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12. abcd is a parallelogram. dce is an isosceles triangle where dc = de…

Question

  1. abcd is a parallelogram. dce is an isosceles triangle where dc = de. bce is a straight line. find \\(\angle adc\\).
  1. pqrs is a trapezium where pq // sr. rst is a right-angled triangle. pst is a straight line. find \\(\angle qrt\\).
  1. abc is an equilateral triangle and aed is a right-angled triangle. ace is a straight line. find \\(\angle eda\\).
  1. wxyz is a rhombus. find \\(\angle yxz\\).

Explanation:

Question 12: Find \(\angle ADC\)

  • Identify given information:
  • \(ABCD\) is a parallelogram.
  • \(DCE\) is an isosceles triangle where \(DC = DE\).
  • \(BCE\) is a straight line.
  • From the diagram, \(\angle DEC = 53^\circ\).
  • Calculate angles in \(\triangle DCE\):
  • Since \(DC = DE\), the base angles are equal: \(\angle DCE = \angle DEC = 53^\circ\).
  • The sum of angles in \(\triangle DCE\) is \(180^\circ\), so:
$$ \angle CDE = 180^\circ - 53^\circ - 53^\circ = 74^\circ $$
  • Relate to parallelogram \(ABCD\):
  • Since \(BCE\) is a straight line, \(\angle BCD\) and \(\angle DCE\) are supplementary:
$$ \angle BCD = 180^\circ - \angle DCE = 180^\circ - 53^\circ = 127^\circ $$
  • In a parallelogram, consecutive angles are supplementary, so:
$$ \angle ADC = 180^\circ - \angle BCD = 180^\circ - 127^\circ = 53^\circ $$

(Alternatively, since \(AD \parallel BC\), the alternate interior angle relationship gives \(\angle ADC = \angle DCE = 53^\circ\).)

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Question 13: Find \(\angle QRT\)

  • Identify given information:
  • \(PQRS\) is a trapezium where \(PQ \parallel SR\).
  • \(RST\) is a right-angled triangle with \(\angle RST = 90^\circ\).
  • \(PST\) is a straight line.
  • \(\angle SPQ = 90^\circ\) (indicated by the right-angle symbol).
  • \(\angle PQR = 105^\circ\).
  • \(\angle STR = 55^\circ\).
  • Calculate angles in \(\triangle RST\):
  • Since \(\angle RST = 90^\circ\) and \(\angle STR = 55^\circ\):
$$ \angle SRT = 180^\circ - 90^\circ - 55^\circ = 35^\circ $$
  • Calculate angles in trapezium \(PQRS\):
  • Since \(PQ \parallel SR\), consecutive interior angles add up to \(180^\circ\):
$$ \angle QRS = 180^\circ - \angle PQR = 180^\circ - 105^\circ = 75^\circ $$
  • Find \(\angle QRT\):
  • \(\angle QRT = \angle QRS - \angle SRT = 75^\circ - 35^\circ = 40^\circ\).

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Question 14: Find \(\angle EDA\)

  • Identify given information:
  • \(ABC\) is an equilateral triangle, so all its interior angles are \(60^\circ\).
  • \(AED\) is a right-angled triangle with \(\angle AED = 90^\circ\).
  • \(ACE\) is a straight line.
  • From the diagram, \(\angle CAD = 44^\circ\).
  • Calculate \(\angle CAE\):
  • Since \(ABC\) is equilateral, \(\angle BAC = \angle ACB = \angle BCA = 60^\circ\).
  • Since \(ACE\) is a straight line, \(\angle CAE = 180^\circ\).
  • Calculate \(\angle EAD\):
  • From the diagram, \(\angle CAE\) is split into \(\angle CAB\) and \(\angle BAE\). Since \(ACE\) is a straight line, the angle \(\angle CAB = 60^\circ\).
  • The angle \(\angle EAD\) is given by:
$$ \angle EAD = \angle CAB - \angle CAD = 60^\circ - 44^\circ = 16^\circ $$
  • Calculate \(\angle EDA\) in \(\triangle AED\):
  • The sum of angles in right-angled \(\triangle AED\) is \(180^\circ\):
$$ \angle EDA = 180^\circ - \angle AED - \angle EAD = 180^\circ - 90^\circ - 16^\circ = 74^\circ $$

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Question 15: Find \(\angle YXZ\)

  • Identify given information:
  • \(WXYZ\) is a rhombus.
  • \(\angle…

Answer:

Question 12

\(\angle ADC = 53^\circ\)

Question 13

\(\angle QRT = 40^\circ\)

Question 14

\(\angle EDA = 74^\circ\)

Question 15

\(\angle YXZ = 53^\circ\)