MCQ based on areas of similar and nonsimilar triangles
Answers
Answer:
Below are the MCQs for Triangles
1. Which of the following triangles have the same side lengths?
(a)Scalene
(b)Isosceles
(c)Equilateral
(d)None of these
Answer: (c)
Explanation: Equilateral triangles have all its sides and all angles equal.
2. Area of an equilateral triangle with side length a is equal to:
(a)√3/2a
(b)√3/2a2
(c)√3/4 a2
(d)√3/4 a
Answer: c
3.D and E are the midpoints of side AB and AC of a triangle ABC, respectively and BC=6cm. If DE || BC, then the length of DE is:
(a)2.5
(b)3
(c)5
(d)6
Answer: b
Explanation: By midpoint theorem,
DE=½ BC
DE = ½ of 6
DE=3cm
4. The diagonals of a rhombus are 16cm and 12cm, in length. The side of rhombus in length is:
(a)20cm
(b)8cm
(c)10cm
(d)9cm
Answer: c
Explanation: Here, half of the diagonals of a rhombus are the sides of the triangle and side of the rhombus is the hypotenuse.
By Pythagoras theorem,
(16/2)2+(12/2)2=side2
82+62=side2
64+36=side2
side=10cm
5. Corresponding sides of two similar triangles are in the ratio of 2:3. If the area of small triangle is 48 sq.cm, then the area of large triangle is:
(a)230 sq.cm.
(b)106 sq.cm
(c)107 sq.cm.
(d)108 sq.cm
Answer: d
Solution: Let A1 and A2 are areas of the small and large triangle.
Then,
A2/A1=(side of large triangle/side of small triangle)
A2/48=(3/2)2
A2=108 sq.cm.
6. If perimeter of a triangle is 100cm and the length of two sides are 30cm and 40cm, the length of third side will be:
(a)30cm
(b)40cm
(c)50cm
(d)60cm
Answer: a
Solution: Perimeter of triangle = sum of all its sides
P = 30+40+x
100=70+x
x=30cm
7. If triangles ABC and DEF are similar and AB=4cm, DE=6cm, EF=9cm and FD=12cm, the perimeter of triangle is:
(a)22cm
(b)20cm
(c)21cm
(d)18cm
Answer: d
Explanation: ABC ~ DEF
AB=4cm, DE=6cm, EF=9cm and FD=12cm
AB/DE = BC/EF = AC/DF
4/6 = BC/9 = AC/12
BC = (4.9)/6 = 6cm
AC = (12.4)/6 = 8cm
Perimeter = AB+BC+AC
= 4+6+8
=18cm
8. The height of an equilateral triangle of side 5cm is:
(a)4.33
(b)3.9
(c)5
(d)4
Answer: a
Explanation:The height of the equilateral triangle ABC divides the base into two equal parts at point D.
Therefore,
BD=DC= 2.5cm
In triangle ABD, using Pythagoras theorem,
AB2=AD2+BD2
52=AD2+2.52
AD2 = 25-6.25
AD2=18.75
AD=4.33 cm
9. If ABC and DEF are two triangles and AB/DE=BC/FD, then the two triangles are similar if
(a)∠A=∠F
(b)∠B=∠D
(c)∠A=∠D
(d)∠B=∠E
Answer: b
10. Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles are in the ratio
(a)2: 3
(b)4: 9
(c)81: 16
(d)16: 81
Answer: d
Explanation: Let ABC and DEF are two similar triangles, such that,
ΔABC ~ ΔDEF
And AB/DE = AC/DF = BC/EF = 4/9
As the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,
∴ Area(ΔABC)/Area(ΔDEF) = AB2/DE2
∴ Area(ΔABC)/Area(ΔDEF) = (4/9)2 = 16/81 = 16: 81
Step-by-step explanation: