NCERT Solutions for Class 8 Math Chapter 4 - Practical Geometry

Answer:

We cannot construct a unique quadrilateral because we can construct $∆$ABD but not $∆$BCD.
Similarly, neither we can construct $∆$ACD nor $∆$ABC.

Answer:

No, any five measurements cannot determine a quadrilateral uniquely.

Answer:

Yes, we can draw a parallelogram BATS because the opposite sides of a parallelogram are equal in length.

Answer:

Yes, we can draw a rhombus ZEAL because all the four sides of a rhombus are equal in length.

Answer:

The student could not draw it because PL + PY < LY.

Answer:

Given: Four sides of a quadrilateral and one diagonal, as given below:
AB = 4.5 cm, BC = 5.5 cm, CD = 4 cm,
AD = 6 cm, AC = 7 cm.
First, we draw a rough sketch to represent the given data.
Steps of Construction:
(i) Take a line segment AC = 7 cm.
(ii) With A as centre and radius AD = 6 cm draw an arc.
(iii) With C as centre and radius CD = 4 cm draw another arc cutting first arc at D.
(iv) Join AD and CD.
(v) With A as centre and radius AB = 4.5 cm, draw an arc.
(vi) With C as centre and radius BC = 5.5 cm, draw another arc intersecting the first arc at B.
(vii) Join AB and BC.
Then, ABCD is the required quadrilateral.

Answer:

Given: Four sides of a quadrilateral and one diagonal as:
JU = 3.5 cm, UM = 4 cm, MP = 5 cm,
PJ = 4.5 cm, PU = 6.5 cm.
Steps of Construction:
First, draw a rough sketch to represent the above data.

(i) Take a line segment PU = 6.5 cm.
(ii) With P as centre and radius PJ = 4.5 cm, draw an arc.
(iii) With U as centre and radius UJ = 3.5 cm, draw another arc intersecting the first arc at J.
(iv) Join PJ and JU.
(v) With P as centre and radius PM = 5 cm, draw an arc.
(vi) With U as centre and radius UM = 4 cm, drawn another arc intersecting the first arc at M.
(vii) Join PM and UM.
Then, JUMP is the required quadrilateral.

Answer:

Given: Adjacent sides of a parallelogram and one diagonal as:
OR = 6 cm, RE = 4.5 cm and EO = 7.5 cm.
Steps of Construction:
(i) Take a line segment OE = 7.5 cm
(ii) With O as centre and radius OR = 6 cm draw an arc.
(iii) With E as centre and radius ER = 4.5 cm draw another arc cutting first arc at R.
(iv) Join OR and ER.
(v) With O as centre and radius OM = 4.5 cm, draw an arc.
(vi) With E as centre and radius EM = 6 cm draw another arc cutting first arc at M.
(vii) Join OM and EM.
Then, MORE is the required parallelogram.

Answer:

Let the Rhombus to be constructed is BEST,
BE = ES = ST = TB = 4.5 cm and diagonal ET = 6 cm.
First we draw a rough sketch of the given data.
Steps of Construction:
(i) Take a line segment ET = 6 cm.
(ii) With E as centre and radius ES = EB = 4.5 cm, draw two arcs on the either sides of ET.
(iii) With T as centre and radius TS = TB = 4.5 cm, draw arcs intersecting the arcs in step (ii) at S and B respectively.
(iv) Join ES, ST, BE and BT. Then, BEST is the required rhombus.

Answer:

No, we cannot construct a quadrilateral PQRS.
Because in $∆$PQS, PQ + SQ < OS (the sum of the length of any two sides of a triangle must always be greater than the third side).

Answer:

Given: Three sides and two diagonals of the quadrilateral LIFT.
i.e., LI = 4 cm, IF = 3 cm, TL = 2.5 cm
LF = 4.5 cm, IT = 4 cm.
Steps of Construction:
(i) Take a line segment LI = 4 cm.
(ii) With I as centre and radius 3 cm. draw an arc.
(iii) With L as centre and radius 4.5 cm draw an arc, which intersects the previous arc at F.
(iv) Join LF and IF.
(v) With L as centre and radius 2.5 cm, draw an arc.
(vi) With I as centre and radius 4 cm, draw an arc, which cuts the previous arc at T.
(vii) Join LI, IT and FT.
∴ LIFT is the required quadrilateral.

Answer:

Given: Three sides and two diagonals of the quadrilateral GOLD.
i.e., OL = 7.5 cm,
GL = 6 cm, GD = 6 cm, LD = 5 cm,
OD = 10 cm.
Steps of Construction:
(i) Take a line segment OL = 7.5 cm.
(ii) With O as centre and radius 10 cm, draw an arc.
(iii) With L as centre and radius 5 cm, draw an arc, which intersects the previous arc at D.
(iv) Join OD and LD.
(v) With D as centre, and radius 6 cm, draw an arc.
(vi) With L as centre, and radius 6 cm, draw an arc, which intersects the previous arc at G.
(vii) Join GL, GD and OG.
∴ GOLD is the required quadrilateral.

Answer:

Given: Two diagonals of the rhombus BEND.
i.e., BN = 5.6 cm, DE = 6.5 cm.
Steps of Construction:
(i) Take a line segment DE = 6.5 cm.
(ii) Bisect the line segment DE.
(iii) With O as centre, draw arcs

Above and below the line DE, which intersect the bisector at point N and D respectively.
(iv) Join BE, EN, ND and BD.
∴ BEND is the required Rhombus.

Answer:

Yes, we can construct the quadrilateral MIST.
Because the sum of all the four angles is equal to 360°.

Answer:

No, we cannot construct the quadrilateral PLAN.
Because the sum of all the four angles is greater than 360°.

Answer:

Yes, we still need the measure of angles between adjacent sides to construct the parallelogram.

Answer:

Given: Two adjacent sides and three angles of the quadrilateral MORE.
i.e., MO = 6 cm, OR = 4.5 cm,
$\angle$M = 60°, $\angle$O = 105°, $\angle$R = 105°.

Steps of Construction:
(i) Draw MO = 6 cm.
(ii) At O; draw an angle $\angle$XOM = 105°.
(iii) From ray OX, cut OR = 4.5 cm.
(iv) At R; draw an angle $\angle$YRO = 105°.
(v) At M, draw an angle $\angle$ZMO = 60° and let, rays RY and MZ intersect each other at E.
∴ MORE is the required quadrilateral.

Answer:

Given: Two adjacent sides and three angles of the quadrilateral PLAN.
i.e., PL = 4 cm, LA = 6.5 cm, $\angle$P = 90°,
$\angle$A = 110°, $\angle$N = 85°.
Steps of Construction:
(i) Draw LA = 6.5 cm.
(ii) At A, draw an
angle $\angle$XAL = 110°.
(iii) At L, draw an angle $\angle$YLA = 75°.
∴ By angle sum property;
$\angle$P + $\angle$L + $\angle$A + $\angle$N = 360°
90° + $\angle$L + 110° + 85° = 360°
$\angle$L + 285° = 360°
⇒ $\angle$L = 360° – 285° = 75°
(iv) From ray LY, cut LP = 4 cm.
(v) At P draw angle of 90°, which intersects the ray AX at N.
∴ PLAN is the required quadrilateral.

Answer:

Given: Two adjacent sides and one angle of the parallelogram.
i.e., HE = 5 cm, EA = 6 cm and $\angle$R = 85°.
Steps of Construction:
(i) Draw EA = 6 cm.
(ii) At A draw an
angle $\angle$XAE = 95°.
From ray AX cut AR = 5 cm.
(iii) At E draw an angle $\angle$YEA = 85°. From ray EY cut HE = 5 cm.
(iv) Join HR.
∴ HEAR is the required parallelogram.

Answer:

Given: Two adjacent sides of a rectangle OKAY.
i.e., OK = 7 cm and KA = 5 cm.
As each angle of a rectangle is of 90°.
Steps of Construction:
(i) Draw OK = 7 cm.
(ii) At points O and K draw angles of 90° each i.e., $\angle$XKO and $\angle$POK.
(iii) From XK cut, AK =5 cm.
(iv) From PO cut, OY = 5 cm.
(v) Join AY.
∴ OKAY is the required rectangle.

Answer:

Yes, we could have first drawn AB or CD.

Answer:

(i) We used five measurements to draw a quadrilaterals.
(ii) We cannot draw a quadrilaterals using five measurements.
Few more examples,
(a) Draw a quadrilateral ABCD in which AB = 6.5 cm, BC = 4 cm, CD = 4.5 cm, $\angle$B = 60° and AB|| CD.
(b) Draw another quadrilateral ABCD in which AB = 7.5 cm, AD = 5 cm, BC = 5 cm, $\angle$A = 90° and BC|| AD.

Answer:

Given: Three sides and two included angles of the quadrilateral DEAR are given i.e., DE = 4 cm, EA = 5 cm, AR = 4.5 cm, $\angle$E = 60°, $\angle$A = 90° Steps of Construction:
(i) Draw a line segment DE = 4 cm.
(ii) At point E, make an $\angle$DEX = 60° and cut an arc of length 5 cm from E, which intersect EX at A.

(iii) At point A, make an $\angle$EAY = 90°, and cut an arc of length 4.5 cm from A, which intersects AY at R.
(iv) Join DR.
∴ DEAR is the required quadrilateral.

Answer:

Given: Three sides and two included angles of the quadrilateral TRUE are given, i.e., TR = 3.5 cm, RU = 3 cm, UE = 4 cm, $\angle$R = 75° and $\angle$U = 120°.
Steps of Construction:
(i) Draw a line segment RU = 3 cm.
(ii) At point R, make an $\angle$XRU = 75°, and cut an arc of length 3.5 cm from R, which intersects RX at T.
(iii) At point U, make an $\angle$RUY = 120° and cut an arc of length 4 cm from U, which intersects UY at E.

(iv) Join TE.
∴ TRUE is the required quadrilateral.

Answer:

Steps of Construction:
(i) Draw a line segment PQ of given length.
(ii) At Q, draw a ray QX such that $\angle$PQX = 90°
(iii) From ray QX, cut QR of given length.
(iv) Through P, draw a ray PY parallel to QR.
(v) Through R, draw a ray RZ parallel to QP to meet the ray to step (iv) at S. Thus, PQRS is the required quadrilateral.

Answer:

Steps of Construction:
Given AE = EY = 4 cm
AY = 8 cm
∴ AE + EY = AY = 8 cm
Which is not possible because the sum of the length of any two sides of a triangle must be greater than the third side.
∴ Kite EASY cannot be constructed.

Answer:

Steps of Construction:

(i) Take a line segment RE = 5.1 cm.
(ii) At R draw $\angle$XRE = 90$°$ and at E draw $\angle$YER = 90$°$.
(iii) From RX cut off RD = 5.1 cm and also from EY cut off EA = 5.1 cm.
(iv) Join AD.
Then READ is the required square.

Answer:

Steps of Construction:

(i) Draw AC = 6.4 cm
(ii) Construct the perpendicular bisector of AC. Let it meet AC at O.
(iii) Cut off

lengths of either side of the perpendicular bisector.
(iv) Join AB, BC, CD and DA. Then, ABCD is the required rhombus.

Answer:

Steps of Construction:
(i) Take a line segment AB = 5 cm.
(ii) At A, draw $\angle$XAB = 90$°$ and also at B draw $\angle$YBA = 90$°$.
(iii) From AX, cut off AD = 4 cm and also from BY, cut off BC = 4 cm.
(iv) Join DC.
Then ABCD is the the required rectangle.

Answer:

Construction of a || gm with two adjacent sides only is not possible because we need the measurement of included angle or any diagonal between the sides.