Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 5 Geometry Ex 5.3 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 5 Geometry Ex 5.3

Question 1.

In the figure, given that ∠1 = ∠2 and ∠3 ≡ ∠4. Prove that ∆ MUG ≡ ∆TUB.

Answer:

Statements | Reasons |

1. In △MUG and △TUG
Mu = TU |
∠3 = ∠4, opposite sides of equal angles |

2. UG = UB | ∠1 = ∠2
Side opposite to equal angles are equal |

3. ∠GUM = ∠BUT | Vertically opposite angle |

4. ∆MUG ≡ ∆TUG | SAS criteria By 1,2 and 3 |

Question 2.

From the figure, prove that ∆SUN ~ ∆RAY.

Answer:

Proof: from the ∆ SUN and ∆RAY

SU = 10

UN = 12

SN = 14

RA = 5

AY = 6

RY = 7

From (1), (2) and (3) we have

The sides are proportional

∴ ∆SUN ~ ∆RAY

Question 3.

The height of a tower is measured by a mirror on the ground at R by which the top of the tower’s reflection is seen. Find the height of the tower. If ∆PQR ~ ∆STR

Answer:

The image and its reflection make similar shapes

∴ ∆PQR ~ ∆STR

⇒ \(\frac{h}{8}=\frac{60}{10}\)

h = \(\frac{60}{10}\) × 8

= 48 feet

∴ Height of the tower = 48 feet.

Question 4.

Find the length of the support cable required to support the tower with the floor.

Answer:

From the figure, by Pythagoras theorem,

x^{2} = 20^{2} + 15^{2}

= 400 + 225 = 625

x^{2} = 25^{2} ⇒ x = 25ft.

∴ The length of the support cable required to support the tower with the floor is 25ft.

Question 5.

Rithika buys an LED TV which has a 25 inches screen. If its height is 7 inches, how wide is the screen? Her TV cabinet is 20 inches wide. Will the TV fit into the cabinet? Give reason.

Answer:

Take the sides of a right angled triangle ∆ABC as

a = 7 inches

b = 25 inches

c = ?

By Pythagoras theorem,

b^{2} = a^{2} + c^{2}

25^{2} = 7^{2} + c^{2}

⇒ c^{2} = 25^{2} – 7^{2} = 625 – 49 = 576

∴ c^{2} = 24^{2}

⇒ c = 24 inches

∴ Width of TV cabinet is 20 inches which is lesser than the width of the screen ie.24 inches.

∴ The TV will not fit into the cabinet.

Challenging Problems

Question 6.

In the figure, ∠TMA ≡∠IAM and ∠TAM ≡ ∠IMA. P is the midpoint of MI and N is the midpoint of AI. Prove that Δ PIN ~ Δ ATM.

Answer:

proof:

Question 7.

In the figure, if ∠FEG ≡ ∠1 then, prove that DG^{2} = DE.DF.

Answer:

Proof:

Question 8.

The diagonals of the rhombus is 12 cm and 16 cm. Find its perimeter. (Hint: the diagonals of rhombus bisect each other at right angles).

Answer:

Here AO = CO = 8cm

BO = DO = 6cm

(∴the diagonals of rhombus bisect each other at right angles)

∴ In ∆ AOB, AB^{2} = AO^{2} + OB^{2}

= 8^{2} + 6^{2} = 64 + 36

= 100 = 10^{2}

∴ AB = 10

Since it is a rhombus, all the four sides are equal.

AB = BC = CD = DA

∴ Its Perimeter = 10 + 10 + 10 + 10 = 40 cm

Question 9.

In the figure, find AR.

Answer:

∆ AFI, ∆ FRI are right triangles.

By Pythagoras theorem,

AF^{2} = AI^{2} – FI^{2}

= 25^{2} – 15^{2}

= 625 – 225 = 400 = 20^{2}

∴ AF = 20ft.

FR^{2} = RI^{2} – FI^{2}

= 17^{2} – 15^{2} = 289 – 225 = 64 = 8^{2}

FR = 8ft.

∴ AR = AF + FR

= 20 + 8 = 28 ft.

Question 10.

In ∆DEF, DN, EO, FM are medians and point P is the centroid. Find the following.

(i) IF DE = 44, then DM = ?

(ii) IFPD=12, then PN= ?

(iii) IfDO = 8, then PD = ?

(iv) IF 0E = 36 then EP = ?

Answer:

Given DN, EO, FM are medians.

∴ FN = EN

DO = FO

EM = DM

(i) If DE = 44,then

DM = \(\frac{44}{2}\) = 22

DM = 22

(ii) If PD = 12,PN = ?

\(\frac{P D}{P N}=\frac{2}{1}\)

\(\frac{12}{\mathrm{PN}}=\frac{2}{1}\) ⇒ PN = \(\frac{12}{2}\) = 6.

PN = 6

(iii) If DO = 8, then

FD = DO + OF

= 8 + 8

FD = 16

(iv) If OE = 36

PE = 24