8th Maths : Chapter 5 : Geometry : Miscellaneous Practice problems, Challenge Problems, Text Book Back Numerical problems, Exercises Questions with Answers, Solution

**Exercise
5.3**

**Miscellaneous
Practice Problems**

** **

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

**Solution: **

**Proof:**

__Statements:____ Reasons__

**1. In ΔMUG and ΔTUB**

**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. ΔMUG ≡ ΔTUB : **SAS criteria By 1,
2 and 3

** **

**2. From the figure, prove that ∆ SUN**

**Solution: **

**Proof: **from the ΔSUN and ΔRAY

SU = 10

UN = 12

SN = 14

RA = 5

AY = 6

RY = 7

We have

SU/RA = 10/5 = 2/1 …….(1)

UN/AY = 12/6 = 2/1 …….(2)

SN/RY = 14/7 = 2/1 …….(3)

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

SU/RA = UN/AY = SN/RY = 2/1

The sides are proportional

∴ ΔSUN ~ Δ RAY

** **

**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**

**Solution: **

The image and its reflection make similar shapes

Δ PQR ~ Δ STR

PQ/ST = QR/TR =PR/SR

= 48 feet

∴ Height of the tower = 48 feet.

** **

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

**Solution:**

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 25 ft.

** **

**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.**

**Solution:**

Take the sides of a right angled triangle ΔABC 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**

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

**Solution: **

**Proof:**

__Statements:____ Reasons__

**1. ****∠****TMA = ****∠****IAM**

**∠****TAM = ****∠****IMA : **Reasons: given

**2. ****∠****ATM = ****∠****TIM : **Reasons: Remaining angle

By angle sum property

**3. IP = PM ****⇒**** = IP/PM = 1**

**In = NA ****⇒**** = IN/NA = 1 : **Reasons: P is the midpoint of IM and N is the
midpoint of IA

**4. IP / PM = IN /
NA :** Reasons: By 3

**5. PN || MA :** Reasons: By 4

**6. ****∠****IPN = ****∠****IMN**

**∠****INP = ****∠****IAM :** Reasons: By 5

**7. In ΔPIN and ΔATM**

**(i) ****∠****IPN = ****∠****TAM**

**(ii) ****∠****INP = ****∠****TMA**

**(iii) ****∠****ATM = ****∠****PIN : **Reasons: By 1, 2 and 6

**8. ΔPIN ~ ΔATM :** Reasons: By AAA criteria

** **

**7. In the figure, if ****∠***FEG*** ****≡ ****∠****1 then, prove that DG ^{2} **

**Solution: **

**Proof:**

__Statements : ____Reasons__

**1. ****∠****FEG ≡ ****∠****1**

**⇒ ∠****DEG = 180° − ****∠****FEG **

Reasons:

Given

linear pair

**2. ****∠****FDG + ****∠****DFG = 1**

**∠****EDG + ****∠****DFG = 1 **

Reasons: Exterior angle = Sum of interior opposite angles

∵ ∠FDG = ∠EDG

**3. ****∠****DEG = 180° − ****∠****FEG**

**∠****DEG = 180 − ****∠****1 **

Reasons: By 1

**4. In ΔDFG**

**∠****DGF =180° − [****∠****FDG + ****∠****DFG] **

Reasons: Angle sum property

**5. ****∠****DGF= 180° − ****∠****1 **

Reasons: By 2

**6. ****∠****DGF = ****∠****DEG **

Reasons: By 3

**7. ****∠****EDG = ****∠****EDG **

Reasons: Common in ΔFDG and ΔEDG

**8. ****∴ ∠****DGE = ****∠****DFG **

Reasons: Remaining angle by angle sum property and by 6

**9. ****∴**** ΔDGF ~ ΔDEG **

Reasons: By 6, 7, 8

By AAA similarity

**10. DG / DE = GF /EG
= DF / DG**

Reasons: Corresponding sides of similar trianlge are proportional.

**11. DG / DE = DF /
DG**

**DG. DG = DF. DE**

**DG ^{2} =
DE.DF **

Reasons: From 9

** **

**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).**

**Solution:**

Here AO = CO = 8 cm

BO = DO = 6 cm

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

∴ In Δ 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**

** **

**9. In the figure, find AR.**

**Solution:**

Δ AFI, Δ FRI are right triangles.

By Pythagoras theorem,

AF^{2} = AI^{2}
− FI^{2}

= 25^{2} − 15^{2}

= 625 − 225 = 400 = 20^{2}

∴ AF = 20 ft. ….(1)

FR^{2} = RI^{2}
− FI^{2}

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

FR = 8 ft.

∴ AR = AF + FR

= 20 + 8 = **28 ft.**

** **

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

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

**(ii) IF PD = 12, then PN = ?**

**(iii) If DO = 8, then FD = ?**

**(iv) IF OE = 36 then EP = ?**

**Solution:**

Given DN, EO, FM are medians.

∴ FN = EN

DO = FO

EM = DM

(i) If DE = 44, then

DM = 44 / 2 = 22

**DM = 22 **

(ii) If PD = 12, PN = ?

PD / PN = 2/1

12 / PN = 2/1 ⇒ PN = 12/2 = 6.

**PN = 6**

(iii) If DO = 8, then

FD = DO + OF

= 8 + 8

** FD = 16**

(iv) IF OE = 36,

then EP / PO = 2 / 1

EP / 2 = PO

OE = OP + PE

36 = [PE / 2] + PE

36 = [PE / 2] + [2PE / 2]

36 = 3PE / 2

PE = [36 × 2] / 3

**PE = 24 **

**Answer:**

**Exercise 5.3**

**Miscellaneous Practice
Problems**

**3. 48 ft **

**4. 25 ft **

**5. No, the wide of
cabinet is lesser than the wide of TV**

**Challenging Problems**

**8. 40 cm **

** 9. 28 ft **

**10. (i) 24 (ii) 6
(iii) 16 (iv) 24**

Tags : Questions with Answers, Solution | Geometry | Chapter 5 | 8th Maths , 8th Maths : Chapter 5 : Geometry

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8th Maths : Chapter 5 : Geometry : Exercise 5.3 | Questions with Answers, Solution | Geometry | Chapter 5 | 8th Maths

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