Chapter 4: Geometry — Online MCQ Test
MATHS · CLASS 10th · Tamil Nadu State Board
Practice Chapter 4: Geometry with a free chapter-wise online MCQ test.
This chapter covers: Thales theorem Angle Bisector theorem Pythagoras theorem Similar triangles Tangent Circle Construction Concurrent lines Median Altitude Perpendicular bisector Geometric proof.
AI-generated questions from basic to board-exam level, with instant results and explanations.
Chapter 4: Geometry — Important Questions & Answers
In a triangle, a line drawn parallel to one side intersects the other two sides. Which theorem states that these two sides are divided in the same ratio?
- A. Thales theorem
- B. Pythagoras theorem
- C. Angle bisector theorem
- D. Median theorem
Answer: A. Thales theorem
Thales theorem, also called the Basic Proportionality Theorem, states that a line parallel to one side of a triangle divides the other two sides proportionally.
Thales theorem, also called the Basic Proportionality Theorem, states that a line parallel to one side of a triangle divides the other two sides proportionally.
In triangle ABC, AD is the internal angle bisector of angle A meeting BC at D. According to the angle bisector theorem, which relation is true?
- A. AB/AC = BD/DC
- B. AB/BC = AC/DC
- C. BD/AB = DC/BC
- D. AB + AC = BD + DC
Answer: A. AB/AC = BD/DC
The angle bisector theorem says that an internal angle bisector divides the opposite side in the ratio of the adjacent sides.
The angle bisector theorem says that an internal angle bisector divides the opposite side in the ratio of the adjacent sides.
Fill in the blank: In a right triangle with legs a, b and hypotenuse c, Pythagoras theorem is ______.
- A. a + b = c
- B. a² + b² = c²
- C. a² - b² = c²
- D. 2ab = c²
Answer: B. a² + b² = c²
Pythagoras theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Pythagoras theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.
A tangent is drawn to a circle at point P. If O is the centre of the circle, then angle between OP and the tangent at P is:
- A. 45°
- B. 60°
- C. 90°
- D. 180°
Answer: C. 90°
The radius drawn to the point of contact is perpendicular to the tangent. Hence the angle is 90°.
The radius drawn to the point of contact is perpendicular to the tangent. Hence the angle is 90°.
In triangle ABC, D lies on AB and E lies on AC. If DE is parallel to BC, AD = 4 cm, DB = 6 cm and AE = 6 cm, then EC is:
- A. 4 cm
- B. 6 cm
- C. 9 cm
- D. 10 cm
Answer: C. 9 cm
By Thales theorem, AD/DB = AE/EC. So 4/6 = 6/EC, giving EC = 9 cm.
By Thales theorem, AD/DB = AE/EC. So 4/6 = 6/EC, giving EC = 9 cm.
In triangle ABC, AD is the angle bisector of angle A meeting BC at D. If AB = 5 cm, AC = 7 cm and BD = 10 cm, then DC is:
- A. 12 cm
- B. 14 cm
- C. 15 cm
- D. 17 cm
Answer: B. 14 cm
By the angle bisector theorem, AB/AC = BD/DC. Thus 5/7 = 10/DC, so DC = 14 cm.
By the angle bisector theorem, AB/AC = BD/DC. Thus 5/7 = 10/DC, so DC = 14 cm.
In triangle ABC, D lies on AB and E lies on AC. If DE is parallel to BC, AD = 3 cm, DB = 2 cm and AE = 4.5 cm, then AC is:
- A. 6 cm
- B. 7.5 cm
- C. 8 cm
- D. 9 cm
Answer: B. 7.5 cm
Since DE is parallel to BC, AD/AB = AE/AC. Here AB = 5 cm, so 3/5 = 4.5/AC, giving AC = 7.5 cm.
Since DE is parallel to BC, AD/AB = AE/AC. Here AB = 5 cm, so 3/5 = 4.5/AC, giving AC = 7.5 cm.
Two similar triangles have perimeters 18 cm and 30 cm. If the area of the smaller triangle is 72 cm², then the area of the larger triangle is:
- A. 120 cm²
- B. 150 cm²
- C. 180 cm²
- D. 200 cm²
Answer: D. 200 cm²
The ratio of corresponding sides is 18 : 30 = 3 : 5. Hence the area ratio is 9 : 25, so the larger area is 72 × 25/9 = 200 cm².
The ratio of corresponding sides is 18 : 30 = 3 : 5. Hence the area ratio is 9 : 25, so the larger area is 72 × 25/9 = 200 cm².
In triangle ABC, D lies on AB and E lies on AC such that DE is parallel to BC. If AD : DB = 2 : 3 and the area of quadrilateral DBCE is 84 cm², then the area of triangle ADE is:
- A. 12 cm²
- B. 16 cm²
- C. 20 cm²
- D. 24 cm²
Answer: B. 16 cm²
Since AD : AB = 2 : 5, area of triangle ADE : area of triangle ABC = 4 : 25. Thus the quadrilateral is 21 parts = 84, so 1 part = 4 and area of triangle ADE = 16 cm².
Since AD : AB = 2 : 5, area of triangle ADE : area of triangle ABC = 4 : 25. Thus the quadrilateral is 21 parts = 84, so 1 part = 4 and area of triangle ADE = 16 cm².
In right triangle ABC, angle A = 90°. The angle bisector of angle A meets hypotenuse BC at D. If BD : DC = 3 : 4 and BC = 25 cm, then the area of triangle ABC is:
- A. 120 cm²
- B. 135 cm²
- C. 150 cm²
- D. 180 cm²
Answer: C. 150 cm²
By the angle bisector theorem, AB : AC = 3 : 4. Since the hypotenuse is 25 cm, the sides are 15 cm and 20 cm, so area = 1/2 × 15 × 20 = 150 cm².
By the angle bisector theorem, AB : AC = 3 : 4. Since the hypotenuse is 25 cm, the sides are 15 cm and 20 cm, so area = 1/2 × 15 × 20 = 150 cm².