Q1: Which of the following triangles have the same side lengths?
(a) Scalene
(b) Isosceles
(c) Equilateral
(d) None of these
Ans: (c)
Sol: Equilateral triangles have all sides and all angles equal.
Q2: Area of an equilateral triangle with side length a is equal to:
(a) (√3/2)a
(b) (√3/2)a^{2}
(c) (√3/4) a^{2}
(d) (√3/4) a
Ans: (c)
Sol: Area of an equilateral triangle with side length a = √3/4 a^{2}
Q3: D and E are the midpoints of side AB and AC of a triangle ABC, respectively, and BC = 6 cm. If DE  BC, then the length (in cm) of DE is:
(a) 2.5
(b) 3
(c) 5
(d) 6
Ans: (b)
Sol: By midpoint theorem, DE = ½ BC;
DE = ½ of 6; DE = 3 cm
Q4: The diagonals of a rhombus are 16 cm and 12 cm, in length. The side of the rhombus in length is:
(a) 20 cm
(b) 8 cm
(c) 10 cm
(d) 9 cm
Ans: (c)
Sol: Here, half of the diagonals of a rhombus are the sides of the triangle, and the side of the rhombus is the hypotenuse.
By Pythagoras theorem,
(16/2)^{2} + (12/2)^{2} = side^{2}
8^{2} + 6^{2} = side^{2}
side = 10 cm
Q5: Corresponding sides of two similar triangles are in the ratio of 2:3. If the area of the small triangle is 48 sq.cm, then the area of the large triangle is:
(a) 230 sq.cm.
(b) 106 sq.cm
(c) 107 sq.cm.
(d) 108 sq.cm
Ans: (d)
Sol: Let A_{1} and A_{2} be the areas of the small and large triangles.
Then,
A_{2}/A_{1} = (side of large triangle/side of small triangle);
A_{2}/48 = (3/2)^{2}
A_{2} = 108 sq.cm.
Q6: If the perimeter of a triangle is 100 cm and the length of two sides are 30 cm and 40 cm, the length of the third side will be:
(a) 30 cm
(b) 40 cm
(c) 50 cm
(d) 60 cm
Ans: (a)
Sol: Perimeter of the triangle = sum of all its sides;
P = 30 + 40 + x
100 = 70 + x
x = 30 cm
Q7: If triangles ABC and DEF are similar and AB = 4 cm, DE = 6 cm, EF = 9 cm, and FD = 12 cm, the perimeter of triangle ABC is:
(a) 22 cm
(b) 20 cm
(c) 21 cm
(d) 18 cm
Ans: (d) 18 cm
Sol: ABC ~ DEF
AB = 4 cm, DE = 6 cm, EF = 9 cm, and FD = 12 cm;
AB/DE = BC/EF = AC/DF
BC = (4.9)/6 = 6 cm
AC = (12.4)/6 = 8 cm
Perimeter of triangle ABC = AB + BC + AC
= 4 + 6 + 8 = 18 cm
Q8: The height of an equilateral triangle of side 5 cm is:
(a) 4.33 cm
(b) 3.9 cm
(c) 5 cm
(d) 4 cm
Ans: (a)
Sol: The height of the equilateral triangle ABC divides the base into two equal parts at point D. Therefore, BD = DC = 2.5 cm. In triangle ABD, using Pythagoras theorem,
AB^{2} = AD^{2} + BD^{2}
5^{2} = AD^{2} + 2.5^{2}
AD^{2} = 25  6.25
AD^{2} = 18.75
AD = 4.33 cm
Q9: 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
Ans: (b)
Sol: If ABC and DEF are two triangles and AB/DE=BC/FD, then the two triangles are similar if ∠B=∠D.
Q10: 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
Ans: (d)
Sol: Let ABC and DEF be 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) = AB^{2}/DE^{2}
∴ Area(ΔABC)/Area(ΔDEF) = (4/9)^{2} = 16/81 = 16: 81
Q1: The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR.
Ans:
We know that
QR = 6cm
Q2: Determine whether the triangle having sides (b − 1) cm, 2√b cm and (b + 1) cm is a right angled triangle.
Ans: These are sides of the Right angle triangle
Q3: Sides of triangles are given below. Determine which of them are right triangles.In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm,4 cm,5 cm
(iii) 40 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
Ans: i. 25^{2} = 7^{2 }+ 24^{2}
Hence right angle triangle with 25 cm as hypotenuse
ii. 5^{2} = 3^{2} + 4^{2}
Hence right angle triangle with 5 cm as hypotenuse
iii. 100^{2} = 40^{2} + 80^{2}
Hence right angle triangle with 100 cm as hypotenuse
iv. 13^{2} = 12^{2} + 5^{2}
Hence right angle triangle with 13 cm as hypotenuse
Q4: In ΔABC, AD is perpendicular to BC. Prove that:
a. AB^{2} + CD^{2} = AC^{2} + BD^{2}
b. AB^{2} − BD^{2} = AC^{2} − CD^{2}
Ans: In ΔABC, AD is perpendicular to BC
Now ΔABD is an right angle triangle
So from Pythagoras theorem
Similarly ΔADC is an right angle triangle
So from Pythagoras theorem
From (1) and (2)
AB^{2}  BD^{2} = AC^{2 } CD^{2}
Which proved part (b)
Now rearranging,
Which proved part (a).
Q5: Triangle ABC is right angled at B and D is the mid  point of BC.
Prove that: AC^{2 }= 4AD^{2} − 3AB^{2}
Ans:
In right angle triangle ABC
AC^{2 }= AB^{2} − BC^{2}___(1)
In right triangle ABD
Substituting this in equation (1)
AC^{2 }= 4AD^{2} − 3AB^{2}
Q6: ABC is an isosceles triangle, right angled at C. Prove that AB^{2} = 2BC^{2} .
Ans:
Now Since it is an isosceles triangle
AC=BC (1)
Now from Pythagorean theorem
Q7: A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Ans:
AB = 6m
BC= 4 m
Let Height of tower (DE)=h m
EF = 28 m
h = 42m
Q8: The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach?
Ans:
First case is depicted in the figure (a)
Now from Pythagoras theorem
This is the length of the ladder.
Now Second case is depicted in figure (b)
AC=10 cm
BC=8 cm
Now from Pythagoras theorem
Q9: The areas of two similar triangles are 121 cm^{2} and 64 cm^{2} respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other.
Ans: We know that
QR = 8.8 cm
Q10: DEF is an equilateral triangle of side 2b. Find each of its altitudes.
Ans:
In right angle triangle DNF
116 videos420 docs77 tests

1. What are the different types of triangles? 
2. How do you determine the area of a triangle? 
3. What is the Pythagorean theorem? 
4. How do you find the perimeter of a triangle? 
5. Can all angles of a triangle be acute? 

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