Geometry (Lines, Angles, Triangles, Polygons) Questions and Answers

Answer: D

The three altitudes of a triangle are concurrent at the orthocentre.

The other three classical centres come from different cevians: the medians meet at the centroid, the angle bisectors at the incentre, and the perpendicular bisectors of the sides at the circumcentre.

Answer: C

Let the angles be x, x, and 2x.

Sum of angles = x + x + 2x = 180°

4x = 180°

x = 45°

The angles are 45°, 45°, and 2x = 90°.

Since two angles are equal (45°), the triangle is isosceles.

Since one angle is 90°, the triangle is right-angled.

Answer: A

\[ 90^{\circ} - 63^{\circ} = 27^{\circ} \]

The 117° on offer is the supplement, which uses 180° instead.

Answer: C

We check if the side lengths satisfy the Pythagorean theorem. The longest side is AC.

\(AB^2 + BC^2 = 6^2 + 8^2 = 36 + 64 = 100\)

\(AC^2 = 10^2 = 100\)

Since \(AB^2 + BC^2 = AC^2\), the triangle is right-angled, and the right angle is opposite the hypotenuse AC. Therefore, angle B is 90°.

Answer: C

Let the angles be 5x, 3x, and 10x.

5x + 3x + 10x = 180°

18x = 180° => x = 10°

The angles are 50°, 30°, and 100°.

The largest angle is 100° and the smallest is 30°.

The difference is 100° - 30° = 70°.

Answer: C

In an equilateral triangle, the altitude bisects the base. This creates two 30-60-90 right-angled triangles.

The hypotenuse is 'a', the base is 'a/2', and the altitude is 'h'.

Using the Pythagorean theorem: \(h^2 + (a/2)^2 = a^2\)

\(h^2 = a^2 - a^2/4 = \frac{3a^2}{4}\)

\(h = \sqrt{\frac{3a^2}{4}} = \frac{a\sqrt{3}}{2}\).

Answer: C

The formula for the number of diagonals is \(\frac{n(n-3)}{2}\).

\(\frac{n(n-3)}{2} = 27\)

\(n(n-3) = 54\)

We need two numbers with a difference of 3 that multiply to 54. By inspection, 9 and 6 satisfy this condition (9 × 6 = 54). Therefore, n=9. The polygon is a nonagon.

Answer: A

Correct answer: 115°

The angles of \(\triangle ABC\) sum to \(180°\), so

\[\angle C = 180° - 90° - 40° = 50°\]

CD bisects it, giving \(\angle ACD = 25°\).

In \(\triangle ADC\) the angles at A and C are \(90°\) and \(25°\), so \(\angle ADC = 65°\). Since D lies on AB, \(\angle BDC\) is its supplement:

\[\angle BDC = 180° - 65° = 115°\]

The exterior-angle theorem gives the same result directly: in \(\triangle ADC\), \(\angle BDC = \angle A + \angle ACD = 90° + 25° = 115°\).

Answer: A

Correct answer: 120°

First, \(\angle A = 180° - 90° - 30° = 60°\), and AD bisects it, so \(\angle BAD = \angle DAC = 30°\).

In \(\triangle ABD\) the angles at B and A are \(90°\) and \(30°\), so \(\angle ADB = 60°\). As D lies on BC, \(\angle ADC\) is its supplement:

\[\angle ADC = 180° - 60° = 120°\]

Checking against \(\triangle ADC\) directly: its angles \(\angle DAC = 30°\) and \(\angle C = 30°\) leave \(\angle ADC = 120°\).

Answer: C

The exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Here, \(\angle ACD\) is the exterior angle, and \(\angle ABC\) and \(\angle BAC\) are the interior opposite angles.

\(\angle ACD = \angle ABC + \angle BAC\)

120° = 45° + \(\angle BAC\)

\(\angle BAC = 120° - 45° = 75°\).