Quadrilaterals Class 8 Solutions

The first three are closed polygons made from exactly four line segments. Figure (iv) has a curved side, and figure (v) is not a single simple four-sided closed polygon.

Use another 8 cm strip and join the strips at both midpoints. Their angle can be any non-zero, non-straight angle: equal diagonals that bisect each other always form a rectangle.

Compare triangles AOB and COD. Vertically opposite angles at O are equal, and the remaining angle relation gives AAS congruence; hence OA = OC and OB = OD.

No. Those three facts give SSA, which is not a valid general congruence test. Instead, use the angle relations from the rectangle to obtain AAS congruence, or use another already-proved corresponding side with a valid congruence condition.

The four angles at O are 60, 120, 60, and 120 degrees. Each triangle with vertex angle 60 degrees has base angles 60 degrees; each with vertex angle 120 degrees has base angles 30 degrees. Thus every corner of the quadrilateral is 90 degrees.

ABCD is a rectangle. All four vertex angles are 90 degrees, with AB = CD and AD = BC by congruent opposite triangles.

Yes. If one intersection angle is x, the adjacent one is 180 - x. The base angles in the resulting isosceles triangles combine to 90 degrees at every vertex, so the quadrilateral remains a rectangle.

At O the angles are x,x,180^\circ-x,180^\circ-x. The relevant base angles are 90^\circ-\frac{x}{2} and \frac{x}{2}, so each vertex angle is 90 degrees. Also, AB=CD and AD=BC.

No. Drawing a diagonal creates two congruent triangles, forcing both pairs of opposite sides to be equal. Such a quadrilateral is necessarily a rectangle.

BD is common, angles BAD and DCB are both 90 degrees, and the complementary angle argument gives angle ABD = angle BDC. Therefore the triangles are congruent by AAS.

No. The order must match corresponding vertices. The correct statement is triangle BAD congruent to triangle DCB; writing CDB pairs B with D and A with C incorrectly.

AD is a transversal of AB and DC, and angles DAB and ADC are 90 degrees. Their same-side interior sum is 180 degrees, so AB is parallel to DC.

All four are rectangles because each has four right angles. Figure (iv) is also a square, which is a special rectangle.

They must be equal, bisect each other, and meet at 90 degrees. The right-angle condition makes all four sides equal in addition to producing four right vertex angles.

Compare triangles BOA and BOC. They have BO common, OA = OC, and BA = BC, so they are congruent by SSS.

They are equal corresponding angles and form a linear pair. Therefore each is 90 degrees.

Draw AC = 8 cm and construct its perpendicular bisector at midpoint O. Mark B and D on it with OB = OD = 4 cm, then join A-B-C-D-A.

Each is 45 degrees. A diagonal divides the square into isosceles right triangles and therefore bisects the two 90-degree corner angles.

(i) angle ABD = 30, CAD = 60, ADB = 60, BDC = 30, ACD = 30, and ACB = 60 degrees. (ii) angle POS = 110, QOP = ROS = 70, OQR = ORQ = 35, and OQP = OPQ = ORS = OSR = 55 degrees.

Draw one 8 cm diagonal and mark its midpoint O. Through O draw a line at the required angle; mark 4 cm on both sides of O for the second diagonal, then join the four endpoints. Each result is a rectangle; the 90-degree case is a square.

APML is a square. Its diagonals are equal diameters, bisect each other at the centre, and are perpendicular.

Join the equal sticks at their midpoints. Pass a taut thread through their four endpoints to form the quadrilateral. Equal diagonals bisecting each other make a rectangle, so every thread corner is exactly 90 degrees.

No. Those conditions define a parallelogram, which may have oblique angles. A non-rectangular parallelogram is a counterexample.

No. Quadrilateral angles total 360 degrees. Three right angles total 270 degrees, leaving 90 degrees for the fourth.

The two triangles contribute 180 + 180 = 360 degrees. Regrouping those six angles at the four vertices proves that every quadrilateral has angle sum 360 degrees.

Draw two parallel lines, then another pair of parallel lines crossing them at an angle other than 90 degrees. Their four intersections form a non-rectangular parallelogram.

Yes. A rectangle has both pairs of opposite sides parallel, so it satisfies the definition of a parallelogram.

The opposite sides are 4 cm and 5 cm. Opposite angles are equal and adjacent angles are supplementary, so the angles are 30, 150, 30, and 150 degrees.

The opposite angle is also x, and each adjacent angle is 180^\circ-x. Thus P = A = x and R = E = 180 degrees - x.

Draw diagonal BD and compare triangles ABD and CDB. Alternate angles from the two pairs of parallel sides are equal and BD is common, giving congruence by AAS; hence AD = CB and AB = CD.

No. The correct correspondence is triangle ABD congruent to triangle CDB. The order ABD-CBD mismatches A with C but B with B, which is inconsistent with the equal angles used.

No. They are equal in special parallelograms such as rectangles, but a general slanted parallelogram can have diagonals of different lengths.

Yes. Opposite sides and alternate angles make the two triangles across the intersection congruent, so each diagonal is divided into two equal parts.

No. From the marked angles and sides, the correct correspondence is triangle AOE congruent to triangle YOS. The alternative order pairs the wrong vertices.

No. They bisect each other, but their intersection angle varies with the parallelogram's shape.

No. Draw two equal adjacent sides at any angle other than 90 degrees, then locate the fourth vertex at the same distance from both free endpoints. The result is a non-square rhombus.

Opposite angles are equal and adjacent angles sum to 180 degrees. The angles are therefore 50, 130, 50, and 130 degrees.

GE = ME and GA = MA because all rhombus sides are equal, while AE is common. Thus the triangles are congruent by SSS.

Rectangles and rhombuses are overlapping subsets of parallelograms. Their overlap is exactly the set of squares.

No. They are equal only in the special case when the rhombus is a square. They always bisect each other, but generally have different lengths.

They intersect at 90 degrees.

GE = ME, GO = OM because the diagonal is bisected, and EO is common. Thus they are congruent by SSS, making the adjacent angles at O equal; as a linear pair, each is 90 degrees.

(i) E = R = 140 and A = P = 40 degrees. (ii) Q = S = 70 and R = 110 degrees. (iii) the diagonal halves the 60-degree acute angles into 30-degree parts, while U = W = 120 degrees. (iv) OEI = AOE = EOI = 20 and A = I = 140 degrees.

Draw a 7 cm diagonal and mark midpoint O. Through O draw a 140-degree line, mark 2.5 cm on both sides for the second diagonal, then join the endpoints in order.

Draw a 5 cm diagonal and its perpendicular bisector at O. Mark 2 cm on both sides of O along the perpendicular, then join the four endpoints.

A square, because the diagonals are equal, bisect each other, and meet at 90 degrees.

A rhombus. The diagonals remain perpendicular and bisect each other, but are no longer equal.

A rhombus. Its outer boundary has four sides of 8 cm, and its opposite angles are equal.

Joining along the 6 cm bases gives a rhombus with four 8 cm sides. Joining along an 8 cm side gives a kite with adjacent pairs 8-8 and 6-6.

Joining equal corresponding sides can produce three kites: the remaining boundary pairs are 6-6 and 9-9, 6-6 and 12-12, or 9-9 and 12-12. Other flipped arrangements can produce general parallelograms when opposite corresponding sides align.

Triangles ABD and CBD are congruent by SSS: AB = CB, AD = CD, and BD is common. Hence BD bisects angles ABC and ADC. The two smaller triangles on either side are congruent, giving AO = OC and equal adjacent angles at O; since they form a linear pair, BD is perpendicular to AC.

Angles on the same leg are supplementary because the bases are parallel: S = 180 degrees - P and R = 180 degrees - Q.

Draw parallel bases UV and XW, placing X and W so UX = VW. Drop perpendiculars XY and WZ to UV. Rectangle XWZY and congruent right triangles UXY and VWZ show U = V; similarly X = W, while angles along each leg are supplementary.

The quadrilateral is a rhombus with four sides of 4 cm. Its angles are 60, 120, 60, and 120 degrees.

Draw PQ = 6 cm and construct its perpendicular bisector at O. On that perpendicular, mark R and S so that OR = OS = 4 cm, then join P-R-Q-S-P. The result is a rhombus, which is also a kite, with diagonals 6 cm and 8 cm.

(i) With bases parallel, R = 75 and S = 45 degrees. (ii) The equal legs make an isosceles trapezium, so A = B = 80 degrees while C = D = 100 degrees.

Rhombuses lie in both the kite and parallelogram sets; squares are also rectangles inside that overlap. A non-square rectangle is not a kite. Every rhombus is a kite, but not every kite is a rhombus.

Angle IOD is 30 degrees. The equal diagonal/isosceles-triangle angle relations in the two rectangles transfer the marked 30-degree angle to the angle at O.

Draw the 6 cm diagonal and construct its perpendicular bisector with a compass. From midpoint O mark 3 cm in both perpendicular directions, then join the four endpoints.

UVWX is a square. If the outer side is s, each inner side is \sqrt{(s/2)^2+(s/2)^2}=s/\sqrt2. The four sides are equal, and adjacent 45-degree parts combine to make each inner angle 90 degrees.

Yes. It is a rhombus, so opposite angles are equal and adjacent angles are supplementary. One 90-degree angle forces all four angles to be 90 degrees, making it a square.

It is a parallelogram. A diagonal gives two SSS-congruent triangles; equal alternate angles then prove both pairs of opposite sides parallel.

Yes. A diagonal divides it into two triangles; counting the interior angles of both triangles gives 180 + 180 = 360 degrees, including the reflex interior angle correctly.

(i) False: the figure may be a non-square rectangle. (ii) True: the fourth angle is also 90 degrees. (iii) True: bisecting diagonals characterise a parallelogram. (iv) False: a kite can have perpendicular diagonals. (v) True. (vi) True: each equal angle is 90 degrees. (vii) False: an isosceles trapezium need not have both pairs of opposite sides parallel.

Opening the sheet produces a symmetric quadrilateral pattern; the central closed figure is a rhombus (a square in the special equal-diagonal fold).

Align the intended vertices or edges while the paper is quarter-folded, make the diagonal crease through the folded corner, and reopen. Reflection across the two earlier folds reproduces the crease symmetrically in all four quarters.

Make the new crease at 45 degrees to both existing perpendicular fold lines and at equal distance from their intersection. Reflections of that crease across the two folds form four equal sides and four right angles.