Every Book Back multiple-choice question from Geometry (10th Standard Mathematics, Samacheer Kalvi) — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
If in triangles ABC and EDF, \(\frac{AB}{DE} = \frac{BC}{FD}\) then they will be similar, when
- A. \(\angle B = \angle E\)
- B. \(\angle A = \angle D\)
- C. \(\angle B = \angle D\)Correct
- D. \(\angle A = \angle F\)
Explanation. For triangles to be similar by SAS criterion, the angle included between the proportional sides must be equal. The included angle for sides AB and BC is \(\angle B\), and for DE and FD is \(\angle D\).
Q2
In \(\Delta LMN\), \(\angle L = 60^{\circ}\), \(\angle M = 50^{\circ}\). If \(\Delta LMN \sim \Delta PQR\) then the value of \(\angle R\) is
- A. \(40^{\circ}\)
- B. \(70^{\circ}\)Correct
- C. \(30^{\circ}\)
- D. \(110^{\circ}\)
Explanation. Since the triangles are similar, their corresponding angles are equal, so \(\angle R = \angle N\). In \(\Delta LMN\), the sum of angles is \(180^{\circ}\), so \(\angle N = 180^{\circ} - (60^{\circ} + 50^{\circ}) = 70^{\circ}\).
Q3
If \(\Delta ABC\) is an isosceles triangle with \(\angle C = 90^{\circ}\) and \(AC = 5\) cm, then \(AB\) is
- A. 2.5 cm
- B. 5 cm
- C. 10 cm
- D. \(5\sqrt{2}\) cmCorrect
Explanation. Since the triangle is isosceles and right-angled at C, \(AC = BC = 5\) cm. By Pythagoras theorem, the hypotenuse \(AB = \sqrt{AC^2 + BC^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}\) cm.
Q4
In a given figure \(ST \parallel QR\), \(PS = 2\) cm and \(SQ = 3\) cm. Then the ratio of the area of \(\Delta PQR\) to the area of \(\Delta PST\) is
- A. 25 : 4Correct
- B. 25 : 7
- C. 25 : 11
- D. 25 : 13
Explanation. Since \(ST \parallel QR\), \(\Delta PST \sim \Delta PQR\). The ratio of their areas is equal to the ratio of the squares of their corresponding sides, which is \(\frac{PQ^2}{PS^2} = \frac{(2+3)^2}{2^2} = \frac{25}{4}\).
Q5
The perimeters of two similar triangles \(\Delta ABC\) and \(\Delta PQR\) are 36 cm and 24 cm respectively. If \(PQ = 10\) cm, then the length of \(AB\) is
- A. \(6\frac{2}{3}\) cm
- B. \(10\frac{2}{3}\) cm
- C. \(66\frac{2}{3}\) cm
- D. 15 cmCorrect
Explanation. The ratio of the perimeters of two similar triangles is equal to the ratio of their corresponding sides. Therefore, \(\frac{AB}{PQ} = \frac{\text{Perimeter}(\Delta ABC)}{\text{Perimeter}(\Delta PQR)} \implies \frac{AB}{10} = \frac{36}{24}\), which gives \(AB = 15\) cm.
Q6
If in \(\Delta ABC\), \(DE \parallel BC\). \(AB = 3.6\) cm, \(AC = 2.4\) cm and \(AD = 2.1\) cm then the length of \(AE\) is
- A. 1.4 cmCorrect
- B. 1.8 cm
- C. 1.2 cm
- D. 1.05 cm
Explanation. By Basic Proportionality Theorem, since \(DE \parallel BC\), we have \(\frac{AD}{AB} = \frac{AE}{AC}\). Substituting the given values, \(\frac{2.1}{3.6} = \frac{AE}{2.4}\), which yields \(AE = 1.4\) cm.
Q7
In a \(\Delta ABC\), \(AD\) is the bisector of \(\angle BAC\). If \(AB = 8\) cm, \(BD = 6\) cm and \(DC = 3\) cm. The length of the side \(AC\) is
- A. 6 cm
- B. 4 cmCorrect
- C. 3 cm
- D. 8 cm
Explanation. According to the Angle Bisector Theorem, \(\frac{AB}{AC} = \frac{BD}{DC}\). Substituting the given lengths gives \(\frac{8}{AC} = \frac{6}{3} = 2\), which simplifies to \(AC = 4\) cm.
Q8
In the adjacent figure \(\angle BAC = 90^{\circ}\) and \(AD \perp BC\) then
- A. \(BD \cdot CD = BC^2\)
- B. \(AB \cdot AC = BC^2\)
- C. \(BD \cdot CD = AD^2\)Correct
- D. \(AB \cdot AC = AD^2\)
Explanation. By similarity of right triangles, \(\Delta ABD \sim \Delta CAD\). The ratio of corresponding sides gives \(\frac{BD}{AD} = \frac{AD}{CD}\), which simplifies to \(AD^2 = BD \cdot CD\).
Q9
Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m, what is the distance between their tops?
- A. 13 mCorrect
- B. 14 m
- C. 15 m
- D. 12.8 m
Explanation. Using Pythagoras theorem on the right triangle formed by the difference in heights of the poles (5 m) and the distance between their feet (12 m), the distance between their tops is \(\sqrt{12^2 + 5^2} = 13\) m.
Q10
In the given figure, \(PR = 26\) cm, \(QR = 24\) cm, \(\angle PAQ = 90^{\circ}\), \(PA = 6\) cm and \(QA = 8\) cm. Find \(\angle PQR\)
- A. \(80^{\circ}\)
- B. \(85^{\circ}\)
- C. \(75^{\circ}\)
- D. \(90^{\circ}\)Correct
Explanation. In right \(\Delta PAQ\), \(PQ = \sqrt{6^2+8^2} = 10\) cm. In \(\Delta PQR\), the sides are 10, 24, and 26. Since \(10^2+24^2=26^2\), it satisfies Pythagoras theorem, making \(\angle PQR = 90^{\circ}\).
Q11
A tangent is perpendicular to the radius at the
- A. centre
- B. point of contactCorrect
- C. infinity
- D. chord
Explanation. By standard geometric properties of circles, a tangent is perpendicular to the radius at the point of contact.
Q12
How many tangents can be drawn to the circle from an exterior point?
- A. one
- B. twoCorrect
- C. infinite
- D. zero
Explanation. From any point outside a circle, exactly two distinct tangent lines can be drawn to touch the circle.
Q13
The two tangents from an external point P to a circle with centre at O are PA and PB. If \(\angle APB = 70^{\circ}\) then the value of \(\angle AOB\) is
- A. \(100^{\circ}\)
- B. \(110^{\circ}\)Correct
- C. \(120^{\circ}\)
- D. \(130^{\circ}\)
Explanation. The angles subtended by the tangents at the centre and between the tangents are supplementary. Thus, \(\angle AOB = 180^{\circ} - \angle APB = 180^{\circ} - 70^{\circ} = 110^{\circ}\).
Q14
In figure CP and CQ are tangents to a circle with centre at O. ARB is another tangent touching the circle at R. If CP = 11 cm and BC = 7 cm, then the length of BR is
- A. 6 cm
- B. 5 cm
- C. 8 cm
- D. 4 cmCorrect
Explanation. Since tangents from an external point are equal, \(CQ = CP = 11\) cm. Since \(BQ\) and \(BR\) are tangents from B, \(BR = BQ = CQ - BC = 11 - 7 = 4\) cm.
Q15
In figure if PR is tangent to the circle at P and O is the centre of the circle, then \(\angle POQ\) is
- A. \(120^{\circ}\)Correct
- B. \(100^{\circ}\)
- C. \(110^{\circ}\)
- D. \(90^{\circ}\)
Explanation. Based on the printed answer key, the value of \(\angle POQ\) is \(120^{\circ}\).
