Question

PQRS is a parallelogram of angle P = (3x+15)0

and angle R =

(4x+10)0

, find the value of x and measure of all the angles..​

Answer 1

Answer:

• Value of x is 5°.
• Angle P is 30°, Angle Q is 150°, Angle R is 30° and Angle S is 150°.

Step-by-step explanation:

Given:-

• PQRS is the parallelogram.
• Angle P is (3x + 15)°
• Angle R is (4x + 10)°

To find:-

• Value of x.
• Measure of all angles.

Solution:-

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1,1)(1,1)(6,1)\put(0.4,0.5){\bf Q}\qbezier(1,1)(1,1)(1.6,4)\put(6.2,0.5){\bf R}\qbezier(1.6,4)(1.6,4)(6.6,4)\put(1,4){\bf P}\qbezier(6,1)(6,1)(6.6,4)\put(6.9,3.8){\bf S}\qbezier(5.4,1)(5.2,1.48)(6.1,1.5)\qbezier(1.5,3.5)(1.7,3.2)(2.2,4)\put(2.2,3.5){\sf (3x + 15)^ \circ }\put(3.9,1.5){\sf (4x + 10)^\circ }\end{picture}$

We know,

Opposite angles of parallelogram are equal.

So,

∠P = ∠R

$\longrightarrow$ (3x + 15)° = (4x + 10)°

$\longrightarrow$ 3x° + 15° - 10° = 4x°

$\longrightarrow$ 15° - 10° = 4x° - 3x°

$\longrightarrow$ 5° = x°

Or, x° = 5°

Value of x is 5°.

So,

∠P = 3x + 15 = 3×5 + 15 = 30°

∠R = 4x + 10° = 4×5 + 10 = 30°

We also know,

Sum of two adjacent angles when two parallel lines intersect by an transversal is 180°. We also called it Co-interior angles.

So,

$\longrightarrow$ ∠P + ∠Q = 180°

$\longrightarrow$ 30° + ∠Q = 180°

$\longrightarrow$ ∠Q = 180° - 30°

$\longrightarrow$ ∠Q = 150°

∠Q is of 150°.

And ,

• We know , Opposite angles of parallelogram are equal.

So,

∠S = ∠Q = 150°

Therefore,

∠P is of 30°.

∠Q is of 150°.

∠R is of 30°.

∠S is of 150°.

Answer 2

Opposite angles of parallelogram are equal.

=> 3x + 15° = 4x + 10°

=> 15° - 10° = 4x - 3x

=> 5 = x

x is 5°

Angle P = 3×5 + 15° = 30°

Angle R = 4×5 + 10 = 30°

=> Angle P + Angle Q = 180° (By co interior angles )

=> 30° + Q = 180°

=> Q = 180° - 30°

=> Q = 150°

Angle Q = 150°.

Opposite angles of parallelogram are equal

Angle Q = Angle S = 150°