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In the parallelogram ABCD, the bisectors of angles A and B meet at P. ∴ ∠APB is ______. - Mathematics

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प्रश्न

In the parallelogram ABCD, the bisectors of angles A and B meet at P. ∴ ∠APB is ______.

विकल्प

  • 60°

  • 45°

  • 90°

  • 120°

MCQ
रिक्त स्थान भरें
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उत्तर

In the parallelogram ABCD, the bisectors of angles A and B meet at P. ∴ ∠APB is 90°.

Explanation:


Step 1: Use the property of a parallelogram

In a parallelogram, consecutive angles are supplementary, meaning their sum is 180°. 

Therefore, in parallelogram ABCD, we have ∠A + ∠B = 180°

Step 2: Consider the angle bisectors

Since AP bisects ∠A and BP bisects ∠B, the angles within the triangle APB can be expressed as:

`∠PAB = 1/2 ∠A`

`∠PBA = 1/2 ∠B`

Step 3: Find the sum of the angles in triangle APB

The sum of the interior angles of any triangle is 180°. 

In ΔAPB, we can write:

∠PAB + ∠PBA + ∠APB = 180°

Substitute the expressions from Step 2:

`1/2 ∠A + 1/2 ∠B + ∠APB = 180^circ`

Factor out `1/2`:

`1/2 (∠A + ∠B) + ∠APB = 180^circ`

Step 4: Substitute the sum of angles A and B

From Step 1, we know that ∠A + ∠B = 180°.

Substitute this value into the equation from Step 3:

`1/2 (180^circ) + ∠APB = 180^circ`

`90^circ + ∠APB = 180^circ`

Step 5: Solve for the angle APB

Subtract 90° from both sides of the equation:

∠APB = 180° – 90°

∠APB = 90°

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अध्याय 12: Rectilinear Figures (Theorems on Parallelograms and Construction of Polygons) - MULTIPLE CHOICE QUESTIONS [पृष्ठ १५०]

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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 2. | पृष्ठ १५०
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