Lines, Triangles and Circles for CUET General Aptitude

AnswerA · 36°

1. Complementary angles add up to 90°.

2. Let the smaller angle be $x$. Then $x + (x + 18^\circ) = 90^\circ$, so $2x = 72^\circ$ and $x = 36^\circ$.

AnswerB · 71°

1. Let the angle be $x$. Its supplement is $180^\circ - x$.

2. $x = (180^\circ - x) - 38^\circ$, so $2x = 142^\circ$ and $x = 71^\circ$.

3. Check: the supplement is 109°, and $109^\circ - 71^\circ = 38^\circ$.

AnswerD · 110°

1. $\angle AOC$ and $\angle BOD$ are vertically opposite, so they are equal: $3x + 10 = 5x - 30$, giving $x = 20$.

2. So $\angle AOC = 3 \times 20 + 10 = 70^\circ$.

3. $\angle BOC$ sits next to $\angle AOC$ on the straight line AB, so $\angle BOC = 180^\circ - 70^\circ = 110^\circ$.

AnswerC · 116°

1. Interior angles on the same side of a transversal (co-interior angles) add up to 180° when the lines are parallel.

2. Other angle = $180^\circ - 64^\circ = 116^\circ$.

AnswerB · 85°

1. Corresponding angles between parallel lines are equal: $2x + 15 = 3x - 20$.

2. So $x = 35$.

3. Each angle = $2 \times 35 + 15 = 85^\circ$.

AnswerC · 67.5°

1. Let the angle be $x$. Supplement = $180^\circ - x$, complement = $90^\circ - x$.

2. $180 - x = 5(90 - x) = 450 - 5x$.

3. $4x = 270$, so $x = 67.5^\circ$.

AnswerC · 95°

1. Draw a line through E parallel to AB (and so to CD). It splits $\angle BED$ into two parts.

2. The upper part equals $\angle ABE = 40^\circ$ (alternate angles).

3. The lower part equals $\angle CDE = 55^\circ$ (alternate angles).

4. $\angle BED = 40^\circ + 55^\circ = 95^\circ$.

AnswerB · 90°

1. The two co-interior angles are 124° and $180^\circ - 124^\circ = 56^\circ$.

2. Their halves are 62° and 28°.

3. These halves are two angles of the triangle formed at P, so the angle at $P = 180^\circ - 62^\circ - 28^\circ = 90^\circ$.

AnswerA · 45°

1. The three angles together make the straight angle $XOY = 180^\circ$.

2. Total parts = $3 + 4 + 5 = 12$, so one part = $180^\circ \div 12 = 15^\circ$.

3. $\angle XOP = 3 \times 15^\circ = 45^\circ$.

AnswerD · 110°

1. Co-interior angles between parallel lines add up to 180°.

2. Let the smaller be $x$; the larger is $2x - 30$. So $x + 2x - 30 = 180$, giving $x = 70$.

3. Larger angle = $2 \times 70 - 30 = 110^\circ$.

AnswerA · 70°

1. Complement + supplement = $(90^\circ - x) + (180^\circ - x) = 270^\circ - 2x$.

2. $270 - 2x = 130$, so $2x = 140$ and $x = 70^\circ$.

3. Check: complement 20°, supplement 110°, total 130°.

AnswerB · 70°

1. Since E is to the right, $\angle ABE$ is measured from BA (pointing left) round to BE, so the angle between BE and the rightward part of AB is $180^\circ - 150^\circ = 30^\circ$.

2. Similarly at D, the angle between DE and the rightward part of CD is $180^\circ - 140^\circ = 40^\circ$.

3. Draw a line through E parallel to AB. By alternate angles, $\angle BED = 30^\circ + 40^\circ = 70^\circ$.

4. Equivalently, $\angle BED = 360^\circ - 150^\circ - 140^\circ = 70^\circ$.

AnswerC · 75°

1. The three angles $\angle AOP$, $\angle POQ$, $\angle QOB$ make the straight angle: $\angle POQ = 180^\circ - 40^\circ - 70^\circ = 70^\circ$.

2. OR bisects it, so $\angle POR = 35^\circ$.

3. $\angle AOR = \angle AOP + \angle POR = 40^\circ + 35^\circ = 75^\circ$.

AnswerC · $80^\circ$

1. The angles of any triangle add to $180^\circ$.

2. The ratio has $2 + 3 + 4 = 9$ parts, so one part is $180^\circ \div 9 = 20^\circ$.

3. The largest angle has 4 parts: $4 \times 20^\circ = 80^\circ$.

4. Check: $40^\circ + 60^\circ + 80^\circ = 180^\circ$.

AnswerB · 65°

1. The angles of a triangle add up to 180°.

2. $\angle R = 180^\circ - 48^\circ - 67^\circ = 65^\circ$.

AnswerD · 70°

1. The two base angles are equal and add up to $180^\circ - 40^\circ = 140^\circ$.

2. Each base angle = $140^\circ \div 2 = 70^\circ$.

AnswerC · 76°

1. The two equal angles total $2 \times 52^\circ = 104^\circ$.

2. Third angle = $180^\circ - 104^\circ = 76^\circ$.

AnswerD · 4 cm, 7 cm, 10 cm

1. In a triangle, the two shorter sides must add up to MORE than the longest side.

2. $3 + 4 = 7$ is less than 8; $5 + 6 = 11$ is less than 12; $2 + 9 = 11$ only equals 11, so those sticks lie flat.

3. $4 + 7 = 11$ is more than 10, so 4 cm, 7 cm and 10 cm make a triangle.

AnswerB · $65^\circ$

1. The exterior angle of a triangle equals the sum of the two interior angles not adjacent to it.

2. The exterior angle at C is therefore $A + B$, so $110^\circ = 45^\circ + B$.

3. This gives $B = 65^\circ$.

4. Check by the longer route: the interior angle at C is $180^\circ - 110^\circ = 70^\circ$, and $45^\circ + 65^\circ + 70^\circ = 180^\circ$.

AnswerB · 40°

1. Angle sum: $(x + 15) + (2x - 10) + (3x + 25) = 180$.

2. $6x + 30 = 180$, so $x = 25$.

3. $\angle B = 2 \times 25 - 10 = 40^\circ$. ($\angle A$ is also 40° and $\angle C$ is 100°.)

AnswerD · 69°

1. The exterior angle equals the sum of the two opposite interior angles: $\angle A + \angle B = 115^\circ$.

2. Split 115° in the ratio $3 : 2$: one part = $115^\circ \div 5 = 23^\circ$.

3. $\angle A = 3 \times 23^\circ = 69^\circ$.

AnswerA · YZ

1. $\angle Z = 180^\circ - 72^\circ - 45^\circ = 63^\circ$.

2. The largest angle is $\angle X = 72^\circ$.

3. The longest side is the one opposite the largest angle, which is YZ.

AnswerA · 36°

1. $AB = AC$, so $\angle ABC = \angle ACB = (180^\circ - 36^\circ) \div 2 = 72^\circ$.

2. In triangle BDC, $BD = BC$, so $\angle BDC = \angle BCD = 72^\circ$.

3. $\angle DBC = 180^\circ - 72^\circ - 72^\circ = 36^\circ$.

AnswerC · 125°

1. $\angle B + \angle C = 180^\circ - 70^\circ = 110^\circ$, so half of each together = 55°.

2. In triangle BIC, $\angle BIC = 180^\circ - 55^\circ = 125^\circ$.

3. Shortcut: $\angle BIC = 90^\circ + \frac{1}{2}\angle A = 90^\circ + 35^\circ = 125^\circ$.

AnswerC · $\angle A = \angle P$, $\angle B = \angle Q$ and $\angle C = \angle R$

1. Three equal sides (SSS), two sides with the included angle (SAS), and two angles with a side (ASA or AAS) each prove congruence.

2. Three equal angles only fix the SHAPE, not the size: a small and a large equilateral triangle have the same angles.

3. So $\angle A = \angle P$, $\angle B = \angle Q$ and $\angle C = \angle R$ is not enough.

AnswerB · 55°

1. The exterior angle at B is $180^\circ - \angle B$ and at C is $180^\circ - \angle C$. Their halves are $90^\circ - \tfrac{1}{2}\angle B$ and $90^\circ - \tfrac{1}{2}\angle C$.

2. In triangle BJC, $\angle BJC = 180^\circ - (90^\circ - \tfrac{1}{2}\angle B) - (90^\circ - \tfrac{1}{2}\angle C) = \tfrac{1}{2}(\angle B + \angle C)$.

3. $\angle B + \angle C = 110^\circ$, so $\angle BJC = 55^\circ$.

4. Shortcut: $\angle BJC = 90^\circ - \frac{1}{2}\angle A = 90^\circ - 35^\circ = 55^\circ$.

AnswerC · 19°

1. $\angle A = 180^\circ - 74^\circ - 36^\circ = 70^\circ$, so $\angle BAE = 35^\circ$.

2. In right triangle ABD, $\angle BAD = 90^\circ - 74^\circ = 16^\circ$.

3. D lies between B and E, so $\angle DAE = \angle BAE - \angle BAD = 35^\circ - 16^\circ = 19^\circ$.

4. Shortcut: $\angle DAE = \tfrac{1}{2}(\angle B - \angle C) = \tfrac{1}{2}(74^\circ - 36^\circ) = 19^\circ$.

AnswerA · 11

1. The third side must be greater than $11 - 6 = 5$ and less than $11 + 6 = 17$.

2. So it can be 6, 7, 8, …, 16 cm.

3. That is $16 - 6 + 1 = 11$ lengths.

AnswerC · 26 m

1. The pole, the ground and the wire form a right triangle with the wire as the hypotenuse.

2. Wire = $\sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26$ m.

AnswerD · 20, 21, 29

1. Check whether the squares of the two smaller numbers add up to the square of the largest.

2. $20^2 + 21^2 = 400 + 441 = 841 = 29^2$, so 20, 21, 29 works.

3. The others miss by one: $9^2 + 40^2 = 41^2$, $11^2 + 60^2 = 61^2$ and $12^2 + 16^2 = 20^2$, so 42, 62 and 21 are wrong.

AnswerC · 7.5 cm

1. By the basic proportionality theorem, $AD : DB = AE : EC$.

2. $\frac{4}{6} = \frac{5}{EC}$, so $EC = \frac{5 \times 6}{4} = 7.5$ cm.