# Frank Learning Maths Class 6 Solutions Chapter 11

## Frank Learning Maths Class 6 Solutions Chapter 11 Basic Geometrical Ideas

Welcome to NCTB Solution. Here with this post we are going to help 6th class students for the Solutions of Frank Learning Maths Class 6 Book, Chapter 11 Basic Geometrical Ideas. Here students can easily find step by step solutions of all the problems for Basic Geometrical Ideas.

Here students will find solutions for Exercise 11.1, 11.2, 11.3, 11.4, 11.5 and 11.6. Exercise wise proper solutions. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum.

Basic Geometrical Ideas Exercise 11.1 Solution

Question no – (2)

Solution :

(a) Name the rays starting from point A.

(b) How many line segments are there? Name them.

= AP, AB, AQ, PQ, PB, AD, AC, RD, RC, RS, DC, CS, DS, BC, BC

(c) Name the parallel lines.

= PQ and RS

(d) Which lines are not parallel?

(e) Name any four pairs of intersecting lines and their points of intersection.

= (PQ, AC – A), (PQ, BC – B), (PQ, AD – A) (RS, AD – A)

(f) Name the lines containing point C.

= (RS, AC, BC)

(g) Name the lines on which point B lies.

= (PQ, BC)

(h) Name the lines passing through point A.

Question no – (3)

Solution :

(a) Infinitely many lines can be drawn passing through one point.

(b) Only one lines can be drawn passing through two point.

Question no – (4)

Solution :

(a) AE + EC = AC

(b) AC – EC = AE

(c) BD – BE = ED

(d) BD DE = BE

Question no – (5)

Solution :

All the line segments are AB, AC, AD, AE, BC, BD, BE, CD CE, DE

The total number of line segments formed,

10 segments.

Question no – (6)

Solution :

Since, x and y are midpoints of PQ and RR

PX = 1/2 PQ = 1/2 × 15 = 7.5 cm

PY = 1/2 PR = 1/2 × 20 – 10 cm

XQ = 1/2 PQ = 1/2 × 15 = 7.5

YR = 1/2 PR = 1/2 × 20 = 10 CM

Question no – (7)

Solution :

(a) Name the marked points in the figure.

= A, B, C, D, E, F, G, H

(b) How many line segments are there? Name them.

= AB, BC, CD, DA, AE, BF, CG, DH, EF, FG, GH, HE

= Total 12 segments

(c) Name the line segments meeting at point F.

= EF , BF, GF

(d) Name the line segments meeting at point B.

= AB, BC, BF

(e) Name the groups of any four parallel line segments.

= (AB, DC) (AD, BC) (AE, BF) (EF, HG)

Question no – (8)

Solution :

(a) 3 collinear points,

= A, C, E

(b) 3 non-collinear points,

= A, E, B

(c) 3 concurrent lines,

Question no – (9)

Solution :

(a) Name the plane parallel to ABCD.

= EFGH

(b) Name the plane parallel to BCGF.

(c) Name any one plane intersecting ABFE.

= FGHE

Try These 11 (i) :

Question no – (1)

Solution :

Unlimited number of rays can be drawn with a given initial point.

Question no – (2)

Solution :

The intersection of two lines can have only one point.

Question no – (3)

Solution :

The correct option is alternative – (c)

An infinite number of lines can be drawn passing through a given point.

Basic Geometrical Ideas Exercise 11.2 Solution

Question no – (1)

Solution :

Figure – (a) is Closed.

Figure – (b) is Closed.

Figure – (c) is Open.

Figure – (d) is closed.

Figure – (e) is Open.

Figure – (f) is Open.

Figure – (g) is Closed.

Figure – (h) is Open.

Question no – (2)

Solution :

In the question figure (a) and (d) are simple closed curves.

Question no – (4)

Solution :

In the given figures the (a), (c) and (g) figures are polygon.

Question no – (5)

Solution :

A polygon having at least six sides and mark at least three points in the interior, on the boundary and in the exterior of the polygon.

The figure : Question no – (7)

Solution :

(a) A polygon is a simple closed curve formed by more than 2 line segments.

(b) A polygon having three line segments is called a Triangle.

(c) A polygon having four line segments is called a Quadrilateral.

Question no – (9)

Solution :

(a) A polygon having 8 sides, (b) A polygon having 10 sides, Basic Geometrical Ideas Exercise 11.4 Solution

Question no – (1)

Solution :

In the given figures, the (b) and (c) figures are triangles.

Question no – (2)

Solution :

Figure – (a) ABC, BCA, CAB

Figure – (b) = POR, QRP, RPQ

Figure – (c) = LMN, MNL, NLM,

Figure – (d) = DEF, EFD, FDE

Question no – (3)

Solution :

Points – A, B, C

Sides – AB, BC, CA

Question no – (4)

Solution :

Figure – (a) = ABC, ABD, ADC

Figure – (b) = PSU, SQT, TUR, STU, PQR

Figure – (c) = HOG, HOE, EOF, FOG, HEF, EFG, FGH, GHE

Figure – (d) = ABC, ABD, ABE, ACD, ACE, ADE

Question no – (5)

Solution :

(a) P Q and R on AXYZ. (b) A, B and C in the interior of AXYZ. (c) S, T and U in the exterior of AXYZ. Question no – (6)

Solution :

(a) how many line segments are there? Name them.

= Six segments → PO, PS, PR, QS, SR, QR

(b) how many angles are there? Name them.

= 7 Angles → ∠QPS, ∠SPR, ∠QPR, ∠PQS, ∠PSQ, ∠PSR, ∠PRS

Question no – (7)

Solution :

The triangles which have ∠R as common in the figure provided in question 6.

= △PSR and △PQR

Question no – (8)

Solution :

(a) J

= KL

(b) K

= JL

(c) L

= JK

Question no – (9)

Solution :

(a) LM

= N

(b) MN

= L

(c) LN

= M

Question no – (10)

Solution :

(a) Name any eight triangles.

= △ABC, △BDC, BDE, △DEF, △DFG, △DGC, △EDG, △EGC

(b) name the line segment which is parallel to side BC.

= EG

(c) which two line segments are parallel to side DF?

= BE, GG

(d) which line segment is parallel to side EC?

= AB

(e) name the points lying on △DEG.

= D, E, F, G

Basic Geometrical Ideas Exercise 11.5 Solution

Question no – (1)

Solution : = (DA, AB) (AB, BC) (BC, CD) (CD, DA)

(b) opposite sides

(d) opposite angles

Question no – (2)

Solution :

Difference between convex and concave quadrilaterals are :

 Subject Convex concave Angle Each angle is less than 1800 One angle is more than 1800 Intersection of diagonals Intersecting point of diagonals lies inside the quadrilateral Intersecting point of diagonals lies outside the quadrilateral

Question no – (3)

Solution :

(a) in the interior of quadrilateral ABCD.

= K, L, M

(b) in the exterior of quadrilateral ABCD.

= X, W, Z

= A, R, B, Q, C, P, D, S

= A, R, B, Q, C, P, D, S, K, L, M

Question no – (5)

Solution :

No, because in any quadrilateral, none of the sides intersect each other.

Question no – (6)

Solution :

Since, 185° > 180° therefore the quadrilateral is Concave.

Question no – (8)

Solution :

(a) The diameters of a circle intersect at Centre.

(b) The polygon having the least number of sides is a 3.

(c) The opposite edges of a notebook give an example of Parallel lines.

(d) The area of a rectangle having sides as 5 cm and 4 cm is 20.

(e) The perimeter of a triangle having sides as 15 cm, 12 cm and 20 cm is 47.

(f) Diameter = 2 × radius.

Basic Geometrical Ideas Exercise 11.6 Solution

Question no – (1)

Solution :

(a) Chord :

The line segment which joins two points on the circle, is called chord.

(b) Diameter :

The chord which touches the centre, is called Diameter.

(c) Sector :

A region in the interior of a circle, formed by an arc and two radii joining the end points of the arc, is called a sector.

(d) Segment :

A chord of a circle divides the circular region into two parts. Each of them is called Segments.

(e) Semicircle :

A diameter of a circle divides the circle into two parts. Each of them is called Semicircle.

(f) Circumference :

The perimeter of a circle is known as its Circumference.

Question no – (2)

Solution :

Required figure :

(a) centre

= o is centre

(c) diameter

= BC is diameter

(d) sector

= Are AC Forms a sector.

(e) segment

= Arc PRQ forms a segment.

(f) are

= RPQ is an arc.

Question no – (3)

Solution :

(a) interior

= ‘o’, p

(b) exterior

= R

(c) on the circle

= Q

Question no – (4)

Solution :

Question no – (5)

Solution :

(a) the centre of the circle

= O

= OA, OB, OC, OD

(c) two diameters

= AB, CD

(d) four chords

(e) points in the circular region

= A, B, C, D, O, R, Q, P

(f) a sector

(g) an arc

= AFC

Question no – (6)

Solution :

As per the given question,

A circle with centre O. Draw a chord AB in the same circle and shade its major segment and minor segment differently.

Required figure :

Question no – (7)

Solution :

(a) is every diameter of a circle also a chord?

= Yes

(b) Is every chord of a circle also a diameter?

= No

(c) Is the diameter a part of the semicircle?

= Yes

(d) Is the centre a part of the circle?

= Yes

(e) Is the radius a part of the circle?

= Yes

Question no – (8)

Solution :

As we know that diameter = 2 × radius

(a) 4 cm

= 2 × 4

= 8 cm

(b) 5 cm

= 2 × 5

= 10

(c) 7.5 cm

= 2 × 7.5

= 15

(d) 14.5 cm

= 2 × 14.5

= 29

Question no – (9)

Solution :

Here we solve it using the formula, r = d/2

(a) 6 cm

= r = d/2

= 6/2

= 3 cm

(b) 8 cm

= 8/2

= 4

(c) 18 cm

= 18/2

= 9

(d) 8.5 cm

= 8.5/2

= 4.25

Question no – (11)

Solution :

Akhila’s mother baked a circular cake. She fixed a ribbon around it for decoration.

= It denotes the circumference of the cake.

Question no – (12)

Solution :

Neha put a thread around a CD to measure its length. Then she measured the length of the thread with a ruler

= This length is known as Circumference.

Previous Chapter Solution :

Updated: June 6, 2023 — 7:26 am