NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 4 — Figure it Out (Page 94)#

Angles at A. The angle of the rectangle at A is $90°$:

$\angle CAD = 90° - \angle BAC$
$\angle CAD = 90° - 30°$
$\angle CAD = 60°$

Angles at B. The diagonals bisect each other and are equal, so $OA = OB$. In $\triangle OAB$, the angles opposite equal sides are equal:

$\angle ABD = \angle BAC = 30°$

$\angle DBC = 90° - \angle ABD$
$\angle DBC = 90° - 30°$
$\angle DBC = 60°$

Angles at C and D. In the same way, $OC = OD$ and $OB = OC$, and $AB \parallel DC$ gives equal alternate angles:

$\angle ACD = \angle BAC = 30°$ (alternate angles)

$\angle ACB = 90° - 30° = 60°$

$\angle BDC = \angle ACD = 30°$ (since $OC = OD$)

$\angle ADB = 90° - 30° = 60°$

∴ $\angle CAD = 60°$, $\angle ABD = 30°$, $\angle DBC = 60°$, $\angle ACB = 60°$, $\angle ACD = 30°$, $\angle BDC = 30°$ and $\angle ADB = 60°$.

Angles at O. Vertically opposite angles are equal, and angles on a line add up to $180°$:

$\angle POS = \angle QOR = 110°$

$\angle QOP = 180° - 110°$
$\angle QOP = 70°$

$\angle ROS = \angle QOP = 70°$

Angles in $\triangle QOR$. The diagonals are equal and bisect each other, so $OQ = OR$, and the base angles are equal:

$\angle OQR + \angle ORQ = 180° - 110°$
$\angle OQR + \angle ORQ = 70°$

$\Rightarrow \angle OQR = \angle ORQ = 35°$

Angles in $\triangle QOP$. Here $OQ = OP$:

$\angle OQP + \angle OPQ = 180° - 70°$
$\angle OQP + \angle OPQ = 110°$

$\Rightarrow \angle OQP = \angle OPQ = 55°$

The other two triangles are the same: $\triangle POS$ gives $\angle OPS = \angle OSP = 35°$, and $\triangle ROS$ gives $\angle ORS = \angle OSR = 55°$.

Check: at Q, $\angle OQR + \angle OQP = 35° + 55° = 90°$. ✓

∴ $\angle POS = 110°$, $\angle QOP = \angle ROS = 70°$, $\angle OQR = \angle ORQ = \angle OPS = \angle OSP = 35°$ and $\angle OQP = \angle OPQ = \angle ORS = \angle OSR = 55°$.

Each diagonal is 8 cm and they bisect each other, so each half is 4 cm. The steps are the same for every angle; only the angle in step 3 changes.

1. Draw a line segment $AB = 8$ cm.
2. Mark its midpoint O, so that $OA = OB = 4$ cm.
3. At O, draw a ray making the given angle ($30°$, $40°$, $90°$ or $140°$) with OB, and extend it backwards through O to make a line.
4. With centre O and radius 4 cm, cut this line at C on one side and at D on the other side, so $CD = 8$ cm.
5. Join A to C, C to B, B to D and D to A.

Because the diagonals are equal and bisect each other, each quadrilateral is a rectangle. For (iii) $90°$ the diagonals are also perpendicular, so the quadrilateral is a square.

The diagonals of APML are the two diameters PL and AM.

• They are equal, because every diameter of a circle has the same length (twice the radius).
• They bisect each other, because both pass through the centre O, and $OA = OM = OP = OL$ (all radii).
• They are perpendicular, as given.

A quadrilateral whose diagonals are equal and bisect each other is a rectangle, and if they are also perpendicular, all its sides are equal too.

∴ APML is a square.

1. Use the thread to find the midpoint of each stick: hold the thread along a stick, fold it in half, and mark the middle.
2. Lay the two sticks AB and CD across each other so that their midpoints meet at O. The angle between the sticks does not matter.
3. Tie a thread from end A to end C, and on from C to end B.

The sticks are the diagonals of the quadrilateral ACBD. They are equal (equal sticks) and bisect each other (crossed at their midpoints), so ACBD is a rectangle. The thread therefore makes an exact $90°$ at C.

No. A quadrilateral can have its opposite sides parallel and equal without having any right angle. Such a figure is a parallelogram, and most parallelograms are not rectangles.

So “opposite sides parallel” (or “parallel and equal”) is true of every rectangle, but it does not pick out only rectangles. To be a rectangle, the quadrilateral must also have a right angle.