मराठी

PQRS is a parallelogram, QP is extended to T so that ∠STP = 90°. Find x and y. - Mathematics

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प्रश्न

PQRS is a parallelogram, QP is extended to T so that ∠STP = 90°. Find x and y.

बेरीज
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उत्तर

Given:

  • PQRS is a parallelogram.
  • QP is extended to point T such that ∠STP = 90°.
  • The marked angles on the figure are ∠SPT = 7x°, ∠PQR = 3x° and ∠PSR = y°.
  • We need to find the values of (x) and (y).

Stepwise Calculation:

Step 1: Use the property of adjacent angles in a parallelogram

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

Therefore, for parallelogram PQRS, ∠SPQ + ∠PQR = 180°.

Given ∠SPQ = 7x and ∠PQR = 3x, we can write the equation: 7x + 3x = 180°

Step 2: Solve for x

Combine the terms and solve the equation from Step 1:

10x = 180°

`x = 180^circ/10`

x = 18

Step 3: Find the measure of ∠SPQ 

Substitute the value of x back into the expression for ∠SPQ:

∠SPQ = 7x = 7 × 18 = 126°

Step 4: Use the angle sum property of triangle STP 

In triangle STP, the sum of interior angles is 180°. 

We are given ∠STP = 90° and ∠TSP = y.

∠STP  and ∠SPQ form a linear pair on the straight line TQ, so ∠SPT + ∠SPQ = 180°.

We found ∠SPQ = 126°, so: ∠SPT = 180° – 126° = 54°

Now, apply the angle sum property to triangle STP:

∠STP + ∠TSP + ∠SPT = 180°

90° + y + 54° = 180°

Step 5: Solve for y

Combine the constant terms and solve for y:

144° + y = 180°

y = 180° – 144°

y = 36

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पाठ 12: Rectilinear Figures (Theorems on Parallelograms and Construction of Polygons) - EXERCISE 12A [पृष्ठ १४१]

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बी निर्मला शास्त्री Mathematics [English] Class 9 ICSE
पाठ 12 Rectilinear Figures (Theorems on Parallelograms and Construction of Polygons)
EXERCISE 12A | Q 23. (i) | पृष्ठ १४१
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