The document NCERT Solutions(Part - 2) - Practical Geometry Class 7 Notes | EduRev is a part of the Class 7 Course Mathematics (Maths) Class 7.

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**Exercise 10.4 **

**Question 1: **

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

**Answer 1: **

**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°.**

**Answer 2: **

**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 = 80°. Justify your answer.**

**Answer 3: **

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.**

**Answer 1: **

**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. **

**Answer 2: **

**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.

**Answer 3: **

**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 | |||

| ΔABC | m∠A | m∠B | AB |

2. | ΔPQR |
| m∠R | QR = 4.7 cm |

3. | ΔABC |
| m∠B = 50° | AC |

4. | ΔLMN | m∠L | m∠N | LM |

5. | ΔABC | BC | AB | AC |

| ΔPQR | PQ | QR | PR |

7. | ΔXYZ | XY | YZ | XZ |

3. | ΔDEF | DE | EF | DF |

**Answer 1: **

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

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

**Answer 2: **

**To construct:** ΔPQR where m∠Q = 30^{o} , m∠R = 60^{o} 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.

**Answer 3:**

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.

**Answer 4: **

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.

**Answer 5: **

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,

**Answer 6: **

**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.

**Answer 7: **

**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.

**Answer 8: **

**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.

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