the area

Answer: Area(ABD)/Area(ACD) = AB/AC, because both triangles have the same height from A and BD/DC = AB/AC

Triangles ABD and ACD share the same vertex A and their bases BD, DC lie on the same line BC, so both have the same height h (the perpendicular distance from A to BC). Therefore Area(ABD)/Area(ACD) = (1/2 * BD * h)/(1/2 * DC * h) = BD/DC. By the angle bisector theorem, BD/DC = AB/AC. Hence Area(ABD)/Area(ACD) = AB/AC.

Q6. In triangle ABC, the two equal sides satisfy AB = AC = 6. If the triangle's circumradius is 5, find the length of the base BC.

1. 25/3
2. 9
3. 48/5
4. 10

Answer: 48/5

For the isosceles triangle, express the circumradius in terms of the sides and the area (found via the altitude to the base), then solve for the base a = BC.

Q7. ABC is a right triangle with the right angle at A. A circle is inscribed in the triangle. If the two legs containing the right angle are 6 cm and 8 cm, find the radius of the inscribed circle.

1. 2 cm
2. 1 cm
3. 3 cm
4. 2.5 cm

Answer: 2 cm

Legs 6 and 8, hypotenuse = sqrt(36 + 64) = 10. For a right triangle, r = (leg1 + leg2 - hypotenuse)/2 = (6 + 8 - 10)/2 = 4/2 = 2 cm. (Equivalently r = area/s = 24/12 = 2.)