Question

drag the tiles to the correct boxes to complete the pairs.
match the statements with their values.
( m angle a b c + m angle b a c + m angle a c b ) when ( \triangle a b c ) is an isosceles triangle with ( overline { a b } = overline { a c } )
( m angle a b c ) when ( m angle b a c = 70 ^ { circ } ) and ( \triangle a b c ) is an isosceles triangle with ( overline { a b } = overline { a c } )
( m angle q p r ) when ( m angle q r p = 30 ^ { circ } ) and ( \triangle p q r ) is an isosceles triangle with ( overline { p q } = overline { q r } )
( m angle b d e ) when ( m angle b a c = 45 ^ { circ } ) and points ( d ) and ( e ) are the midpoints of ( overline { a b } ) and ( overline { b c } ), respectively, in ( \triangle a b c )

Explanation:

Step1: Find \(m\angle ABC + m\angle BAC + m\angle ACB\)

By the triangle - angle sum theorem, the sum of the interior angles of a triangle is \(180^{\circ}\). So \(m\angle ABC + m\angle BAC + m\angle ACB=180^{\circ}\)

Step2: Find \(m\angle ABC\) when \(m\angle BAC = 70^{\circ}\) and \(\triangle ABC\) is isosceles with \(\overline{AB}=\overline{AC}\)

Since \(\overline{AB}=\overline{AC}\), \(\angle ABC=\angle ACB\). Using the triangle - angle sum theorem \(m\angle ABC + m\angle BAC + m\angle ACB = 180^{\circ}\). Substitute \(m\angle BAC = 70^{\circ}\) and \(m\angle ABC=m\angle ACB\). Let \(x = m\angle ABC\), then \(x + 70^{\circ}+x=180^{\circ}\), \(2x=180^{\circ}- 70^{\circ}=110^{\circ}\), \(x = 55^{\circ}\)

Step3: Find \(m\angle QPR\) when \(m\angle QRP = 30^{\circ}\) and \(\triangle PQR\) is isosceles with \(\overline{PQ}=\overline{QR}\)

Since \(\overline{PQ}=\overline{QR}\), \(\angle QPR=\angle QRP\). So \(m\angle QPR = 30^{\circ}\)

Step4: Find \(m\angle BDE\) when \(m\angle BAC = 45^{\circ}\) and \(D\) and \(E\) are mid - points of \(\overline{AB}\) and \(\overline{BC}\) respectively in \(\triangle ABC\)

By the mid - segment theorem, \(DE\parallel AC\). Then \(\angle BDE=\angle BAC\) (corresponding angles). So \(m\angle BDE = 45^{\circ}\)

Answer:

• \(m\angle ABC + m\angle BAC + m\angle ACB\) matches with \(180^{\circ}\)
• \(m\angle ABC\) when \(m\angle BAC = 70^{\circ}\) and \(\triangle ABC\) is isosceles with \(\overline{AB}=\overline{AC}\) matches with \(55^{\circ}\)
• \(m\angle QPR\) when \(m\angle QRP = 30^{\circ}\) and \(\triangle PQR\) is isosceles with \(\overline{PQ}=\overline{QR}\) matches with \(30^{\circ}\)
• \(m\angle BDE\) when \(m\angle BAC = 45^{\circ}\) and \(D\) and \(E\) are mid - points of \(\overline{AB}\) and \(\overline{BC}\) respectively in \(\triangle ABC\) matches with \(45^{\circ}\)