NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 4 — Figure it Out (Pages 107–109)#

Join the equilateral triangles ABD and BCD along their common side BD.

Sides. Every side of each triangle is 4 cm, so:

$AB = BC = CD = DA = 4$ cm

Angles. Every angle of an equilateral triangle is $60°$:

$\angle A = 60°$ and $\angle C = 60°$

$\angle B = \angle ABD + \angle DBC$
$\angle B = 60° + 60°$
$\angle B = 120°$

$\angle D = \angle ADB + \angle BDC$
$\angle D = 60° + 60°$
$\angle D = 120°$

∴ All four sides are 4 cm, and the angles are $\angle A = \angle C = 60°$ and $\angle B = \angle D = 120°$. The quadrilateral is a rhombus.

The diagonals of a kite are perpendicular, and the longer one bisects the shorter one.

1. Draw a line segment $PQ = 6$ cm.
2. Draw the perpendicular bisector of PQ with a compass. It cuts PQ at its midpoint T.
3. On this bisector, mark S on one side of T and R on the other side so that $SR = 8$ cm. For example, $TS = 2$ cm and $TR = 6$ cm. (If $TS = TR$, the kite would be a rhombus.)
4. Join P to S, S to Q, Q to R and R to P.

PSQR is the required kite: $SP = SQ$ and $RP = RQ$, because S and R lie on the perpendicular bisector of PQ.

SR ∥ PQ, so the angles on the same side of each slanting side add up to $180°$ (co-interior angles):

$\angle S = 180° - \angle P$
$\angle S = 180° - 135°$
$\angle S = 45°$

$\angle R = 180° - \angle Q$
$\angle R = 180° - 105°$
$\angle R = 75°$

∴ $\angle S = 45°$ and $\angle R = 75°$. Check: $135° + 105° + 75° + 45° = 360°$. ✓

AB ∥ DC, so $\angle A$ and $\angle D$ are co-interior angles:

$\angle A = 180° - \angle D$
$\angle A = 180° - 100°$
$\angle A = 80°$

$AD = BC$, so ABCD is an isosceles trapezium, and the angles at each end of a parallel side are equal:

$\angle B = \angle A = 80°$
$\angle C = \angle D = 100°$

∴ $\angle A = \angle B = 80°$ and $\angle C = 100°$.

• Rectangles and rhombuses are both inside parallelograms.
• Rhombuses are exactly the quadrilaterals that are both kites and parallelograms.
• Squares are exactly the quadrilaterals that are both rhombuses and rectangles.

A kite has two pairs of equal adjacent sides, and a parallelogram has equal opposite sides. Together, all four sides must be equal, so the quadrilateral is a rhombus (and a square is a special rhombus).

Yes, a square. A rectangle has equal opposite sides; a kite needs equal adjacent sides as well, so all four sides are equal. A rectangle with all sides equal is a square. No other rectangle is a kite.

No. A kite only needs two pairs of equal adjacent sides, like the kite in question 2 (sides 3.6 cm and 6.7 cm), so its four sides need not all be equal.

The correct relationship is the other way round: every rhombus is a kite, but not every kite is a rhombus.

In $\triangle ROI$, $\angle RIO = 90°$ (a corner of rectangle PAIR) and $\angle ORI = 30°$:

$\angle ROI = 180° - (90° + 30°)$
$\angle ROI = 60°$

At O, $\angle ROD = 90°$ (a corner of rectangle RODS). The angles $\angle ROI$ and $\angle IOD$ together make up $\angle ROD$:

$\angle IOD = \angle ROD - \angle ROI$
$\angle IOD = 90° - 60°$
$\angle IOD = 30°$

∴ $\angle IOD = $ 30°.

The diagonals of a square are equal, bisect each other and are perpendicular. So both diagonals are 6 cm, and the halves are 3 cm.

1. Draw a line segment $AB = 6$ cm.
2. With a compass, draw the perpendicular bisector of AB: with centres A and B and the same radius (more than 3 cm), draw arcs above and below AB, and join the two points where they cross. This line cuts AB at its midpoint O at $90°$.
3. With centre O and radius 3 cm, cut the bisector at C and D.
4. Join A to C, C to B, B to D and D to A.

ADBC is the required square. No protractor is needed, because the $90°$ comes from the perpendicular bisector.

Let each side of the square be $x$. Each midpoint cuts a side into two halves of $\dfrac{x}{2}$.

Sides. The four corner triangles CUV, AXU, SWX and EVW are right triangles with both legs $\dfrac{x}{2}$, so they are congruent (SAS). Their hypotenuses are equal:

$UV = UX = XW = WV$

Angles. Each corner triangle is a right isosceles triangle, so its other two angles are $45°$ each. At U, the angles on the line CA add up to $180°$:

$\angle VUX = 180° - (45° + 45°)$
$\angle VUX = 90°$

In the same way, the angles at X, W and V are $90°$.

∴ UXWV has four equal sides and four right angles, so it is a square. Measuring the sides and angles of a construction confirms this.

Mark a point on each side at the same distance $a$ from a corner, going round the square in the same direction (clockwise), and join the four points.

The four corner triangles again have legs $a$ and $x - a$ with a right angle between them, so they are congruent (SAS). Their hypotenuses, the sides of the inner figure, are equal, and the same angle argument as above gives right angles. So the inner figure is always a square, for any $a$ between 0 and $x$. Taking $a = \dfrac{x}{2}$ gives the midpoint square of part (a).

Yes. A quadrilateral with four equal sides is a rhombus, and a rhombus is a parallelogram. So:

• opposite angles are equal: the angle opposite the $90°$ angle is also $90°$;
• adjacent angles add up to $180°$: each angle next to it is $180° - 90° = 90°$.

All four angles are $90°$ and all four sides are equal, so it is a square. Constructing one (four equal sides with one corner drawn at $90°$) and measuring the other angles gives $90°$ each.

Let ABCD have $AB = CD$ and $AD = BC$. Draw the diagonal AC.

In $\triangle ABC$ and $\triangle CDA$:

• $AB = CD$ (given)
• $BC = DA$ (given)
• $AC = CA$ (common side)

So $\triangle ABC \cong \triangle CDA$ (SSS). Their matching angles are equal:

• $\angle 1 = \angle 4$, that is $\angle BAC = \angle DCA$. These are alternate angles for AB and CD with AC as a transversal, so $AB \parallel CD$.
• $\angle 3 = \angle 2$, that is $\angle BCA = \angle DAC$. These are alternate angles for BC and AD, so $BC \parallel AD$.

∴ Both pairs of opposite sides are parallel, so the quadrilateral is a parallelogram.

Yes. In this arrowhead (dart) ABCD, the angle at D is a reflex angle (more than $180°$) measured inside the figure.

Join BD. It splits ABCD into two triangles, ABD and CBD:

$\angle 1 + \angle 2 + \angle ADB = 180°$ (angles of $\triangle ABD$)
$\angle 3 + \angle 4 + \angle CDB = 180°$ (angles of $\triangle CBD$)

Adding the two:

$\angle 1 + (\angle 2 + \angle 3) + \angle 4 + (\angle ADB + \angle CDB) = 360°$

Here $\angle 2 + \angle 3 = \angle B$, and $\angle ADB + \angle CDB$ is the reflex angle at D. So:

$\angle A + \angle B + \angle C + \angle D = 360°$

Measuring a constructed dart gives the same result, as long as the reflex angle at D is measured (360° minus the smaller angle at D).

False. Equal diagonals that bisect each other give a rectangle. It is a square only if the diagonals are also perpendicular, as in question 2 of page 94, where a $30°$ angle gives a rectangle that is not a square.

True. The angles add up to $360°$, so the fourth angle is:

$\angle D = 360° - 3 \times 90°$
$\angle D = 90°$

All four angles are right angles, so it is a rectangle.

True. If the diagonals AC and BD bisect each other at O, then $\triangle AOB \cong \triangle COD$ (SAS, with vertically opposite angles at O). This gives $\angle OAB = \angle OCD$, which are alternate angles, so $AB \parallel CD$. In the same way, $AD \parallel BC$.

False. A kite also has perpendicular diagonals, but its sides are not all equal (for example, the kite in question 2). The diagonals must also bisect each other for the quadrilateral to be a rhombus.

True. If $\angle A = \angle C$ and $\angle B = \angle D$, then:

$2\angle A + 2\angle B = 360°$
$\angle A + \angle B = 180°$

Co-interior angles on AB add up to $180°$, so $AD \parallel BC$. In the same way, $\angle A + \angle D = 180°$ gives $AB \parallel DC$.

True. Four equal angles adding up to $360°$ are $90°$ each, so it is a rectangle.

False. An isosceles trapezium has only one pair of parallel sides; its equal sides are slanted towards each other (like ABCD in question 3 (ii)), so they are not parallel.