Samacheer Class 8 Maths - Geometry

Statements
Reasons
1. CI = CO
∵ CIP ≡ COP, by CPCTC
2. IP = OP
By CPCTC
3. CP = CP
By CPCTC
4. Also HI = HO
CPCTC ∆HIP ≡ HOP given
5. IP = OP
By CPCTC and (4)
6. ∴ IP ≡ OP
By (2) and (4)

Statements
Reasons
1. OD = ED
D is the midpoint OE (given)
2. DC = DC
Common side
3. ∠CDE = ∠CDO = 90°
Linear pair and given ∠CDE = 90°
4. △ODC ≡ △EDC
By RHS criteria

Given △EAT ≡ △BUN
∴ Corresponding angles are equal
∴ ∠E = ∠B …… (1)
∠A = ∠U …… (2)
∠T = ∠N ……. (3)
∠E = x°
∠A = 2x°
Sum of three angles of a triangle = 180°
In △ EAT, x + 2x + ∠T = 180°
∠T = 180° – (x° + 2x°)
∠T = 180° – 3x°
Also in △BUN
(x + 40)° + x° + ∠U = 180°
x + 40° + x + ∠U = 180°
2x° + 40° + ∠U = 180°
∠U = 180° – 2x – 40° = 140° – 2x°
Now by (2)
∠A = ∠U
2x = 140° – 2x°
2x + 2x = 140°
4x = 140°
x = \(\frac{140}{4}\) = 35°
∠A = 2x° = 2 × 35° = 70°
∠N = x + 40° = 35° + 40° = 75°
∴ ∠T = ∠N = 75°
∠E = ∠8 = 35°
∠A = ∠U = 70°

Statements
Reasons
1. ∠CUB = ∠CBU
∵ In △CUB, CU = CB
2. ∠CUB = ∠CAB
∵ UB || AT, Corresponding angle if CA is the transversal.
3. ∠CBU = ∠CTA
CT is transversal UB || AT,
Corresponding angle commom angle.
4. ∠UCB = ∠ACT
Common angle
5. △CUB ~ △CAT
By AAA criteria
6. CA = CT
∵ ∠CAT = ∠CTA
7. Also △CAT is isoceles
By 1, 2 and 3 and sides opposite to equal angles are equal.

Objective Type Questions

Take a = 8 b = 15 and c = 17
Now a 2 + b 2 = 8 2 + 15 2 = 64 + 225 = 289
172 = 289 = c 2
∴ a 2 + b 2 = c 2
By the converse of Pythagoras theorem, the triangle with given measures is a right angled triangle.
∴ Ans: yes

(ii) 12, 13, 15
(ii) 12, 13. 15
Take a = 12,b = 13 and c = 15
Now a 2 + b 2 = 12 2 + 13 2 = 144 + 169 = 313
15 2 = 225 ≠ 313
By the converse of Pythagoras theorem, the triangle with given measures is not a right angled triangle.
∴ Ans: No.
(iii) 30, 40, 50
Take a = 30, b = 40 and c = 50
Now a 2 + b 2 = 30 2 + 40 2 = 900 + 1600 = 2500
C 2 = 50 2 = 2500
∴ a 2 + b 2 = c 2
By the converse of Pythagoras theorem, the triangle with given measures is a right
angled triangle.
∴ Ans: yes
(iv) 9, 40, 41
Take a = 9, b = 40 and c = 41
Now a 2 + b 2 = 9 2 + 40 2 = 81 + 1600 = 1681
c 2 = 41 2 = 1681
∴ a 2 + b 2 = c 2
By the converse of Pythagoras theorem, the triangle with given measures is a right
angled triangle.
∴ Ans: yes

(v) 24, 45, 51
Take a = 24,b = 45 and c = 51
Now a 2 + b 2 = 24 2 + 45 2 = 576 + 2025 = 2601
c 2 = 51 2 = 2601
∴ a 2 + b 2 = c 2
By the converse of Pyhtagoreas theorem, the triangle with given measure is a right angled triangle.
∴ Ans: yes

In an isosceles triangle the altitude dives its base into two equal parts.
Now in the figure, ∆ABC is an isosceles triangle with AD as its height
In the figure, AD is the altitude and ∆ABD is a right triangle.
By Pythagoras theorem,
AB 2 = AD 2 + BD 2
⇒ AD 2 = AB 2 – BD 2
= 13 2 – 12 2 = 169 – 144 = 25
AD 2 = 25 = 5 2
Height: AD = 5cm

Given A is the midpoint of YZ.
∴ ZA = AY
G is the centroid of XYZ centroid divides each median in a ratio 2 : 1 ⇒ XG : GA = 2:1
\(\frac{\mathrm{XG}}{\mathrm{GA}}=\frac{2}{1}\)
\(\frac{\mathrm{XG}}{3}=\frac{2}{1}\)
XG = 2 × 3
XG = 6 cm
XA = XG + GA = 6 + 3 ⇒ XA = 9cm

Since I is the incentre of AXYZ
∠IYZ = 30° ⇒ ∠IYX = 30°
∠IZY = 40° ⇒ ∠IZX = 40°
∴ ∠XYZ = ∠XYI + ∠IYZ = 30° + 30°
∠XYZ = 60°

∠XYZ = 80°
By angle sum property of a triangle
∠XZY + ∠XYZ + ∠YXZ = 180°
80° + 60° + ∠YXZ = 180°
140° + ∠YXZ = 180°
∠YXZ = 180° – 140°
∠YXZ = 40°

Objective Type Questions

1. A. 28 cm
2. B. 20 cm
3. C. 24 cm
4. D. 21 cm

(B) 20 cm
Hint:
Side take a = 12 cm
b = 16cm
The hypotenuse c 2 = a 2 + b 2
= 12 2 + 16 2
= 144 + 256
c 2 = 400 ⇒ c = 20 cm

1. A. 609 cm 2
2. B. 580 cm 2
3. C. 420 cm 2
4. D. 210 cm 2

(C) 420 cm 2

length = 21 cm
diagonal = 29 cm
By the converse of Pythagoras theorem,
AB 2 + BC 2 = AC 2
21 2 + x 2 = 29 2
x 2 = 841 – 441 400 = 20 2
x = 20 cm
Now area of the rectangle = length × breadth.
ie AB × BC = 21 cm × 20 cm = 420 cm 2

1. A. 25, 36, 59
2. B. 10, 24, 26
3. C. 36, 39, 45
4. D. 20, 48, 52

(D) 20,48,52
Hint:
The sides of a right angled triangle are in the ratio 5 : 12 : 13
Take the three sides as 5a, 12a, 13a
Its perimeter is 5a + 12a + 13a = 30a
It is given that 30a = 120 units
a = 4 units
∴ the sides 5a = 5 × 4 = 20 units
12a = 12 × 4 = 48 units
13a = 13 × 4 = 52 units
Posted in Class 8 on September 18, 2024 September 19, 2024
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Take the sides of a right angled triangle ∆ABC as
a = 7 inches
b = 25 inches
c = ?
By Pythagoras theorem,
b 2 = a 2 + c 2
25 2 = 7 2 + c 2
⇒ c 2 = 25 2 – 7 2 = 625 – 49 = 576
∴ c 2 = 24 2
⇒ c = 24 inches
∴ Width of TV cabinet is 20 inches which is lesser than the width of the screen ie.24 inches.
∴ The TV will not fit into the cabinet.

Challenging Problems

Here AO = CO = 8cm
BO = DO = 6cm
(∴the diagonals of rhombus bisect each other at right angles)
∴ In ∆ AOB, AB 2 = AO 2 + OB 2
= 8 2 + 6 2 = 64 + 36
= 100 = 10 2
∴ AB = 10
Since it is a rhombus, all the four sides are equal.
AB = BC = CD = DA
∴ Its Perimeter = 10 + 10 + 10 + 10 = 40 cm

Given ABCD, AB = 5 cm, BC = 4.5 cm, CD = 3.8 cm, DA = 4.4 cm and AC = 6.2 cm.

Steps:
Draw a line segment AB = 5 cm
With A and B as centers drawn arcs of radii 6.2 cm and 4.5cm respectively and let them cut at C.
Joined AC and BC.
With A and C as centrers drawn arcs of radii 4.4cm and 3.8 cm respectively and let them at D.
Joined AD and CD.
ABCD is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral ABCD = \(\frac { 1 }{ 2 }\) × d × (h 1 1 + h 2 ) sq. units
= \(\frac { 1 }{ 2 }\) × 6.2 × (2.6 + 3.6)cm 2 = 3.1 × 6.2 = 19.22 cm 2

Given PLAY, PL = 7cm, LA = 6cm, AY = 6cm, PA = 8cm and LY = 7cm

Steps:
Drawn a line segment PL = 7 cm
With P and L as centers, drawn arcs of radii 8 cm and 6 cm respectively, let them cut at A.
Joined PA and LA.
With L and A as centers, drawn arcs of radii 7 cm and 6 cm respectively and let them cut at Y.
Joined LY, PY and AY.
PLAY is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral PLAY = \(\frac { 1 }{ 2 }\) × d × (h 1 + h 2 ) sq. units = \(\frac { 1 }{ 2 }\) × 8 × (5.1 + 1.4) cm 2
\(\frac { 1 }{ 2 }\) × 8 × 6.5 cm 2 = 26 cm 2
Area of the quadrilateral = 26 cm 2

Given PQ = QR = 3.5 cm, RS = 5.2 cm, SP = 5.3 cm and ∠Q = 120°

Steps:
Draw a line segment PQ = 3.5 cm
Made ∠Q = 120°. Drawn the ray QX.
With Q as centre drawn an arc of radius 3.5 cm. Let it cut the ray QX at R.
With R and P as centres drawn arcs of radii 5.2cm and 5.5 cm respectively and let them cut at S.
Joined PS and RS.
PQRS is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral PQRS = \(\frac { 1 }{ 2 }\) × d × (h 1 + h 2 ) sq. units
= \(\frac { 1 }{ 2 }\) × 6 × (4.3 + 17)cm 2
= 3 × 6cm 2
= 18 cm 2
Area of the quadrilateral PQRS = 18 cm 2

Given MI = 3.6 cm, ND = 4 cm, MD = 4 cm, ∠M = 50° and ∠D = 100°

Steps:
Drawn a line segment MI = 3.6 cm
At M on MI made an angle ∠IMX = 500
Drawn an arc with center M and radius 4 cm let it cut MX it D
At D on DM made an angle ∠MDY = 100°
With I as center drawn an arc of radius 4 cm, let it cut DY at N.
Joined DN and IN.
MIND is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral MIND = \(\frac { 1 }{ 2 }\) × d × (h 1 + h 2 ) sq. units = \(\frac { 1 }{ 2 }\) × 3.2 × (2.7 + 33) cm 2
= \(\frac { 1 }{ 2 }\) × 3.2 × 6 cm 2 = 9.6 cm 2
Area of the quadrilateral = 9.6 cm 2

AG = 4.5 cm, GR = 3.8 cm, ∠A = 60°, ∠G = 110° and ∠R = 90°.

Steps:
Draw a line segment AG = 4.5 cm
At G on AG made ∠AGX = 110°
With G as centre drawn an arc of radius 3.8 cm let it cut GX at R.
At R on GR made ∠GRZ = 90°
At A on AG made ∠GAY = 90°
AY and RZ meet at I.
AGRI is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral AGRI = \(\frac { 1 }{ 2 }\) × d × (h 1 + h 2 ) sq. units = × 6.8 × (2.9 + 2.4) cm 2
= \(\frac { 1 }{ 2 }\) × 6.8 × 5.3 × cm 2
Area of the quadrilateral = 18.02 cm 2

Given AI = 6 cm, IM = 5 cm
AM = 9 cm, and \(\overline{\mathrm{AI}}\) || \(\overline{\mathrm{SM}}\)
MS = 6.5 cm
Rough Diagram


Construction:
Steps:
Draw a line segment AI = 6cm.
With A and I as centres, draw arcs of radii 9 cm and 5 cm respectively and let them cut at M
Join AM and IM.
Draw MX parallel to AI
With M as centre, draw an arc of radius 6.5 cm cutting MX at S.
Join AS AIMS is the required trapezium.
Calculation of Area
Area of the trapezium AIMS = \(\frac { 1 }{ 2 }\) × h × (a + b) sq.units
= \(\frac { 1 }{ 2 }\) × 4.6 × (6 + 6.5)
= \(\frac { 1 }{ 2 }\) × 4.6 × 12.5
= 28.75 Sq.cm

Given: In the trapezium CUTE,
CU = 7cm, ∠UCE = 80°,
CE = 6cm,TE = 5cm and \(\overline{\mathrm{CD}}\) || \(\overline{\mathrm{ET}}\)
Rough Diagram


Construction:
Steps:
Draw a line segment CU = 7cm.
Construct an angle ∠UCE = 80° at C
With C as centre, draw an arc of radius 6 cm cutting CY at E
Draw EX parallel to CU
With E as centre, draw an arc 7 cm of radius 5 cm cutting EX at T
Join UT. CUTE is the required trapezium.
Calculation of area:
Area of the trapezium CUTE = \(\frac { 1 }{ 2 }\) × h × (a + b) sq. units
= \(\frac { 1 }{ 2 }\) × 5.9 × (7 + 5)sq. units
= 35.4 sq.cm

Given: In the trapezium ARMY
AR = 7cm,RM = 6.5cm,
∠RAY = 100° and ARM 60°, \(\overline{\mathrm{AR}}\) || \(\overline{\mathrm{YM}}\)
Rough Diagram


Construction:
Steps:
Draw a line segment AR = 7 cm.
Construct an angle ∠RAX = 100° at A
Construct an angle ∠ARN = 60° at R
With R as centre, draw an arc of radius 6.5 cm cutting RN at M
Draw MY parallel to AR
ARMY is the required trapezium.
Calcualtion of Area:
Area of the trapezium ARMY = \(\frac { 1 }{ 2 }\) × h × (a + b) sq. units
= \(\frac { 1 }{ 2 }\) × 5.6 × (7 + 4.8) sq. units
= \(\frac { 1 }{ 2 }\) × 5.6 × 1.18
= 33.04 sq.cm

Given: In the trapezium CITY,
CI = 7cm,IT = 5.5cm, TY = 4cm
YC = 6cm, and \(\overline{\mathrm{CI}}\) || \(\overline{\mathrm{YT}}\)
Rough Diagram


Construction:
Steps:
Draw a line segment CI = 7 cm.
Mark a point D on CI such that CD = 4cm
With D and I as centres, draw arcs of radii 6 cm and 5.5 cm respectively.
Let them cut at T. Join DT and IT.
With C as centre, draw an arc of radius 6 cm.
Draw TY parallel to Cl. Let the line cut the previous arc at Y.
Join CY. CITY is the required trapezium.
Construction of area:
Area of the trapezium CITY = \(\frac { 1 }{ 2 }\) × h × (a + b) sq. units
= \(\frac { 1 }{ 2 }\) × 5.5 × (7 + 4)sq.units
= \(\frac { 1 }{ 2 }\) × 5.5 × 11
= 30.25 sq.cm
Posted in Class 8 on September 20, 2024 September 21, 2024
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Given : In the Parallelogram ARTS,
AR = 6 cm, RT = 5 cm, and ∠ART = 70°
Rough Diagram


Construction:
Steps:
Draw a line segment AR = 6 cm.
Make an angle ∠ART = 70° at R on AR
With R as centre, draw an arc of radius 5 cm cutting RX at T
Draw a line TY parallel to AR through T.
With T as centre, draw an arc of radius 6 cm cutting TY at S. Join AS
ARTS is the required parallelogram.
Calculation of area:
Area of the parallelogram ARTS = b × h sq. units
= 6 × 4.7 = 28.2 sq.cm

Given : In the parallelogram CAMP,
CA = 6 cm, AP = 8 cm, and CP = 5.5cm
Rough Diagram


Construction:
Steps:
Draw a line segment CA = 6 cm.
With C as centre, draw an arc of length 5.5 cm
With A as centre, draw an arc of length 8 cm
Mark the intersecting point of these two arcs as P
Draw a line PX parallel to CA
With P as centre draw an arc of radius 6 cm cutting PX at M. Join AM
CAMP is the required parallelogram.
Calculation of area:
Area of the Parallelogram CAMP = b × h sq. units
= 6 × 5.5 = 33 sq.cm

Given: In the parallelogram EARN,
ER = 10 cm, AN = 7 cm, and LEOA = 1100
Where \(\overline{\mathrm{ER}}\) and \(\overline{\mathrm{AN}}\) intersect at 0
Rough diagram


Construction:
Steps:
Draw a line segment PX. Mark a point O on PX
Make an angle ∠EOA = 1100 on PX at O
Draw arcs of radius 3.5 cm with O as centre on either side of PX. Cutting YZ on A and N
With A as centre, draw an arc of radius 10 cm, cutting PX at E. Join AE
Draw a line parallel to AE at N cutting PX at R. Join EN and AR
EARN is the required parallelogram
Calculation of area:
Area of the Parallelogram EARN = b × h sq. units
= 10 × 5.5 = 55 sq.cm

Given : In the parallelogram GAIN,
GA = 7.5 cm, GI = 9 cm, and ∠GAI = 100°


Construction:
Steps:
Draw a line segment GA = 7.5 cm.
Make an angle GAI = 100° at A.
With G as centre, draw an arc of radius 9 cm cutting AX at I. Join GI.
Draw a line IY parallel to GA through I.
With I as centre, draw an arc of radius 7.5 cm on IY cutting at N. Join GN
GAIN is the required parallelogram.
Construction of area:
Area of the Parallelogram GAIN = b × h sq. units
= 7.5 × 39 = 29.25 sq. cm

Given FA = 6 cm and FC = 8 cm
Rough Diagram


Steps:
Drawn a line segment FA = 6 cm.
With F and A as centres, drawn arcs of radii 8 cm and 6 cm respectively and let them cut at C.
Joined FC and AC.
With F and C as centres, drawn arcs of radius 6 cm each and let them cut at E.
Joined FE and EC.
FACE is the required rhombus.
Calculation of Area :
Area of the rhombus = \(\frac { 1 }{ 2 }\) × d 1 × d 2 sq.units = \(\frac { 1 }{ 2 }\) × 8 × 9 sq.units = 36 cm

(ii) CAKE, CA = 5 cm and ∠A = 65°
Given CA = 5 cm and ∠A = 65°
Rough Diagram


Steps:
Drawn a line segment CA = 5 cm.
At A on AC, made ∠CAX = 65°
With A as centre, drawn arc of radius 5 cm. Let it cut AX at K.
With K and C as centres, drawn arcs of radius 5 cm each and let them cut at E.
Joined KE and CE.
CAKE is the required rhombus.
Calculation of Area:
Area of the rhombus = \(\frac { 1 }{ 2 }\) × d 1 × d 2 sq.units
= \(\frac { 1 }{ 2 }\) × 54 × 85cm 2
= 22.95 cm 2

(iii) LUCK, LC = 7.8 cm and UK = 6 cm
Given LC = 7.8 cm and UK = 6 cm
Rough Diagram


Steps:
Drawn a line segment LC = 7.8 cm.
Drawn the perpendicular bisector XY to LC. Let it cut LC at ‘O’
With O as centres, drawn arc of radius 3 cm on either side of O which cut OX at K and OY at U.
Joined LU, UC, CK and LK.
UCK is the required rhombus.
Calculation of Area:
Area of the rhombus = \(\frac { 1 }{ 2 }\) × d 1 × d 2 sq.units
= \(\frac { 1 }{ 2 }\) × 7.8 × 6 cm 2 = 23.4 cm 2

(iv) PARK, PR = 9 cm and ∠P = 70°
Given PR = 9 cm and ∠P = 70°
Rough Diagram


Steps:
Drawn a line segment PR = 9 cm.
At P, made ∠RPX ∠RPY = 35° on either side of PR.
At R, made ∠PRQ = ∠PRS = 35° on either side of PR
Let PX and RQ cut at A and PY and RS at K.
PARK is the required rhombus
Constructon of Area:
Area of the rhombus = \(\frac { 1 }{ 2 }\) × d 1 × d 2 sq.units = \(\frac { 1 }{ 2 }\) × 9 × 6.2 cm 2
= 27.9 cm 2

Given HA = 7cm and AN = 4 cm
Rough Diagram


Steps:
Drawn a line segment HA = 7 cm.
At H, constructed HX ⊥ HA.
With H as centre, drawn an arc of radius 4 cm and let it cut at HX at D.
With A and D as centres, drawn arcs of radii 4 cm and 7 cm respectively and let them cut at N.
Joined AN and DN.
HAND is the required rectangle.
calculation of’ area :
Area of the rectangle HAND = l × b sq.units
= 7 × 4 cm 2
= 28 cm 2

(ii) LAND, LA = 8cm and AD = 10 cm
Given LA = 8cm and AD = 10 cm


Sleps :
Drawn a line segment LA = 8 cm.
At L, constructed LX ⊥ LA.
With A as centre, drawn an arc of radius 10 cm and let it cut at LX at D.
With A as centre and LD as radius drawn an arc. Also with D as centre and LA as radius drawn another arc. Let then cut at N.
Joined DN and AN.
LAND is the required rectangle.
Calcualtion of arca :
Area of the rectangle LAND = l × b sq.units
= 8 × 5.8 cm 2
= 46.4 cm 2

Given side = 6.5 cm
Rough diagram


Steps:
Drawn a line segment EA = 6.5 cm.
At E, constructed EX⊥ EA.
With E as centre, drawn an arc of radius 6.5 cm and let it cut EX at T.
With A and T as centre drawn an arc of radius 6.5 cm each and let them cut at S.
Joined TS and AS.
EAST is the required square.
Calcualtion of Area:
Area of the square EAST = a 2 sq.units
= 6.5 × 6.5 cm 2
= 42.25 cm 2

(ii) WEST, WS = 7.5 cm
Given diagonal = 7.5 cm
Rough Diagram


Steps:
Drawn a line segment WS = 7.5 cm.
Drawn the perpendicular bisector XY to WS. Let it bisect BS at O.
With O as centre, drawn an arc of radius 3.7 cm on either side of O which cut OX at T and OY at E
Joined BE, ES, ST and BT.
WEST is the required square.
Calculation of Area:
Area of the square WEST = a 2 sq.units
= 5.3 × 53 cm 2
= 28.09 cm 2 .
Posted in Class 8 on September 20, 2024 September 21, 2024
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30
Hints:
= \(\frac { 1 }{ 2 }\) × h × (a + b) sq. units
= \(\frac { 1 }{ 2 }\) × 5 × (7 + 5) = \(\frac { 1 }{ 2 }\) × 5 × 12 = 30 sq. cm

10 cm
Hint:
Area of the trapezium = \(\frac { 1 }{ 2 }\) × h (a + b)
60 = \(\frac { 1 }{ 2 }\) × h × (12)
∴ h = \(\frac{60 \times 2}{12}\) = 10cm

Square, rectangle, parallelogram, rhombus.
(b) In which the diagonals are perpendicular bisectors of each other.
Rhombus and square.
(c) Which have diagonals of different lengths.
Parallelogram and Rhombus
(d) Which have equal diagonals.
Rectangle, square.
(e) Which have parallel opposite sides.
Square, Rectangle. Rhombus, parallelogram.
(f) In which opposite angles are equal.
Square, rectangle. rhombus, parallelogram