1. In the given figure, m1 is parallel to m2, AC and BC are angle bisectors. Find the measure of ∠ACB.
a) 45° b) 60° c) 75° d) 90°
Ans: d;
2x = 180° − 2y (Interior angles)
⇒ 2x + 2y = 180° ⇒ x + y = 90°
In ∆ABC, ∠ ACB + x + y = 180°
⇒ ∠ ACB = 180° − (x + y) = 90°.
2. Find the value of x in the figure given below.
a) 42° b) 128° c) 108° d) 148°
Ans: c; Assume ∠ AFG = a
∠ ABC = ∠ BFE = 128° (Corresponding angles)
a = 180° − 128° = 52°.
In ∆AGF, ∠AFG + 20° + x = 180°
⇒ x = 108°.
3. Find the values of x and y respectively in the figure given below.
a) 140°, 40° b) 20°, 160° c) 40°, 140° d) 160°, 20°
Ans: c;
∠2 = 90° (Alternate angles),
∠1 = x (Corresponding angles)
∠1 + ∠2 = 3x + 10°
⇒ x + 90° = 3x + 10° ⇒ 2x = 80° ⇒ x = 40°
⇒ y = 180° − ∠1 (Adjacent angles)
⇒ y = 180° − 40° = 140°
4. Find the ratio of the area of an equilateral triangle drawn with the side of a square as its base to the area of an equilateral triangle described on the diagonal of the square.
a) 1 : 2 b) 2 : 1 c) 2 : 3 d) 3 : 2
Ans: a; Let the side of the square be a. Its diagonal is a√2
Area of the first equilateral triangle = √3/4 × a2
Area of the second equilateral triangle √3/4 × (a√2 )2
Required ratio = 1 : 2
5. If QR is 40% larger than AB and PB = 10 cm, then what is the length of PQ?
a) 10 cm b) 14 cm c) 16 cm d) 16.2 cm
Ans: b; ∠PBA = 180° − 95° = 85°
∠QPR = 180° − (85° + 65°) = 30°
∴ In ∆PAB, ∠PAB = 180° − (85° + 30°) = 65°
Thus, ∆PAB and ∆PRQ are similar
6. ABCD is a trapezium with AB || CD. If AC and BD intersect at E such that BE : ED = 2 : 3, then find the ratio of the areas of ∆AEB and ∆CED.
a) 2 : 3 b) 3 : 2 c) 4 : 9 d) 9 : 4
Ans: c;
BE : DE = 2 : 3 and AE : EC = 2 : 3 (By property)
∆AEB and ∆CED are similar.
The ratio of the proportional sides is 2 : 3
⇒ BD2 = AD × DC = 8 × 2
⇒ BD2 = 16 ⇒ BD = 4 cm
7. ∆ABC is a right-angled triangle right angled at B. BD is perpendicular to AC. If AD = 8 cm and DC = 2 cm, then find the length of BD.
a) 4 cm b) 4.5 cm c) 5 cm d) 5.5 cm
Ans: a; ∆ADB ∼ ∆BDC
8. ABC is a triangle. The medians CD and BE intersect each other at O. Then Area of ∆ODE : Area of ∆ABC is
a) 1 : 3 b) 1 : 4 c) 1 : 6 d) 1 : 12
Ans: d;
9. The length of the sides of a right-angled triangle are proportional to the numbers 3, 4 and 5. The largest side of the triangle exceeds the smallest side by 4 cm. The area (in cm2) of the triangle is
a) 24 b) 12 c) 10 d) 15 10.
Ans: a;
Let the sides of the triangle be 3x, 4x and 5x.
Then, 5x − 3x = 4 ⇒ x = 2
∴ The sides of the triangle = 6 cm, 8 cm, 10 cm
which is a Pythogorean triplet
∴ Area of the triangle = 1/2 × 6 × 8 = 24 cm2
10. In the figure given above AB = 2 cm, BD = 12 cm and AC = 4 cm. ∠ADC = ∠AEB, then what is the value of the side CE?
a) 24 cm b) 12 cm c) 6 cm d) 3 cm
Ans: d; ∆ADC ∼ ∆AEB Let x be the length of the side CE.
11. In the following figure, AD = BD = DC and ∠ABD = 30°. What is the value of the ∠ACD?
a) 90° b) 55° c) 45° d) 60°
Ans: d;
In ∆ABD, AD = BD ⇒ ∠ABD = ∠DAB = 30°
∴ ∠ADB = 180° − (∠ABD + ∠DAB) = 120°
⇒ ∠ADC = 180° − ∠ADB = 60°
Let ∠DAC = ∠DCA = θ
∴ 60° + 2θ = 180° ⇒ θ = 60°
12. What is the value of ‘d’ in the given figure?
a) 150° b) 60° c) 105° d) 90°
Ans: c; Sum of the angles of a triangle is 180°.
In ∆ABC ∴ 2a + 2b + 30° = 180° ⇒ 2(a + b) = 150°
⇒ a + b = 150°/2 = 75°
and in ∆ACD
a + b + d = 180° ⇒ d = 180° − 75° = 105°
13. In the following figure, if LM || NC, then ∠x is
a) 105° b) 115° c) 125° d) 135°
Ans: d; In ∆ABC, ∠NBA = 135° by exterior angle property.
Since LM || NC
∴ ∠x = ∠NBA = 135°.
14. In the figure given below, ABCD is a square and triangle ABE is an isosceles triangle with AB = AE. If ∠BAE = 40°, then ∠AED is equal to
a) 45° b) 55° c) 65° d) 75°
Ans: c;
∠BAE = 40° ⇒ ∠EAD = 90° − 40° = 50°
AB = AE = AD
∴ ∆AED is an isosceles triangle.
∴ ∠AED = ∠ADE = 180° − 50°/2 = 65°
15. If the angles of a triangle ABC are in the ratio 2 : 3 : 1, then the angles ∠A, ∠B and ∠C are
a) ∠A = 60°, ∠B = 90°, ∠C = 30° b) ∠A = 40°, ∠B = 120°, ∠C = 20°
c) ∠A = 20°, ∠B = 60°, ∠C = 60° d) ∠A = 45°, ∠B = 90°, ∠C = 45°
Ans: a; The sum of angles in a triangle is 180°.
∠A = 2x°, ∠B = 3x°, ∠C = x°
⇒ 2x° + 3x° + x° = 180° ⇒ 6x° = 180°
⇒ x = 180/6 = 30
∴ ∠A = 2 × 30° = 60°
∠B = 3x × 30° = 90°; ∠C = x = 30°
16. In ∆ABC, if 2∠A = 3∠B = 6∠C, value of ∠B is
a) 60° b) 30° c) 45° d) 90°
Ans: a; 2∠A = 3∠B = 6∠C