Ex 10.4 & 10.5 NCERT Solutions: Practical Geometry- 2

# NCERT Solutions for Class 8 Maths Chapter 10 - Ex 10.4 and 10.5 Practical Geometry- 2

Exercise 10.4

Question 1:

Construct ΔABC, given m∠A = 60:, m∠B = 30° and AB = 5,8 cm.

To construct: ΔABC, given m∠A = 60:, m∠B = 30° and AB = 5,8 cm.

Steps of construction:

(a) Draw a line segment AB = 5.8 cm.
(b) At point A, draw an angle Z YAB = 6ti= with the help of a compass.
(c) At point B, draw Z XBA = 30" with the help of a compass.
(d) AY and BX intersect at the point C.
It is the required triangle ABC

Question 2:

Construct Δ PQR if PQ = 5 cm, m∠PQR = 105° and m∠.QRP = 40°.

Given:

m∠ PQR = 105° and m∠ QRP = 40°
We know that sum of angles of a triangle is 180°.
∴ m∠PQR + m∠QRP + m∠ QPR = 180°
⇒ 105° + 40° + m∠QPR= 180°
⇒ 145° + m∠QPR= 180°
⇒  m∠QPR= 180° - 145°
⇒ m∠QPR = 35°

To construct: ΔPQR where m∠P = 35°, m∠Q = 105° and PQ = 5 cm.

Steps of construction:

(a) Draw a line segment PQ = 5 cm,
(b) At point P, draw ∠ XPQ = 35° with the help of protractor.
(c) At point Q, draw ∠ YQP = 105° with the help of protractor.
(d) XP and YQ intersect at point R.
It is the required triangle PQR.

Question 3:

Examine whether you can construct ΔDEF such that EF = 7.2 cm, m∠ E= 110° and m∠F

Given: In ΔDEF, m∠E = 110° and m∠T = 80°
Using angle sum property of triangle
∠D + ∠E + ∠F = 180°
⇒∠D + 110°+ 80° = 180°
⇒ ∠D + 190° = 180°
⇒ ∠D = 180° - 190° = -10°

Which is not possible.

Exercise 10.5

Question 1:

A right angled triangle PQR where m∠Q = 9ff, QR= 8 cm and PQ = 10 cm.

To construct: Construct the right angled Δ PQR, where m∠Q = 90", QR = 8 cm and PR = 10 cm.

Steps of construction:

(a)  Draw a line segment QR = 8 cm,
(b) At point Q, draw QX ⊥ QR,
(c) Taking R as centre, draw an arc of radius 10 cm.
(d) This arc cuts QX at point P.
(e) foin PQ.
It is the required right-angled triangle PQR,

Question 2:

Construct a right-angled triangle whose hypotenuse is 6 cm long and one the legs is 4 cm long.

To construct: A right-angled triangle DEF where DF = 6 cm and EF = 4 cm

Steps of construction:

(a) Draw a line segment EF = 4 cm.
(b) At point Q, draw EX ⊥ EF.
(c) Taking F as centre and radius 6 cm, draw an arc. (Hypotenuse)
(d) This arc cuts the EX at point D.
(e) Join DF.
It is the required right-angled triangle DEF.

Question 3:

Construct an isosceles right angled triangle ABC, where m∠ACB = 90° and AC = 6 cm.

To construct:  An isosceles right angled triangle ABC where m∠ C = 90% AC = BC = 6 cm.

Steps of construction:

(a) Draw a line segment AC = 6 cm.
(b) At point C, draw XC ⊥ CA.
(c) Taking C as centre and radius 6 cm, draw an arc.
(d) This arc cuts CX at point B.
(e) Join BA.
It is the required isosceles right-angled triangle ABC.

Miscellaneous Questions

Questions:

Below are given the measures of certain sides and angles of triangles. Identify those which cannot be constructed and say why you cannot construct them. Construct rest of the triangle.

 Triangle Given measurements 1. ΔABC m∠A = 85° ; m∠B = 115° ; AB = 5 cm 2. ΔPQR m∠ Q = 30° ; m∠R = 60° ; QR = 4.7 cm 3. ΔABC m∠ A = 70° ; m∠B = 50°; AC = 3 cm 4. ΔLMN m∠L = 60° ; m∠N = 120° ; LM = 5 cm 5. ΔABC BC = 2 cm; AB = 4 cm; AC = 2 cm 6. ΔPQR PQ = 3.5 cm; QR = 4 cm; PR = 3.5 cm 7. ΔXYZ XY = 3 cm; YZ = 4 cm; XZ = 5 cm 3. ΔDEF DE = 4.5 cm; EF = 5.5 cm; DF = 4 cm

In ΔABC, m∠A = 85o , m∠B = 115o , AB = 5 cm

Construction of ΔABC is not possible because m∠A = 85o + m∠B = 200o, and we know that the sum of angles of a triangle should be 180o

To construct: ΔPQR where m∠Q = 30o , m∠R = 60o and QR = 4.7 cm.

Steps of construction:

(a) Draw a line segment QR = 4.7 cm.
(b) At point Q, draw ∠XQR = 30' with the help of a compass.
(c) At point R, draw ∠YRQ = 60' with the help of a compass.
(d) QX and RY intersect at point P.
It is the required triangle PQR.

We know that the sum of angles of a triangle is 180°.
m∠ A + m∠B + m∠C = 180°
∴ 70° + 50° + m∠C = 180°
⇒  120° + m∠C = 180°
⇒  m∠C = 180° - 120°
⇒  m∠C = 60°

To construct: ΔABC where m∠A = 70°, m∠C = 60° and AC = 3 cm.

Steps of construction:

(a) Draw a line segment AC = 3 cm.
(b) At point C, draw ∠YCA = 60°.
(c) At point A, draw ∠XAC = 70°
(d) Rays XA and YC intersect at point B

It is the required triangle ABC.

In ΔLMN , m∠L - 60° m∠N = 120°, LM = 5 cm
This ΔLMN is not possible to construct because m∠L + m∠N = 60° + 120° = 180° which forms a linear pair.

Δ ABC, BC = 2 cm, AB = 4 cm and AC = 2 cm

This Δ ABC is not possible to construct because the condition is

Sum of lengths of two sides of a triangle should be greater than the third side.

AB< BC + AC
⇒ 4 < 2 + 2
⇒ 4 = 4,

To construct: ΔPQR where PQ = 3.5 cm, QR = 4 cm and PR = 3.5 cm

Steps of construction:

(a) Draw a line segment QR = 4 cm.
(b) Taking Q as centre and radius 3.5 cm, draw an arc.
(c) Similarly, taking R as centre and radius 3.5 cm, draw another arc which intersects the first arc at point P.
it is the required triangle QPR.

To construct: A triangle whose sides are XY = 3 cm, YZ = 4 cm and XZ = 5 cm.

Steps of construction:

(a) Draw a line segment ZY = 4 cm.
(b) Taking Z as center and radius 5 cm, draw an arc.
(c) Taking Y as centre and radius 3 cm, draw another arc.
(d) Both arcs intersect at point X.
It is the required triangle XYZ.

To construct: A triangle DEF whose sides are DE = 4.5 cm, EF = 5.5 cm and DF = 4 cm.

Steps of construction:

(a) Draw a line segment EF = 5.5 cm.
(b) Taking E as center and radius 4.5 cm, draw an arc.
(c) Taking F as center and radius 4 cm, draw another arc which intersects the first arc at point D.

It is the required triangle DEF.

The document NCERT Solutions for Class 8 Maths Chapter 10 - Ex 10.4 and 10.5 Practical Geometry- 2 is a part of the Class 7 Course NCERT Textbooks & Solutions for Class 7.
All you need of Class 7 at this link: Class 7

417 docs

## FAQs on NCERT Solutions for Class 8 Maths Chapter 10 - Ex 10.4 and 10.5 Practical Geometry- 2

 1. What are the basic tools required for practical geometry?
Ans. The basic tools required for practical geometry are a ruler, compass, protractor, divider, pencil, and eraser.
 2. How can I construct an equilateral triangle using a compass and ruler?
Ans. To construct an equilateral triangle using a compass and ruler, follow these steps: 1. Draw a line segment of any length. 2. With the compass, open to the same length as the line segment, draw arcs from both ends of the line segment. 3. Where the arcs intersect, mark the point. 4. Using the ruler, draw lines connecting the marked point with both ends of the line segment. 5. The triangle formed by these lines is an equilateral triangle.
 3. What is the difference between a line segment and a ray in geometry?
Ans. In geometry, a line segment is a part of a line with two endpoints, whereas a ray is a part of a line with one endpoint and extends infinitely in one direction.
 4. How can I construct a perpendicular bisector of a line segment?
Ans. To construct a perpendicular bisector of a line segment, follow these steps: 1. Draw a line segment. 2. With the compass, open to more than half the length of the line segment, draw arcs from both ends of the line segment. 3. Without changing the compass width, draw arcs from the other end of the line segment. 4. The point where the arcs intersect is the midpoint of the line segment. 5. Connect the midpoint with the endpoints of the line segment using a ruler. 6. The line drawn is the perpendicular bisector of the line segment.
 5. How can I construct a triangle when its base, altitude, and one angle are given?
Ans. To construct a triangle when its base, altitude, and one angle are given, follow these steps: 1. Draw the given base of the triangle. 2. From one endpoint of the base, draw a perpendicular line (altitude) using a protractor. 3. Measure the given angle on the altitude and mark the point. 4. Join the marked point with the other endpoint of the base. 5. The triangle formed is the required triangle.

## NCERT Textbooks & Solutions for Class 7

417 docs

### Up next

 Explore Courses for Class 7 exam

### Top Courses for Class 7

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;