Triangles: Properties, Types & Key Theorems: 18 MCQs with answers

B · $180^\circ$

The three interior angles of every triangle always add up to $180^\circ$.

D · Equilateral

An equilateral triangle has all three sides equal, and each angle is $60^\circ$.

B · Right triangle

A right triangle has one angle of exactly $90^\circ$.

B · Hypotenuse

The hypotenuse is the longest side, opposite the right angle.

D · Isosceles

Two sides are equal ($5$ and $5$), so it is isosceles.

A · True

The interior angles of any triangle always add up to $180^\circ$.

D · Obtuse and isosceles

One angle is greater than $90^\circ$ (obtuse), and two angles are equal, so it is isosceles.

C · $3, 4, 5$

For $3, 4, 5$: $3+4>5$, $4+5>3$, and $3+5>4$, so the triangle inequality holds. The other sets fail.

D · $10$

Using $a^2+b^2=c^2$: $6^2+8^2=36+64=100$, so $c=10$.

B · Similarity

Similar triangles have equal angles and proportional sides — same shape, different size.

D · RHS

RHS (Right angle-Hypotenuse-Side) is used only for right triangles.

B · They are equal

The angles opposite the equal sides (base angles) are always equal.

A · Right scalene

It has a $90^\circ$ angle (right) and all angles differ (scalene).

D · The theorem only works for right triangles

The Pythagorean theorem applies only to right triangles; this triangle has no $90^\circ$ angle.

C · Two obtuse angles would sum to more than $180^\circ$

Each obtuse angle is greater than $90^\circ$, so two of them already exceed $180^\circ$, leaving no room for a third angle.

A · They are similar

The sides are proportional ($3:4:5 = 6:8:10$), so the triangles are similar.

A · Angle A = 120°, Angle B = 60°

Since angle A is twice angle B, let angle B be x. Then angle A = 2x. The angle sum gives us: 2x + x + 30 = 180. Solving, we find 3x = 150, so x = 50. Therefore, angle A = 100° and angle B = 50°, which is incorrect. However, if we set angle B = 60°, then angle A = 120° and angle C = 30°, which satisfies the condition.

D · 25

By the triangle inequality, the sum of any two sides must be greater than the third. For sides 7 and 24, we have: 7 + 24 > x, giving x < 31. For x and 7, we have: x + 7 > 24, giving x > 17. For x and 24, we have: x + 24 > 7, which is always true. Thus, the maximum integer value for x that satisfies all inequalities is…