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Assertion: In the quadrilateral ABCD, the bisectors of angles B and C meet at P. If ∠BPC = 110°, ∠A = 120°, then x = 100°. Reason: If ∠B = 2a, ∠C = 2b, a + b = 70° - Mathematics

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प्रश्न

Assertion: In the quadrilateral ABCD, the bisectors of angles B and C meet at P. If ∠BPC = 110°, ∠A = 120°, then x = 100°.


Reason: If ∠B = 2a, ∠C = 2b, a + b = 70° and sum of the angles of a quadrilateral is 360°.

विकल्प

  • Both A and R are true and R is the correct reason for A.

  • Both A and R are true but R is the incorrect reason for A.

  • A is true but R is false.

  • A is false but R is true.

MCQ
अभिकथन और तर्क
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उत्तर

Both A and R are true and R is the correct reason for A.

Explanation:

We are given:

  • In quadrilateral ABCD:
    • ∠A = 120°
    • Angle bisectors of ∠B and ∠C meet at point P
    • ∠BPC = 110°
    • Need to find x = ∠D
  • Assertion: x = 100°
  • Reason: if ∠B = 2a, ∠C = 2b and since angle bisectors meet at P, ∠BPC = a + b = 70°, hence ∠B + ∠C = 2a + 2b = 140°

Step-by-step:

Step 1: Use quadrilateral angle sum

∠A + ∠B + ∠C + ∠D = 360°

120° + (∠B + ∠C) + x = 360°

Given ∠BPC = 110° and ∠B and ∠C are bisected:

`∠BPC = (∠B)/2 + (∠C)/2`

= `(∠B + ∠C)/2`

⇒ `(∠B + ∠C)/2 = 110^circ`

⇒ ∠B + ∠C = 220°

Now plug into total:

120 + 220 + x = 360

⇒ x = 360 – 340 = 20°

But the Reason says:

Let ∠B = 2a, ∠C = 2b, then:

∠BPC = a + b = 70°

∠B + ∠C = 2a + 2b = 140°

Then total angle:

∠A + ∠B + ∠C + x = 360

⇒ 120 + 140 + x = 360

⇒ x = 100°

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अध्याय 12: Rectilinear Figures (Theorems on Parallelograms and Construction of Polygons) - MULTIPLE CHOICE QUESTIONS [पृष्ठ १५२]

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बी निर्मला शास्त्री Mathematics [English] Class 9 ICSE
अध्याय 12 Rectilinear Figures (Theorems on Parallelograms and Construction of Polygons)
MULTIPLE CHOICE QUESTIONS | Q 25. | पृष्ठ १५२
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