# Frank Learning Maths Class 5 Solutions Chapter 10

## Frank Learning Maths Class 5 Solutions Chapter 10 Geometry

Welcome to NCTB Solution. Here with this post we are going to help 5th class students for the Solutions of Frank Learning Maths Class 5 Book, Chapter 10 Geometry. Here students can easily find step by step solutions of all the problems for Geometry. Here students will find solutions for Exercise 10.1, 10.2, 10.3, 10.4, 10.5, 10.6, 10.7, 10.8 and 10.9. Exercise wise proper solutions for every problems. 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.

Geometry Exercise 10.1 Solution

Question number – (1)

Solution :

Figure – (a) parallel.

Figure – (b) intersecting.

Figure – (c) intersecting.

Figure – (d) parallel.

Figure – (e) intersecting.

Figure – (f) perpendicular.

Figure – (g) perpendicular.

Figure – (h) intersecting.

Question no – (2)

Solution :

Figure – (a) it’s a line

Figure – (b) it’s a ray

Figure – (c) it’s a line

Figure – (d) it’s a ray

Question number – (3)

Solution :

In finger – (a)

line → lm, ms, jk

In finger – (b)

Line → qo, or

Rays → pq, rs

In finger – (c)

Rays → ab, cd

In finger – (d) ray → bc, qp

Line → ab, al, lm, ms

In finger – (e)

Line → km, mp

Rays → kl, mn, ps

Question number – (5)

Solution :

(a) Line segment with A as endpoint are AB, AC.

(b) In the given figure the line is BC.

(c) In the given figure the ray is DH.

Geometry Exercise 10.2 Solution

Question number – (2)

Solution :

Figure – (a) Angles in two different ways – ∠AOB, ∠AOB

Figure – (b) Angles in two different ways – ∠PQR, ∠RQP

Figure – (c) Angles in two different ways – ∠XYZ, ∠ZYX

Figure – (d) Angles in two different ways – ∠LMN, ∠NML

Question number – (3)

Solution :

Figure – (a) 3 ANGLES

→ ∠AOB, ∠BOC, ∠AOC

Figure – (b) 6 ANGLES

→ ∠POQ, ∠QOR, ∠ROS, ∠POR, ∠QOS, ∠POS

Question number – (4)

Solution :

(a) Name the angles in the given figure.

= ∠ABC ON ∠EBF

(b) Name the points lying in the interior of the angle.

= X,Y, Z POINTS

(c) Name the points lying in the exterior of the angle.

= L, N, M POINTS

(d) Name the points lying on the angle.

= A, E, B, F, C POINTS

Geometry Exercise 10.3 Solution

Question number – (1)

Solution :

 Measure of the Angle (a) Acute (<900) → 720, 450, 380, 40, 68, 890, 550, 120, 800 (b) Right (900) → 900 (c) Obtuse (<900) → 1250, 1750, 1360, 1790, 1640, 1500, 1300 (d) Straight (1800) → 1800

Question number – (2)

Solution :

Figure – (a) is a acute angle.

Figure – (b) is a obtuse angle.

Figure – (c) is a acute angle.

Figure – (d) is a obtuse angle.

Figure – (e) is a right angle.

Figure – (f) is a obtuse angle.

Geometry Exercise 10.4 Solution

Question no – (1)

Solution :

(a) A right angle always measures 900.

(b) A straight angle always measures 1800.

(c) An Obtuse angle is greater than a Right angle but less than a Straight angle.

(d) An 1500 angle is an Obtuse angle.

(e) The angle where the two perpendicular line segments met to form is called Vertex.

Geometry Exercise 10.5 Solution

Question no – (4)

Solution :

(a) A right-angled triangle can have only 2 acute angles.

(b) In scalene triangle, there is no equal side.

So, a scalene triangle have no equal side.

(c) A triangle can have only 1 obtuse angle.

(d) A triangle can have 1 right angle

(e) An Isosceles triangle can have 2 equal side.

Geometry Exercise 10.6 Solution

Question no – (1)

Solution :

(a) If a triangle has one of its angles obtuse, then the other two angles will be Acute angle.

(b) In a right triangle, the sum of other two angles is always 90°

(c) All angles of an equilateral triangle are of measure 60°

(d) An acute triangle has all its angles smaller than 90°

Question number – (2)

Solution :

(a) Given angles,

∠A = 60°,

∠B = 65°,

∠C = 55°

(60 + 65 + 55)

= 180° (possible)

(b) Given angles,

∠A = 90°,

∠B = 50°,

∠C = 50°

(90 + 50 + 50)

= 1900 ≠ 1800 ( not possible)

(c) Given angles,

∠P = 40°,

∠Q = 40°,

∠R = 40°

(40 + 40 + 40)

= 1200 ≠ 1800 (not possible)

(d) Given angles,

∠X = 145°,

∠Y = 15°,

∠Z = 20°

(145 + 15 + 20)

= 1800 (possible)

Question no – (4)

Solution :

(a) From the question we get,

∠A = 55°,

∠B = 72°,

∠C = ?

Now,

= ∠C = 180 – (72 + 55)

= 180 – 127

= 53°

So, the measure of ∠C is 53°

(b) From the question we get,

∠P = 37°,

∠Q = 82°,

∠R = ?

Now,

∠R = 180 – (82 + 37)

= 180 – 119

= 61°

So, the measures of ∠R is 61°

(c) In the given question we get,

∠X = 105°,

∠Y = 30°,

∠Z = ?

Now,

∴ ∠Z = 180 – (105 + 30)

= 180 – 135

= 45°

Thus, the measure of ∠Z is 45°

(d) From the question we get,

∠D = 86°,

∠E = 50°,

∠F = ?

Now,

∴ ∠F = 180 – (86 + 50)

= 180 – 136

= 44°

So, the measure of ∠F is 44°

Geometry Exercise 10.7 Solution

Question no – (1)

Solution :

(a) A rhombus is Always a parallelogram.

(b) A square is Always a quadrilateral.

(c) A rhombus is Sometime a square.

(d) A trapezium Never has all four sides equal.

(e) Quadrilaterals with two pairs of parallel sides are Always parallelograms.

(f) Polygons with all right angles are Sometime squares.

(g) A polygon with three equal sides is Always a triangle.

(h) The sum of all the angles of a triangle is Never 170°.

(i) The sum of all the angles of a quadrilateral is Always 360°.

Question no – (2)

Solution :

(a) A parallelogram can never be a square.

= False

(b) A square is always a rectangle.

= True

(c) A rhombus can never be a square.

= False

(d) A trapezium is a parallelogram.

= False

(e) A parallelogram has only one set of parallel sides.

= False

(f) A rectangle has four right angles.

= True

(g) A rhombus always has four equal sides.

= True

(h) The sum of the interior angles of a quadrilateral is 180°

= False

Geometry Exercise 10.8 Solution

Question no – (1)

Solution :

In the given adjoining figure :

(a) The centre of the circle is O

(b) An arc of the circle is APQ

(c) The diameter of the circle is AB

(d) A chord of the circle is XY

(e) The three radii of the circle is CO, AO, and OB

(f) The biggest chord of this circle is AB

Question no – (3)

Solution :

(a) In a circle, the distance is same from the Centre to all points on the circle.

(b) A line segment joining the centre of a circle and any point on the circle is called its Radius.

(c) Diameter = Two times times the radius.

(d) Radius = 1/2 of the diameter.

(e) The Diameter is the longest chord of a circle.

Question no – (4)

Solution :

(a) All radii in the circle are of equal length.

= True.

(b) All diameters in a circle are of equal length.

= True.

(c) All chords in the circle are of equal length.

= False

(d) All arcs in the circle are of equal length.

= False

(e) Diameter is double the radius.

= True.

Geometry Exercise 10.9 Solution

Question no – (1)

Solution :

As we know that,

radii = 1/2 × diameter so,

(a) 8 cm

= 8/2

= 4 cm

So, the radii of the circle is 4 cm.

(b) 10 cm

= 10/2

= 5 cm

So, the radii of the circle is 5 cm.

(c) 22 cm

= 22/2

= 11 cm

So, the radii of the circle is 11 cm.

(d) 48 cm

= 48/2

= 24 cm

So, the radii of the circle is 24 cm.

Question no – (2)

Solution :

As we all know that,

(a) 2.5 cm

= 2.5 × 2

= 5 cm

So, the diameter of the circle is 5 cm.

(b) 3 cm

= 3 ×2

= 6 cm

So, the diameter of the circle is 6 cm.

(c) 4.5 cm

= 4.5 × 2

= 9 cm

So, the diameter of the circle is 9 cm.

(d) 7 cm

= 7 ×2

= 14 cm

So, the diameter of the circle is 14 cm.

Question number – (3)

Solution :

(a) Given, diameter is 6 cm.

= 3 cm

Now, Circumference,

= 2 × π × 3 …(we know, Circumference = 2 π r)

= 18.84

≃ 19 cm

So, the circumference is 19 cm.

(b) Given, diameter is 7 cm.

Now, Circumference,

= 2 × π × 7/2 …(we know, Circumference = 2 π r)

= 21.99

≃ 22

So, the circumference is 22 (approx.)

(c) Given, diameter is 10 cm.

Now, Circumference,

= 2 × π × 10/2 …(we know, Circumference = 2 π r)

= 31.41

≃ 31

So, the circumference is 31 cm (approx.)

(d) Given, diameter is 15 cm.

Now,

Circumference,

= 2 × π × 15/2 …(we know, Circumference = 2 π r)

= 47.12

≃ 47

So, the circumference is 47 cm (approx.)

Question no – (4)

Solution :

As we know that,

Circumference = 2 π r

(a) Given radii = 3.5 cm

= r = 3.5 cm
Circumference,

= 2 π × 3.5

= 21.99

≃ 22

So, the circumference of the circle is 22 cm (approx.)

(b) Given radii = 6 cm

Circumference,

= 2 π × 6

= 37.69

≃ 38

So, the circumference of the circle is 38 cm (approx.)

(c) Given radii = 7.5 cm

Circumference,

= 2 π × 7.5

≃ 47

So, the circumference of the circle is 47 cm (approx.)

(d) Given radii = 9 cm

Circumference,

= 2 π × 9

= 56.54

≃ 56

So, the circumference of the circle is 56 cm (approx.)

Previous Chapter Solution :

Updated: June 7, 2023 — 5:39 am