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Properties of Triangles

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 × a
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 cm

 

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

Posted Date : 21-04-2021

 

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