Question 15 Marks
Find all the other angles inside the following rectangles.
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Question 25 Marks
Find all the other angles inside the following rectangles.
Question 35 Marks
Will the sum of the angles in a quadrilateral such as the following one also be $360^{\circ}$ ? Find the answer using geometric reasoning as well as by constructing this figure and measuring.
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Question 45 Marks
What type of quadrilateral is one in which the opposite sides are equal? Justify your answer.
Answer
View full question & answer→If a quadrilateral has opposite sides equal, then it is a parallelogram.
Geometric reasoning using a diagonal :
Given : Quadrilateral ABCD with $AB = CD$ and $BC = DA$.
Draw diagonal $A C$.
In $\triangle A B C$ and $\triangle C D A$,
$AB = CD$ (given)
$BC = DA$ (given)
$AC = AC$ (common side)
By SSS congruence, $\triangle ABC \cong \triangle CDA$.
From congruence, corresponding angles are equal:
$\angle BAC =\angle DCA$ and $\angle ACB =\angle CAD$.
But these are alternate interior angles.
$\therefore A B \| D C$ and $A D \| B C$.
Hence, ABCD is a parallelogram.
Geometric reasoning using a diagonal :
Given : Quadrilateral ABCD with $AB = CD$ and $BC = DA$.
Draw diagonal $A C$.
In $\triangle A B C$ and $\triangle C D A$,
$AB = CD$ (given)
$BC = DA$ (given)
$AC = AC$ (common side)
By SSS congruence, $\triangle ABC \cong \triangle CDA$.
From congruence, corresponding angles are equal:
$\angle BAC =\angle DCA$ and $\angle ACB =\angle CAD$.
But these are alternate interior angles.
$\therefore A B \| D C$ and $A D \| B C$.
Hence, ABCD is a parallelogram.
Question 55 Marks
If a quadrilateral has four equal sides and one angle of $90^{\circ}$, will it be a square? Find the answer using geometric reasoning as well as by construction and measurement.
Answer
View full question & answer→Reasoning :
A rhombus is a quadrilateral with four equal sides.
If a rhombus has one angle of $90^{\circ}$, then:
Its opposite angle is also $90^{\circ}$ (opposite angles of a rhombus are equal).
Each adjacent angle must also be $90^{\circ}$ (sum of adjacent angles in a parallelogram/rhombus is $180^{\circ}$ ).
Thus, all four angles are $90^{\circ}$.
Since the quadrilateral has all sides equal and all angles right angles, it is a square.
Construction and measurement :
Steps of construction :
(i) Draw a line segment PQ of length 5 cm .
(ii) At point PP , construct a perpendicular line to PQ .
(iii) On this perpendicular, mark point S such that $PS =5 cm$.
(iv) With S as centre and radius 5 cm , draw an arc to the right of PS .
(v) With Q as centre and radius 5 cm , draw an arc above PQ to intersect the arc from step (4) at point R .
Join $Q - R , R - S$, and $S - P$ to complete the square PQRS .
Verification by measurement :
All sides: $PQ = QR = RS = SP =5 cm$
All angles: $\angle P=\angle Q=\angle R=\angle S=90^{\circ}$.
Conclusion: The figure constructed is a square.
A rhombus is a quadrilateral with four equal sides.
If a rhombus has one angle of $90^{\circ}$, then:
Its opposite angle is also $90^{\circ}$ (opposite angles of a rhombus are equal).
Each adjacent angle must also be $90^{\circ}$ (sum of adjacent angles in a parallelogram/rhombus is $180^{\circ}$ ).
Thus, all four angles are $90^{\circ}$.
Since the quadrilateral has all sides equal and all angles right angles, it is a square.
Construction and measurement :
Steps of construction :
(i) Draw a line segment PQ of length 5 cm .
(ii) At point PP , construct a perpendicular line to PQ .
(iii) On this perpendicular, mark point S such that $PS =5 cm$.
(iv) With S as centre and radius 5 cm , draw an arc to the right of PS .
(v) With Q as centre and radius 5 cm , draw an arc above PQ to intersect the arc from step (4) at point R .
Join $Q - R , R - S$, and $S - P$ to complete the square PQRS .
Verification by measurement :
All sides: $PQ = QR = RS = SP =5 cm$
All angles: $\angle P=\angle Q=\angle R=\angle S=90^{\circ}$.
Conclusion: The figure constructed is a square.
Question 65 Marks
Find the remaining angles in the following trapeziums -
Answer
Since $A B\ \|\ D C$, and $A D$ is a tranversal, then
$\angle A +\angle D =180^{\circ}$ $\qquad$ (Sum of angles on the same side of the transversal)
$135^{\circ}+\angle D =180^{\circ}$
$\angle D =180^{\circ}-135^{\circ}$
$\angle D =45^{\circ}$
Also, since $AB\ \|\ DC$, and BC is a tranversal, then
$\angle B +\angle C =180^{\circ}$ $\qquad$ (Sum of angles on the same side of the transversal)
$\begin{array}{l}105^{\circ}+\angle C=180^{\circ} \\\angle C=180^{\circ}-105^{\circ} \\\angle C=75^{\circ}\end{array}$
Since $PQ \| SR$, and PS is a tranversal, then
$\angle P +\angle S =180^{\circ}$ $\qquad$ (Sum of angles on the same side of the transversal)
$\angle P+100^{\circ}=180^{\circ}$
$\angle P =180^{\circ}-100^{\circ}=80^{\circ}$.
$\angle S =\angle R =100^{\circ}$ $\qquad$ (Angles opposite to equal sides are equal)
Also, since $PQ \| SR$, and QR is a tranversal, then
$\angle Q +\angle R =180^{\circ}$ $\qquad$ (Sum of angles on the same side of the transversal)
$\angle Q+100^{\circ}=180^{\circ}$
$\angle Q=180^{\circ}-100^{\circ}=80^{\circ}$.
View full question & answer→Since $A B\ \|\ D C$, and $A D$ is a tranversal, then
$\angle A +\angle D =180^{\circ}$ $\qquad$ (Sum of angles on the same side of the transversal)
$135^{\circ}+\angle D =180^{\circ}$
$\angle D =180^{\circ}-135^{\circ}$
$\angle D =45^{\circ}$
Also, since $AB\ \|\ DC$, and BC is a tranversal, then
$\angle B +\angle C =180^{\circ}$ $\qquad$ (Sum of angles on the same side of the transversal)
$\begin{array}{l}105^{\circ}+\angle C=180^{\circ} \\\angle C=180^{\circ}-105^{\circ} \\\angle C=75^{\circ}\end{array}$
Since $PQ \| SR$, and PS is a tranversal, then
$\angle P +\angle S =180^{\circ}$ $\qquad$ (Sum of angles on the same side of the transversal)
$\angle P+100^{\circ}=180^{\circ}$
$\angle P =180^{\circ}-100^{\circ}=80^{\circ}$.
$\angle S =\angle R =100^{\circ}$ $\qquad$ (Angles opposite to equal sides are equal)
Also, since $PQ \| SR$, and QR is a tranversal, then
$\angle Q +\angle R =180^{\circ}$ $\qquad$ (Sum of angles on the same side of the transversal)
$\angle Q+100^{\circ}=180^{\circ}$
$\angle Q=180^{\circ}-100^{\circ}=80^{\circ}$.