Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 5 Geometry Ex 5.1 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 5 Geometry Ex 5.1

Question 1.

Fill in the blanks with the correct term from the given list.

(in proportion, similar, corresponding, congruent, shape, area, equal)

(i) Corresponding sides of similar triangles are _________ .

Answer:

in proportion

(ii) Similar triangles have the same ________ but not necessarily the same size.

Answer:

Shape

(iii) In any triangle _______ sides are opposite to equal angles.

Answer:

equal

(iv) The symbol is used to represent _________ triangles.

Answer:

congruent

(v) The symbol ~ is used to represent _________ triangles.

Answer:

similar

Question 2.

In the figure, ∠CIP ≡ ∠COP and ∠HIP ≡ ∠HOP. Prove that IP ≡ OP.

Answer:

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) |

Question 3.

In the given figure, AC ≡ AD and ∠CBD ≡ ∠DEC. Prove that ∆ BCF ≡ ∆ EDF.

Answer:

Statements | Reasons |

1. ∠BCF = ∠EFD | Vertically opposite angles |

2. ∠CBD = ∠DEC | Angles on the same base given |

3. ∠BCF = ∠EDF | Remaining angles of ∆BCF and ∆EDF |

4. ∆BCF ≡ ∆EDF | By (1) and (2) AAA criteria |

Question 4.

In the given figure, △ BCD is isosceles with base BD and ∠BAE ≡∠DEA. Prove that AB ≡ ED.

Answer:

Statements | Reasons |

1. ∠BAE ≡ ∠DEA | Given |

2. AC = EC | By (1) sides opposite to equal angles are equal |

3. BC = DC | Given BCD is isosceles with base BD |

4. AC – BC = EC – DC | 2 – 3 |

5. AB ≡ ED | By 4 |

Midpoint Calculator is a free online tool that displays the midpoint of the line segment.

Question 5.

In the given figure, D is the midpoint of OE and ∠CDE = 90°. Prove that △ODC ≡ △EDC

Answer:

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 |

Question 6.

Is △PRQ ≡ △QSP? Why?

Answer:

In △PRQ and △PSQ

∠PRQ = ∠PSQ = 90° given

PR = QS = 3 cm given

PQ = PQ = 5 cm common

It satisfies RHS criteria

∴ △PRQ congruent to △QSP.

Question 7.

From the given figure, prove that △ABC ~ △EDF

Answer:

From the △ABC, AB = AC

It is an isosceles triangle

Angles opposite to equal sides are equal

∴ ∠B = ∠C = 65°

∴ ∠B + ∠C = 65° + 65°

= 130°

We know that sum of three angles is a triangle = 180°

∠A + ∠B + ∠C = 180°

∠A + 130° = 180°

∠A = 180° – 130°

∠A = 50°

From △EDF, ∠E = 50°

∴ Sum of Remaining angles = 180° – 50° = 130°

DE = FD

∴ ∠D = ∠F

From △ABC and △EDF ∴ △D = \(\frac{130}{2}\) = 65°

∠A = ∠E = 50°

∠B = ∠D = 65°

∠C = ∠F = 65°

∴ By AAA criteria △EDF ~ △ABC

Question 8.

In the given figure YH || TE. Prove that △WHY ~ △WET and also find HE and TE.

Answer:

Statements | Reasons |

1. ∠EWT = ∠HWY | Common angle |

2. ∠ETW = ∠HYW | Since YH || TE, corresponding angles |

3. ∠WET = ∠WHY | Since YH || TE corresponding angles |

4. △WHY ~ △WET | By AAA criteria |

Also △ WHY ~ △WET

∴ Corresponding sides are proportionated

⇒ 6+HE = \(\frac{6}{4}\) × 16

⇒ 6 + HE = 24

∴ HE = 24 – 6

HE = 18

Again \(\frac{4}{\mathrm{ET}}=\frac{4}{16}\)

ET = \(\frac{4}{4}\)

ET = 16

Question 9.

In the given figure, if △EAT ~ △BUN, find the measure of all angles.

Answer:

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°

Question 10.

In the given figure, UB || AT and CU ≡ CB Prove that △CUB ~ △CAT and hence △CAT is isosceles.

Answer:

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

Question 11.

Two similar triangles will always have _______ angles

(A) acute

(B) obtuse

(C) right

(D) matching

Answer:

(D) matching

Question 12.

If in triangles PQR and XYZ, \(\frac{\mathrm{PQ}}{\mathrm{XY}}=\frac{\mathrm{QR}}{\mathrm{ZX}}\) then they will be similar if

(A) ∠Q = ∠Y

(B) ∠P = ∠X

(C) ∠Q = ∠X

(D) ∠P ≡∠Z

Ans:

(C) ∠Q = ∠X

Question 13.

A flag pole 15 m high casts a shadow of 3m at 10 a.m. The shadow cast by a building at the same time is 18.6m. The height of the building is

(A) 90 m

(B) 91 m

(C) 92 m

(D) 93 m

Answer:

(D) 93 m

Question 14.

If ∆ABC – ∆PQR in which ∠A = 53° and ∠Q = 77°, then ∠R is

(A) 50°

(B) 60°

(C) 70°

(D) 80°

Answer:

(A) 50°

Question 15.

In the figure, which of the following statements is true?

(A) AB = BD

(B) BD < CD

(C) AC = CD

(D) BC = CD

Answer:

(C) AC = CD