TRIANGLES :
Triangles are closed figures containing three angles and three sides.
General Properties of Triangles:
1. The sum of the two sides is greater than the third side: a + b > c, a + c > b, b + c > a
The sum of the three angles of a triangle is equal to 180°: In the triangle below ∠A + ∠B + ∠C =
180°
Also, the exterior angle α is equal to sum the two opposite interior angle A and B, i.e. α = ∠A + ∠B.
Area of a Triangle:
Area of a triangle = 1/2 x base x height = 1/2 x a xh
Area of a triangle = 1/2 bc sin A = 1/2ab sinC = 1/2ac sinB
Types of triangles:
Scalene Triangle
No side equal.
All the general properties
of triangle apply
Equilateral Triangle
Each side equal
Each angle = 60°
Length of altitude = √3a/2
Area =√3a^2/4
Inradius = a/2√3
Circumradius = a/√3
Isosceles Triangle
Two sides equal.
The angles opposite to
the opposite sides are
equal
Right Triangle
One of the angles is a
right angle, i.e. 90°.
Area = 1/2 x AB x BC
AC^2 = AB^2 + BC^2
Altitude BD = (AB x BC)/AC
The midpoint of the
hypotenuse is equidistant
from all the three
vertices, i.e. EA = EB = EC
Some Questions based on above concept:
1.A, B, C, are the three angles of a ∆ . If A - B = 15° and B - C = 30° , then ∠A is equal to :
(a) 65°
(b) 80°
(c) 75°
(d) 85°
2.In the given figure, the side BC of a ∆ABC is produced on both sides, then ∠1+ ∠2 is equal to :
(a) ∠A+180°
(b) 180°-∠A
(c) 1/2 (∠A+180°)
(d) ∠A+90°
3.In the given figure, AM = AD, ∠B = 63° and CD is an angle bisector of ∠C, then ∠MAC = ?
(a) 27°
(b) 37°
(c) 63°
(d) none of these
4.In the figure below, if s < 50° < t, then
(a) t < 80
(b) s + t < 130
(c) 50 < t < 80
(d) t > 80
5.ABC is a triangle. It is given that a + c > 90°, then b is
(a) greater than 90°
(b) less than 90°
(c) equal to 90°
(d) can't be said
6.If each angle of a triangle is less than the sum of the other two, then the triangle is :
(a) right - angled
(b) acute - angled
(c) obtuse - angled
(d) None of these
7.The side BC of ∆ is produced to D. If ∠ACD = 108° and ∠B = 1/2 ∠ A then ∠A is :
(a) 36°
(b) 108°
(c) 59°
(d) 72°
8.In the given figure below, if AD = CD = BC, and ∠BCF = 96°, How much is ∠DBC?
(a) 32°
(b) 84°
(c) 64°
(d) can't be determined
9.If ∠A = 44°, BP = BR and CN = RC then ∠PRN = ?
(a) 58°
(b) 78°
(c) 68°
(d) 66°
10.In the given figure, if AD = DE = EC = BC then ∠A : ∠B =
(a) 1 : 3
(b) 2 : 5
(c) 3 : 1
(d) 1 : 2
ANSWERS AND SOLUTION :
1.(b) Since A, B and C are the angles of a ∆.
∴ ∠A+B+C=180°
Now, A - B = 15°, B - C = 30°,
∴ B = C+30°
∴ ∠A = B+15°= C+30°+15°
= C + 45°
∴ A+ B+C =(C+45°)+(C+30°)+C= 180°
3C = 180° - 75° = 105°
C = 35°
∴ ∠A = 35°+45°= 80°
2.(a) ∠1= ∠A+ ∠5 and
∠2= ∠A+ ∠6 (exterior angle)
∠1+ ∠2 = ∠A+(∠A+∠5+ ∠6)
= ∠A+180°
3.(c)
4.(d) s + t + 50° = 180°
=>s + t = 130°
=>t = 130° - s
s < 50°
so t > 130° - 50°
=>t > 80°
5.(b) a + b + c = 180°
so b = 180° - (a+ b)
so b < 90° [a + c > 90°]
6.(b) Let ∠A ,∠B+ ∠C ,then
2 ∠A+ ∠B+ ∠C ⇒2 ∠A<180°
⇒A<90°
Similarly, ∠B<90° and<c<90°
7.(d)
8.(c)
9.(c)
10.(a)
Triangles are closed figures containing three angles and three sides.
1. The sum of the two sides is greater than the third side: a + b > c, a + c > b, b + c > a
The sum of the three angles of a triangle is equal to 180°: In the triangle below ∠A + ∠B + ∠C =
180°
Also, the exterior angle α is equal to sum the two opposite interior angle A and B, i.e. α = ∠A + ∠B.
Area of a Triangle:
Area of a triangle = 1/2 x base x height = 1/2 x a xh
Area of a triangle = 1/2 bc sin A = 1/2ab sinC = 1/2ac sinB
Types of triangles:
Scalene Triangle
No side equal.
All the general properties
of triangle apply
Equilateral Triangle
Each side equal
Each angle = 60°
Length of altitude = √3a/2
Area =√3a^2/4
Inradius = a/2√3
Circumradius = a/√3
Isosceles Triangle
Two sides equal.
The angles opposite to
the opposite sides are
equal
Right Triangle
One of the angles is a
right angle, i.e. 90°.
Area = 1/2 x AB x BC
AC^2 = AB^2 + BC^2
Altitude BD = (AB x BC)/AC
The midpoint of the
hypotenuse is equidistant
from all the three
vertices, i.e. EA = EB = EC
Some Questions based on above concept:
1.A, B, C, are the three angles of a ∆ . If A - B = 15° and B - C = 30° , then ∠A is equal to :
(a) 65°
(b) 80°
(c) 75°
(d) 85°
2.In the given figure, the side BC of a ∆ABC is produced on both sides, then ∠1+ ∠2 is equal to :
(a) ∠A+180°
(b) 180°-∠A
(c) 1/2 (∠A+180°)
(d) ∠A+90°
3.In the given figure, AM = AD, ∠B = 63° and CD is an angle bisector of ∠C, then ∠MAC = ?
(a) 27°
(b) 37°
(c) 63°
(d) none of these
4.In the figure below, if s < 50° < t, then
(a) t < 80
(b) s + t < 130
(c) 50 < t < 80
(d) t > 80
5.ABC is a triangle. It is given that a + c > 90°, then b is
(a) greater than 90°
(b) less than 90°
(c) equal to 90°
(d) can't be said
6.If each angle of a triangle is less than the sum of the other two, then the triangle is :
(a) right - angled
(b) acute - angled
(c) obtuse - angled
(d) None of these
7.The side BC of ∆ is produced to D. If ∠ACD = 108° and ∠B = 1/2 ∠ A then ∠A is :
(a) 36°
(b) 108°
(c) 59°
(d) 72°
8.In the given figure below, if AD = CD = BC, and ∠BCF = 96°, How much is ∠DBC?
(a) 32°
(b) 84°
(c) 64°
(d) can't be determined
9.If ∠A = 44°, BP = BR and CN = RC then ∠PRN = ?
(a) 58°
(b) 78°
(c) 68°
(d) 66°
10.In the given figure, if AD = DE = EC = BC then ∠A : ∠B =
(a) 1 : 3
(b) 2 : 5
(c) 3 : 1
(d) 1 : 2
ANSWERS AND SOLUTION :
1.(b) Since A, B and C are the angles of a ∆.
∴ ∠A+B+C=180°
Now, A - B = 15°, B - C = 30°,
∴ B = C+30°
∴ ∠A = B+15°= C+30°+15°
= C + 45°
∴ A+ B+C =(C+45°)+(C+30°)+C= 180°
3C = 180° - 75° = 105°
C = 35°
∴ ∠A = 35°+45°= 80°
2.(a) ∠1= ∠A+ ∠5 and
∠2= ∠A+ ∠6 (exterior angle)
∠1+ ∠2 = ∠A+(∠A+∠5+ ∠6)
= ∠A+180°
3.(c)
4.(d) s + t + 50° = 180°
=>s + t = 130°
=>t = 130° - s
s < 50°
so t > 130° - 50°
=>t > 80°
5.(b) a + b + c = 180°
so b = 180° - (a+ b)
so b < 90° [a + c > 90°]
6.(b) Let ∠A ,∠B+ ∠C ,then
2 ∠A+ ∠B+ ∠C ⇒2 ∠A<180°
⇒A<90°
Similarly, ∠B<90° and<c<90°
7.(d)
8.(c)
9.(c)
10.(a)
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