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

Q: For the given figure, the values of x and y are respectively

The values are x = 30° and y = 40°.RP passes through the centre, so it is a diameter, and an angle in a semicircle is a right angle. So angle RSP = 90° and angle RQP = 90°.\[ \begin{aligned} \text{In } \triangle RSP: y &= 180^\circ - 90^\circ - 50^\circ = 40^\circ \\ \text{In } \triangle RQP: x &= 180^\circ - 90^\circ - 60^\circ = 30^\circ \end{aligned} \]The first option gives the same two numbers in the wrong order (x = 40°, y = 30°), the commonest mistake here. 50° and 40° reuse the given angle instead of working out x.

Q: If the diagonals of a parallelogram are perpendicular to each other, then it is a

A parallelogram with perpendicular diagonals is a rhombus.The diagonals of a parallelogram bisect each other. If they are also perpendicular, each diagonal is the perpendicular bisector of the other, which makes all four sides equal: the parallelogram is a rhombus.A square also has perpendicular diagonals, but it is a special rhombus with right angles, which the condition does not require; the general answer is rhombus. A rectangle's diagonals are equal but not perpendicular unless it is a square, and a trapezium is not a parallelogram.

Q: In a triangle ABC, &ang;A = 100° and sides AB and AC are equal. Then value of &ang;B is

Angle B is 40°.Since AB = AC, the triangle is isosceles and the base angles B and C are equal. The angles of a triangle sum to 180°:\[ \begin{aligned} \angle B + \angle C &= 180^{\circ} - 100^{\circ} = 80^{\circ} \\ \angle B &= \frac{80^{\circ}}{2} = 40^{\circ} \end{aligned} \]50° comes from halving 100° instead of the remaining 80°, and 60° and 45° would not give a total of 180° with equal base angles.

Q: If the perimeter of a right-angled isosceles triangle is (4√2+4) cm, then length of hypotenuse is?

The hypotenuse is 4 cm.A right-angled isosceles triangle with equal sides \(a\) has hypotenuse \(a\sqrt{2}\), so\[ \begin{aligned} 2a + a\sqrt{2} &= 4 + 4\sqrt{2} \\ a\sqrt{2}(\sqrt{2} + 1) &= 4(1 + \sqrt{2}) \\ a\sqrt{2} &= 4 \end{aligned} \]Since \(a\sqrt{2}\) is the hypotenuse, it is 4 cm, and each leg is \(2\sqrt{2}\) cm. Check: \(2\sqrt{2} + 2\sqrt{2} + 4 = 4\sqrt{2} + 4\).\(2\sqrt{2}\) cm is the length of each equal side, not the hypotenuse. 5 cm and 6 cm do not satisfy the perimeter.

Q: A ladder reaches a window that is 8 m above the ground on one side of the street. Keeping its foot on the same point, the ladder is twined to the other side of the street to reach a window 12 m high. Find the width of the street if the ladder is 13 m.

The street is about 15.2 m wide.The ladder's foot stays at one point, so the street's width is the sum of the two horizontal distances from the foot to each wall:\[ \begin{aligned} \text{side 1} &= \sqrt{13^2 - 8^2} \\ &= \sqrt{105} \approx 10.25 \\ \text{side 2} &= \sqrt{13^2 - 12^2} = 5 \\ \text{width} &\approx 15.25 \text{ m} \end{aligned} \]The paper's 15·2 m is this value cut to one decimal. 14 m and 14·6 m are both shorter than 10.25 + 5, and 12 m is the height of the second window, not a width.

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181.

For the given figure, the values of x and y are respectively

A circle through P, Q, R and S with RP a diameter; angle SRP is 50 degrees, angle PRQ is x degrees, angle RPS is y degrees and angle RPQ is 60 degrees

A.40°, 30°
B.30°, 40°
C.50°, 40°
D.40°, 50°

Exam Year(s)

MPPSC Forest Service - 2024,

Answer: B

The values are x = 30° and y = 40°.

RP passes through the centre, so it is a diameter, and an angle in a semicircle is a right angle. So angle RSP = 90° and angle RQP = 90°.

\[ \begin{aligned} \text{In } \triangle RSP: y &= 180^\circ - 90^\circ - 50^\circ = 40^\circ \\ \text{In } \triangle RQP: x &= 180^\circ - 90^\circ - 60^\circ = 30^\circ \end{aligned} \]

The first option gives the same two numbers in the wrong order (x = 40°, y = 30°), the commonest mistake here. 50° and 40° reuse the given angle instead of working out x.

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182.

If the diagonals of a parallelogram are perpendicular to each other, then it is a

A.Trapezium
B.Rectangle
C.Square
D.Rhombus

Exam Year(s)

MPPSC Forest Service - 2024,

Answer: D

A parallelogram with perpendicular diagonals is a rhombus.

The diagonals of a parallelogram bisect each other. If they are also perpendicular, each diagonal is the perpendicular bisector of the other, which makes all four sides equal: the parallelogram is a rhombus.

A square also has perpendicular diagonals, but it is a special rhombus with right angles, which the condition does not require; the general answer is rhombus. A rectangle's diagonals are equal but not perpendicular unless it is a square, and a trapezium is not a parallelogram.

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183.

In a triangle ABC, ∠A = 100° and sides AB and AC are equal. Then value of ∠B is

A.60°
B.50°
C.40°
D.45°

Exam Year(s)

MPPSC Forest Service - 2024,

Answer: C

Angle B is 40°.

Since AB = AC, the triangle is isosceles and the base angles B and C are equal. The angles of a triangle sum to 180°:

\[ \begin{aligned} \angle B + \angle C &= 180^{\circ} - 100^{\circ} = 80^{\circ} \\ \angle B &= \frac{80^{\circ}}{2} = 40^{\circ} \end{aligned} \]

50° comes from halving 100° instead of the remaining 80°, and 60° and 45° would not give a total of 180° with equal base angles.

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184.

If the perimeter of a right-angled isosceles triangle is (4√2+4) cm, then length of hypotenuse is?

A.4 cm
B.5 cm
C.6 cm
D.2√2 cm

Exam Year(s)

Northern Coalfields Limited - 2020, NCL Technician - 2020,

Answer: A

The hypotenuse is 4 cm.

A right-angled isosceles triangle with equal sides \(a\) has hypotenuse \(a\sqrt{2}\), so

\[ \begin{aligned} 2a + a\sqrt{2} &= 4 + 4\sqrt{2} \\ a\sqrt{2}(\sqrt{2} + 1) &= 4(1 + \sqrt{2}) \\ a\sqrt{2} &= 4 \end{aligned} \]

Since \(a\sqrt{2}\) is the hypotenuse, it is 4 cm, and each leg is \(2\sqrt{2}\) cm. Check: \(2\sqrt{2} + 2\sqrt{2} + 4 = 4\sqrt{2} + 4\).

\(2\sqrt{2}\) cm is the length of each equal side, not the hypotenuse. 5 cm and 6 cm do not satisfy the perimeter.

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185.

A ladder reaches a window that is 8 m above the ground on one side of the street. Keeping its foot on the same point, the ladder is twined to the other side of the street to reach a window 12 m high. Find the width of the street if the ladder is 13 m.

A.15·2 m
B.14 m
C.14·6 m
D.12 m

Exam Year(s)

Punjab Civil Service - 2021,

Answer: A

The street is about 15.2 m wide.

The ladder's foot stays at one point, so the street's width is the sum of the two horizontal distances from the foot to each wall:

\[ \begin{aligned} \text{side 1} &= \sqrt{13^2 - 8^2} \\ &= \sqrt{105} \approx 10.25 \\ \text{side 2} &= \sqrt{13^2 - 12^2} = 5 \\ \text{width} &\approx 15.25 \text{ m} \end{aligned} \]

The paper's 15·2 m is this value cut to one decimal. 14 m and 14·6 m are both shorter than 10.25 + 5, and 12 m is the height of the second window, not a width.

Discuss in Forum

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

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Other Subjects

Tags: Geometry Lines, Angles, Triangles Questions || Polygons MCQ Questions and Answers || Geometry Quantitative Aptitude || Angles and Triangles GK Questions || Quantitative Aptitude Geometry GK

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