Angle Sum Property Of Triangle Class 9
Angle sum property of triangle
∠A + ∠B + ∠C = 180°
Exterior angles of a triangle
Exterior ∠ACD = ∠A + ∠B
Isosceles Triangle
If two sides of a triangle are equal, the angles opposite to them are also equal.
If two angles of a triangle are equal, the sides opposite to them are also equal.de
Altitude
When a perpendicular drawn from a vertex of a triangle to the opposite side, it is called the altitude of the triangle.
Median
When a line segment drawn from a vertex of a triangle to the mid-point of the opposite side, it is called the median of the triangle.
The three medians of a triangle intersect at centroid in the ratio 2:1.
Exercise :
Q 1) In ∆ABC, ∠A - ∠B = 24° and ∠A - ∠B = 24°. Find ∠A .
Solution :
In ∆ABC,
∠A - ∠B = 24°
∴∠A = ∠B + 24° --------(i)
∠B - ∠C = 30°
∴ ∠C = ∠B - 30°----------(ii)
We know that,
∠A + ∠B + ∠C = 180°
∴ ∠B + 24° + ∠B + ∠B - 30° = 180°-----------using (i) and (ii)
∴ 3∠B - 6° = 180°
3∠B = 180° + 6°
∠B = 186/3 = 62°
∠A = ∠B e+ 24°
∠A = 62° + 24° = 86°
Q 2) In the given figure, AB = BC = CD = AC and AD = DE. Find ∠BAE.
Solution :
In the given figure, AB = BC = CD = AC and
AD = DE
To find : ∠BAE
In ∆ABC
AB = BC = CA
∴ ∠A + ∠B + ∠C = 60° ---------------(i)
In ∆ABC,
Ext. ∠ACB = ∠CDA + ∠CAD = 60°
But AC = CD ( Given)
∴ ∠ADC = ∠ CAD = 30° -----------------(ii)
Again in ∆ADC,
Ext. ∠ADC = ∠DAE + ∠DEA = 36°
But AD = DC (Given)
∴ ∠DAE = ∠ DEA = 15° -----------------(iii)
From (i), (ii), (iii)
∴ ∠BAE = ∠BAC + ∠CAD + ∠DAE
= 60° + 30° + 15°
= 105°
Q 3) In ∆ ABC, AC = BC, ∠BAC is bisected by A and AD = AB. Find ∠ACB.
Solution.
In ∆ ABC,
AC = BC,
∠BAC is bisected by A and AD = AB
To find :
∠ACB
In ∆ ABC,
AC = BC,
∠A = ∠ B --------------------------(i)
In ∆ ABD,
AB = AD
∠B = ∠D
In ∆ ADC,
Ext. ∠D = ∠DAC + ∠C = 1/2 ∠A + ∠C
∠B = 1/2 ∠A + ∠C
∠A = 1/2 ∠A + ∠C ---------------------(using i)
∠A = 1/2 ∠A + 2/2∠C
∠A = 1/2 ( ∠A + 2∠C)
2∠A - ∠A = 2∠C
∠A = 2∠C
But
∠A + ∠B + ∠C = 180° (sum of angles of triangles)
2∠C + 2∠C + ∠C = 180°
5∠C = 180°
∠C = 180°/5
∠C = 36°
∠AC = 36°
Q 4) In ∆ ABC, AC = BC, DE is drawn is drawn parallel to BC through A . ∠EAC = 42°. Find angle x.
Solution.
In ∆ ABC, AC = BC
DE is drawn parallel to BC through A.
∠EAC = 42°
To find : angle x
DE ∥ BC
∴ ∠C = ∠EAC = 42°
∠CAB = x
CA = CB ( Given)
∴ ∠B = ∠CAB = x
Now in ∆ ABC, we have
∠A + ∠B + ∠C = 180° (sum of angles of triangles)
x + x + 42° = 180° (sum of angles of triangles)
2x = 180° - 42°
x = 138 /2
x = 69°
Q 5) In ∆PQR, PQ = PR , ∠P = ( 2x +20)° and ∠R = ( x + 10)°. Find the value of x . Assign a special name to the triangle.
Solution.
In ∆PQR,
PQ = PR , ∠P = ( 2x +20)°
and ∠R = ( x + 10)°.
Find the value of x .
In ∆PQR,
PQ = PR ---------------------( Given)
∠Q = ∠R = (x + 10)°
∠P + ∠Q + ∠R = 180° (sum of angles of triangles)
2x +20° + x + 10° + x + 10° = 180°
4x + 40° = 180°
4x = 180° - 40°
x = 140° /4
x = 35°
Now ,
∠P = 2x + 20 = 2* 35 +20 = 70 + 20 = 90°
∴ ∆PQR is a right angled triangle.
Q 6) In the given figure, AB = AC, AD ⊥ BC, ∠PAC = 102°. Find the value of x and y.
Solution
In the given figure,
AB = AC, AD ⊥ BC
∠PAC = 102°
Find the value of x and y
In ∆ABC,
AB = AC -----------------( Given)
∠ B = ∠ C = x
and Ext. ∠PAC = ∠C + ∠B = x + x = 2x
∴ 2x = 102° = x = 102 /2 = 51°
In ∆ABD, ∠D = 90°
∠BAD + ∠ B + ∠ D = 180°
y + 51 + 90 = 180
y + 141 = 180
y = 180 - 141
y = 39°
x = 51° and y = 39°
Q 7) In the given figure, PQ ⊥ AB, AQ = QB, ∠PAC = 42°, ∠ C = 68°. Find ∠ABC.
Solution.
In the figure,
PQ ⊥ AB,
AQ = QB,
∠PAC = 42° and ∠ C = 68°
To find : ∠ABC
In ∆APC; we have
∠PAC + ∠APC + ∠ACP = 180°
42° + ∠APC + 68° = 180° ( angle sum property)
110 + ∠APC = 180
∠APC = 180 - 110 = 70°
Now in ∆ABP,
PQ ⊥ AB and AQ = QB
∴ PQ is the perpendicular bisector of AB
∴ ∠PAB = ∠PBA
and ext. ∠APC = ∠PAB + ∠PBA
70° = ∠PAB + ∠PBA
70° = ∠PBA + ∠PBA
2∠PAB = 70°
∠PAB = 70°/2
∠PAB = ∠ABC = 70° ------(AQ = QB)
Q 8 In the given figure, AB = AD, ∠PAD = 70° and ∠DBC = 90°. Find x, ∠ BDQ and ∠BCD
if AB ∥ DC.
Solution:
In ∆ADB, AB = AD
∴ ∠ABD = ∠ADB = x
and ex. ∠PAD = ∠ABD + ∠ADB
70° = x + x = 2x
x = 70/2 = 35°
x = 35°
Q 9) In the given figure, AB ∥ FC, ∠B = 70° and ∠FDE =148°, Find the values of a and b.
Solution:
AB ∥ FC
∠ABC = ∠BCF = 180° --------( Co -interior angles)
70° + a = 180°
a = 180° - 70° = 110°
∠FDE + ∠FDC = 180°
148 + ∠FDC = 180°
∠FDC = 180° - 148° = 32°
Now in ∆FCD
Ext. ∠FCB = ∠CFD + ∠CDF
110° = b + 32°
b = 110° - 32° = 78°
a = 110°, b = 78°
Q 10) ∠ECF = 3∠ACE and ∠ADB = 72°. Find the value of x.
Solution:
In the given figure
∠ECF = 3∠ACE
∠ADB = 72°, AD ∥ EC
Find the value of x
AD ∥ EC
∴ ∠ECD = ∠ADB = 72° ( corresponding angles)
But ∠ECF = ∠ECD = 180° ( Linear pair)
∠ECF = 72° = 180°
∠ECF = 180°- 72° = 108°
But
∠ECF = 3∠ACE
∴ ∠ACE = 1/3 ∠ECF = 1/3 * 108° = 36°
Since. AD ∥ EC
∠DAC = ∠ACE = 36° ( alternate angles)
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