Question 13 Marks
Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other, and intersect at an angle of $140^{\circ}$
View full question & answer→Questions
3 Marks Question
🎯
Test yourself on this topic
10 questions · timed · auto-graded
Question 23 Marks
Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other, and intersect at an angle of $90^{\circ}$
View full question & answer→Question 33 Marks
Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other, and intersect at an angle of $40^{\circ}$
View full question & answer→Question 43 Marks
Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other, and intersect at an angle of $30^{\circ}$
View full question & answer→Question 53 Marks
CASE is a square. The points U, V, W and X are the midpoints of the sides of the square. What type of quadrilateral is UVWX? Find this by using geometric reasoning, as well as by construction and measurement. Find other ways of constructing a square within a square such that the vertices of the inner square lie on the sides of the outer square, as shown in Figure (b).
Answer
View full question & answer→self
Question 63 Marks
If PAIR and RODS are two rectangles, find $\angle I O D$.
Answer
View full question & answer→Since PAIR and RODS are two triangles.
$\angle RIO =90^{\circ}$ $\qquad$ (Corner angle of a rectangle)
In $\triangle RIO$,
$\angle IRO +\angle IOR +\angle RIO =180^{\circ}$ $\qquad$ (Sum of angles of a triangle)
$30^{\circ}+\angle IOR +90^{\circ}=180^{\circ}$
$120^{\circ}+\angle IOR =180^{\circ}$
$\angle IOR =180^{\circ}-120^{\circ}=60^{\circ}$.
$\therefore \angle IOD =90^{\circ}-\angle IOR =90^{\circ}-60^{\circ}=30^{\circ}$.
$\angle RIO =90^{\circ}$ $\qquad$ (Corner angle of a rectangle)
In $\triangle RIO$,
$\angle IRO +\angle IOR +\angle RIO =180^{\circ}$ $\qquad$ (Sum of angles of a triangle)
$30^{\circ}+\angle IOR +90^{\circ}=180^{\circ}$
$120^{\circ}+\angle IOR =180^{\circ}$
$\angle IOR =180^{\circ}-120^{\circ}=60^{\circ}$.
$\therefore \angle IOD =90^{\circ}-\angle IOR =90^{\circ}-60^{\circ}=30^{\circ}$.
Question 73 Marks
Find the remaining angles in the following quadrilaterals.
Answer
View full question & answer→Here, AEIO is a rhombus (all sides equal).
In $\triangle E A O, A E=A O$, then the angles opposite them are equal.
$\begin{array}{l}\therefore \angle AOE=\angle AEO=20^{\circ} \\\angle AEO=\angle IEO=20^{\circ} \ldots \ldots \ldots \ldots .\end{array}$
(The diagonals of a rhombus bisect its angles)
Also, $\angle AOE =\angle IOE =20^{\circ}$ $\qquad$ (The diagonals of a rhombus bisect its angles)
$\begin{array}{l}\angle E=2 \times \angle A E O=2 \times 20^{\circ}=40^{\circ} \\\angle E=\angle O=40^{\circ} \ldots \ldots \ldots . . \text { (Opposite angles of a rhombus are equal) } \\\angle E+\angle A=180^{\circ} \ldots \ldots \ldots . . \text { (The sum of adjacent angles of a rhombus is } 180^{\circ} \text { ) } \\40^{\circ}+\angle A=180^{\circ} \\\angle A=180^{\circ}-40^{\circ} \\\angle A=140^{\circ} \\\angle A=\angle I=140^{\circ} \ldots \ldots \ldots . . \text { (Opposite angles of a rhombus are equal) }\end{array}$
In $\triangle E A O, A E=A O$, then the angles opposite them are equal.
$\begin{array}{l}\therefore \angle AOE=\angle AEO=20^{\circ} \\\angle AEO=\angle IEO=20^{\circ} \ldots \ldots \ldots \ldots .\end{array}$
(The diagonals of a rhombus bisect its angles)
Also, $\angle AOE =\angle IOE =20^{\circ}$ $\qquad$ (The diagonals of a rhombus bisect its angles)
$\begin{array}{l}\angle E=2 \times \angle A E O=2 \times 20^{\circ}=40^{\circ} \\\angle E=\angle O=40^{\circ} \ldots \ldots \ldots . . \text { (Opposite angles of a rhombus are equal) } \\\angle E+\angle A=180^{\circ} \ldots \ldots \ldots . . \text { (The sum of adjacent angles of a rhombus is } 180^{\circ} \text { ) } \\40^{\circ}+\angle A=180^{\circ} \\\angle A=180^{\circ}-40^{\circ} \\\angle A=140^{\circ} \\\angle A=\angle I=140^{\circ} \ldots \ldots \ldots . . \text { (Opposite angles of a rhombus are equal) }\end{array}$
Question 83 Marks
Find the remaining angles in the following quadrilaterals.
Answer
View full question & answer→Here, XWUV is a rhombus (all sides equal).
In $\triangle VUX , UV = UX$, then the angles opposite them are equal.
$\begin{array}{l}\therefore \angle U X V=\angle U V X=30^{\circ} \\\angle U X V=\angle W X V=30^{\circ} \ldots\end{array}$
$\qquad$ (The diagonals of a rhombus bisect its angles)
Also, $\angle UVX =\angle WVX =30^{\circ}$ $\qquad$ (The diagonals of a rhombus bisect its angles)
$\begin{array}{l}\angle E=2 \times \angle UVX=2 \times 30^{\circ}=60^{\circ} \\\angle V=\angle X=60^{\circ} \ldots \ldots \ldots . . \text { (Opposite angles of a rhombus are equal) } \\\angle V+\angle U=180^{\circ} \ldots \ldots \ldots . . \text { (The sum of adjacent angles of a rhombus is } 180^{\circ} \text { ) } \\
60^{\circ}+\angle U=180^{\circ} \\\angle U=180^{\circ}-60^{\circ} \\\angle U=120^{\circ} \\\angle U=\angle W=120^{\circ} \ldots \ldots \ldots . . \text { (Opposite angles of a rhombus are equal) }\end{array}$
In $\triangle VUX , UV = UX$, then the angles opposite them are equal.
$\begin{array}{l}\therefore \angle U X V=\angle U V X=30^{\circ} \\\angle U X V=\angle W X V=30^{\circ} \ldots\end{array}$
$\qquad$ (The diagonals of a rhombus bisect its angles)
Also, $\angle UVX =\angle WVX =30^{\circ}$ $\qquad$ (The diagonals of a rhombus bisect its angles)
$\begin{array}{l}\angle E=2 \times \angle UVX=2 \times 30^{\circ}=60^{\circ} \\\angle V=\angle X=60^{\circ} \ldots \ldots \ldots . . \text { (Opposite angles of a rhombus are equal) } \\\angle V+\angle U=180^{\circ} \ldots \ldots \ldots . . \text { (The sum of adjacent angles of a rhombus is } 180^{\circ} \text { ) } \\
60^{\circ}+\angle U=180^{\circ} \\\angle U=180^{\circ}-60^{\circ} \\\angle U=120^{\circ} \\\angle U=\angle W=120^{\circ} \ldots \ldots \ldots . . \text { (Opposite angles of a rhombus are equal) }\end{array}$
Question 93 Marks
Find the remaining angles in the following quadrilaterals.
Answer
View full question & answer→Here $PQ \| SR$, and $PS \| QR$
$\therefore PQRS$ is a parallelogram.
$\angle P =\angle R =110^{\circ}$ $\qquad$ (Opposite angles of a parallelogram are equal)
$\angle P +\angle S =180^{\circ}$ $\qquad$ (The sum of the adjacent angles of a parallelogram is $180^{\circ}$ )
$110^{\circ}+\angle S =180^{\circ}$
$\angle S =180^{\circ}-110^{\circ}$
$\angle S =70^{\circ}$.
$\angle S =\angle Q =70^{\circ}$ $\qquad$ (Opposite angles of a parallelogram are equal)
$\therefore PQRS$ is a parallelogram.
$\angle P =\angle R =110^{\circ}$ $\qquad$ (Opposite angles of a parallelogram are equal)
$\angle P +\angle S =180^{\circ}$ $\qquad$ (The sum of the adjacent angles of a parallelogram is $180^{\circ}$ )
$110^{\circ}+\angle S =180^{\circ}$
$\angle S =180^{\circ}-110^{\circ}$
$\angle S =70^{\circ}$.
$\angle S =\angle Q =70^{\circ}$ $\qquad$ (Opposite angles of a parallelogram are equal)
Question 103 Marks
Find the remaining angles in the following quadrilaterals.
Answer
View full question & answer→Here $P R \| E A$, and $P E \| R A$
Therefore, PEAR is a parallelogram.
$\angle P =\angle A =40^{\circ}$ $\qquad$ (Opposite angles of a parallelogram are equal)
$\angle P +\angle R =180^{\circ}$ $\qquad$ (The sum of the adjacent angles of a parallelogram is $180^{\circ}$ )
$\begin{array}{l}40^{\circ}+\angle R=180^{\circ} \\\angle R=180^{\circ}-40^{\circ} \\\angle R=140^{\circ} . \\\angle R=\angle E=140^{\circ} .\end{array}$$\qquad$ (Opposite angles of a parallelogram are equal)
Therefore, PEAR is a parallelogram.
$\angle P =\angle A =40^{\circ}$ $\qquad$ (Opposite angles of a parallelogram are equal)
$\angle P +\angle R =180^{\circ}$ $\qquad$ (The sum of the adjacent angles of a parallelogram is $180^{\circ}$ )
$\begin{array}{l}40^{\circ}+\angle R=180^{\circ} \\\angle R=180^{\circ}-40^{\circ} \\\angle R=140^{\circ} . \\\angle R=\angle E=140^{\circ} .\end{array}$$\qquad$ (Opposite angles of a parallelogram are equal)