JEE Mathematics Geometrical Constructions MCQs Set 01

Answer: (b) AB + AC

Explanation:
1. A is on the perpendicular bisector of CD, so AC = AD.
2. So BD = BA + AD = AB + AC.

Teacher's Note:
1. The perpendicular bisector property turns AC into AD.
2. D is placed at the given sum from B.

Answer: (a) 7.5 cm

Explanation:
1. BD must equal the given sum AB + AC.
2. So BD = 7.5 cm.

Teacher's Note:
1. The base length 6 cm is only used for BC.
2. D is joined to C and its perpendicular bisector gives A.

Answer: (d) the intersection of the perpendicular bisector of CD with BD

Explanation:
1. A must satisfy AC = AD, which means A lies on the perpendicular bisector of CD.
2. It must also lie on ray BD, so it is their intersection.

Teacher's Note:
1. The perpendicular bisector is the locus of points equidistant from C and D.
2. The midpoint of BD generally does not satisfy AC = AD.

Answer: (c) x > a

Explanation:
1. In a triangle the sum of two sides exceeds the third side.
2. So AB + AC > BC, that is x > a.

Teacher's Note:
1. This is the triangle inequality.
2. If x ≤ a, point A cannot be found.

Answer: (b) d < a

Explanation:
1. In a triangle the difference of two sides is less than the third side.
2. So |AB − AC| < BC, that is d < a.

Teacher's Note:
1. Triangle inequality again, written in difference form.
2. The difference is taken as a positive number.

Answer: (a) BD

Explanation:
1. A is on the perpendicular bisector of CD, so AC = AD.
2. AB − AC = AB − AD = BD = d.

Teacher's Note:
1. Here D lies between B and A.
2. The perpendicular bisector again converts AC into AD.

Answer: (d) AC = AD' = AB + BD'

Explanation:
1. A lies on the perpendicular bisector of CD', so AC = AD'.
2. D' is on the opposite side of B, so AD' = AB + BD'. Hence AC − AB = BD' = d.

Teacher's Note:
1. The extension is needed because AC is the longer side.
2. Draw a rough sketch to see why D' lies behind B.

Answer: (c) 30°

Explanation:
1. The rays at the end points of PQ are drawn at half the base angles.
2. At P: 60° ÷ 2 = 30° (and at Q: 45° ÷ 2 = 22.5°).

Teacher's Note:
1. The half angles come from the isosceles triangles formed by the perpendicular bisectors.
2. Do not draw the full base angles.

Answer: (b) PQ

Explanation:
1. BA = BP and CA = CQ because B and C lie on the perpendicular bisectors.
2. AB + BC + CA = PB + BC + CQ = PQ.

Teacher's Note:
1. This is why PQ is drawn equal to the perimeter.
2. The sides AB and AC are 'folded' on to PQ.

Answer: (a) 60°, 45° and 75°

Explanation:
1. Angle ABC is the exterior angle of triangle ABP with angle BPA = angle BAP = 30°, so it is 60°.
2. Likewise angle ACB = 22.5° + 22.5° = 45°, and angle A = 180° − 60° − 45° = 75°.

Teacher's Note:
1. The angles 30° and 22.5° belong to the auxiliary triangle APQ.
2. The required triangle has the angles given in the question.

Answer: (d) \( \frac{32}{7} \) cm

Explanation:
1. Let AB = c, so AC = 10 − c. By the cosine rule (10 − c)² = c² + 36 − 2(6)(c)(1/2).
2. 100 − 20c = 36 − 6c, so 14c = 64 and c = 32/7 cm.

Teacher's Note:
1. The cosine rule connects the three sides with the angle B.
2. Then AC = 38/7 cm and AB + AC = 10 cm.

Answer: (c) 40 cm

Explanation:
1. Let AB = c and AC = c − 3. By the cosine rule (c − 3)² = c² + 49 − 2(7)(c)(1/2).
2. c² − 6c + 9 = c² + 49 − 7c, so c = 40 cm (and AC = 37 cm).

Teacher's Note:
1. The data are valid since 7 + 37 > 40.
2. The cosine rule gives a linear equation because c² cancels.

Answer: (b) 45°

Explanation:
1. Angles of a triangle add up to 180°.
2. Third angle = 180° − 75° − 60° = 45°.

Teacher's Note:
1. The perimeter does not affect the angles.
2. In the construction the rays at the ends of PQ are drawn at 37.5° and 30°.

Answer: (a) E with radius 3.5 cm and B with radius 5 cm

Explanation:
1. A is at distance 3.5 cm from E (the median) and 5 cm from B.
2. So draw arcs centred at E (radius 3.5) and at B (radius 5).

Teacher's Note:
1. First bisect BC to locate E.
2. Each given length becomes one arc.

Answer: (d) \( AC^2 = 7.5 \)

Explanation:
1. By the median formula AB² + AC² = 2(AE² + BE²).
2. 25 + AC² = 2(12.25 + 4) = 32.5, so AC² = 7.5.

Teacher's Note:
1. BE is half of BC, so BE = 2 cm.
2. This is the Apollonius theorem.

Answer: (c) the diagonal BD

Explanation:
1. The line through C parallel to BD meets AB produced at E.
2. Triangles DBC and DBE are on base DB between the same parallels, so they have equal area.

Teacher's Note:
1. The diagonal chosen is the one that does not pass through C.
2. Then D is joined to E.

Answer: (b) triangle AED

Explanation:
1. Area(ABCD) = Area(ABD) + Area(DBC).
2. Area(DBC) = Area(DBE), so the sum is Area(ABD) + Area(DBE) = Area(AED).

Teacher's Note:
1. E lies on AB produced beyond B.
2. The vertex D stays fixed.

Answer: (a) they are on the same base DB and between the parallels DB and CE

Explanation:
1. C and E lie on the line parallel to DB.
2. Triangles on the same base between the same parallels have equal area.

Teacher's Note:
1. They are equal in area, not congruent.
2. This is Theorem 2 from the areas chapter.

Answer: (d) 40 cm²

Explanation:
1. The construction preserves area: Area(AED) = Area(ABCD).
2. So the area is 40 cm².

Teacher's Note:
1. The new triangle always has the same area as the polygon.
2. Do not apply a factor of 1/2 here.

Answer: (c) 2

Explanation:
1. Each step reduces the number of sides by one.
2. Pentagon → quadrilateral → triangle takes 2 steps.

Teacher's Note:
1. An n-gon needs n − 3 steps to become a triangle.
2. A hexagon would need 3 steps.