FINDING THE MISSING ANGLE OF A QUADRILATERAL

Find the measure of the indicated angle in each quadrilateral.

Problem 1 :

Solution :

The sum of interior angles of a quadrilateral is 360°

m ∠A + m ∠B + m ∠C + m ∠D = 360°

Here m ∠B = 87°, m ∠C = 112°, and, m ∠D = 92°

m ∠A + 87° + 112° + 92° = 360° 

m ∠A + 291° = 360° 

m ∠A = 360° - 291°

m ∠A = 69°

Problem 2 :

Solution :

The sum of interior angles of a quadrilateral is 360°

m ∠P + m ∠Q + m ∠R + m ∠S = 360°

Here m ∠P = 124°, m ∠R = 68°, and, m ∠S = 71°

124° + m ∠Q + 68° + 71° = 360°

m ∠Q + 263° = 360°

m ∠Q = 360° - 263°

m ∠Q = 97°

Problem 3 :

Solution :

The sum of interior angles of a quadrilateral is 360°

m ∠W + m ∠X + m ∠Y + m ∠Z = 360°

Here m ∠W = 95°, m ∠X = 74°, and, m ∠Z = 46°

95° + 74° + m ∠Y + 46° = 360°

m ∠Y + 215° = 360°

m ∠Y = 360° - 215°

m ∠Y = 145°

Problem 4 :

Solution :

The sum of interior angles of a quadrilateral is 360°

m ∠J + m ∠K + m ∠L + m ∠M = 360°

Here m ∠J = 72°, m ∠L = 49°, and, m ∠M = 89°

72° + m ∠K + 49° + 89° = 360°

m ∠K + 210° = 360°

m ∠K = 360° - 210°

m ∠K = 150°

Problem 5 :

Solution :

m ∠Q + m ∠R + m ∠S + m ∠T = 360°

Here m ∠Q = 98°, m ∠S = 65°, and, m ∠T = 69°

98° + m ∠R + 65° + 69° = 360°

m ∠R + 232° = 360°

m ∠R = 360° - 232°

m ∠R = 128°

Problem 6 :

Solution :

m ∠B + m ∠C + m ∠D + m ∠E = 360°

Here m ∠B = 77°, m ∠D = 86°, and m ∠E = 123°

77° + m ∠C + 86° + 123° = 360°

m ∠C + 286° = 360°

m ∠C = 360° - 286°

m ∠C = 74°

Problem 7 :

Solution :

m ∠U + m ∠V + m ∠W + m ∠X = 360°

Here m ∠U = 67°, m ∠V = 135°, and m ∠W= 105°

67° + 135° + 105° + m ∠X = 360°

m ∠X + 307° = 360°

m ∠X = 360° - 307°

m ∠X = 53°

Problem 8 :

Solution :

m ∠S + m ∠T + m ∠U + m ∠V = 360°

Here m ∠S = 114°, m ∠U = 92°, and, m ∠V= 66°

114° + m ∠T + 92° + 66° = 360°

m ∠T + 272° = 360°

m ∠T = 360° - 272°

m ∠T = 88°

Problem 9 :

Solution :

m ∠E + m ∠F + m ∠G + m ∠H = 360°

Here m ∠E = 146°, m ∠G = 91°, and, m ∠H= 75°

146° + m ∠F + 91° + 75° = 360°

m ∠F + 312° = 360°

m ∠F = 360° - 312°

m ∠F = 48°

Problem 10 :

If three angles of a quadrilateral are each equal to 75°, the fourth angle is :-

(a) 150°        (b) 135°       (c) 45°         (d) 75°

Solution :

In a quadrilateral, the sum of angles of quadrilateral = 360

Three angles of a quadrilateral = 75 degree. Let x be the fourth angle.

75 + 75 + 75 + x = 360

225 + x = 360

x = 360 - 225

x = 135

So, option b is correct.

Problem 11 :

What is the maximum number of obtuse angles that a quadrilateral can have ?

(a) 1      (b) 2     (c) 3      (d) 4

Solution :

A quadrilateral can have a maximum of three obtuse angles. Since the sum of interior angles in any quadrilateral is 360 degrees, having four obtuse angles (each greater than 90 degrees) would result in a sum greater than 360, according to math sites

Problem 12  :

ABCD is a rhombus such that ∠ACB = 40°. Then ∠ADB is

(a) 40°       (b) 45°       (c) 50°       (d) 60°

Solution :

By observing the rhombus,

∠OAD = 40

∠AOD = 90

∠ADO = x

40 + 90 + x = 180

130 + x = 180

x = 180 - 130

x = 50

Problem 13 :

If PQRS is a parallelogram, then ∠P − ∠R is

(a) 90°       (b) 45°      (c) 60°      (d) 0°

Solution :

So, ∠P = ∠R

Now, ∠P - ∠R = ∠P - ∠P

= 0

Therefore, the required value is 0.

Problem 14 :

ABCD is a square, diagonal AC is joined. Then the measurement of ∠ACB is

(a) 35°           (b) 40°         (c) 45°          (d) 50°

Solution :

∠ACB = ∠BAC = 45 and ∠ABC = 90

∠ABC + ∠ACB + ∠BAC = 180

So, answer is option c.

Problem 15 :

In a parallelogram PQRS, if ∠P = (3x − 5)° and ∠Q = (2x + 15)°, then find the value of x.

Solution :

In a parallelogram PQRS, 

sum of co-interior angles = 180 degree

∠P + ∠Q = 180

3x - 5 + 2x + 15 = 180

5x + 10 = 180

5x = 180 - 10

5x = 170

x = 170/5

x = 34

Problem 16 :

In the figure, ABCD is a rhombus. Find x & y

Solution :

x = 64° due to angle OAB and x being alternate interior angles.

Diagonals of a rhombus are perpendicular bisectors.

So, angle DAC and DCB would be 64 + 64= 128°.

So,in triangle COB, angle O = 90° and angle OCB = 64°.

90 + 64 + y = 180°(angle sum property)

154° + y = 180°

y = 180 - 154 = 26°