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**Class 8 R S Aggarwal Maths
solutions**

**Contents**

**13. Time and Work Ex
13A, **

**14. ****POLYGONS****
Ex 14A**

**15.**** QUADRILATERALS****
Ex 15**

**16. PARALLELOGRAMS****Ex16A**

**18. Area of trapezium and a polygonEx18A,B**

**Q3. ****Time
taken by (A+B) together to finish the work = 6 days.**

** Time taken by A to finish the work =
9 days.**

** Work
done by (A+B) in 1 day = **

** Work done by A
in 1 day = **** **

** B’s
1 day work = { (A+B)’s 1 day work – A’s 1 day work }.**

** = **** - **** = ****
= **** . www.rsmaths99.com**

**Hence, B alone can
finish the work in ****18 days.**

**Q4. ****Time
taken by Raju & Siraj
to finish the work = 6 hrs.**

** Time taken by Raju
to finish the work 15
hrs.**

** Work
done by Raju + Siraj in 1
hr= ****.**

** Work done by Raju
in 1 hr = ****.
www.rsmaths99.com**

** Work
done by Siraj in 1 hr = {( Raju****
& Siraj**** 1 hr works) – Raju’s 1 hr work}**

** = **** - **** = **** = **** .**

**Hence, Siraj alone can finish the work in ****10 hrs.**

**Q5. ****Time
taken by A to finish the work =**** 10 days.**

** ****Time
taken by B to finish the work =**** 12 “**

** ****Time
taken by C to finish the work =**** 15 “**

** Work
done by (A+B+C) in 1 day = (**** )
days.**

**= **** = ****
= **** .
Hence, A+B+C) can finish the work in ****4 days.**

**Q6. ****Time
taken by A to finish the work =**** 24 hrs**

** ****Time
taken by B to finish the work =**** 16 hrs**

** ****Time
taken by A+B+ C to finish the work =**** 8 hrs**

** Work
done by A in 1 hr = **** hrs**

** Work done byB
in 1 hr = **** hrs** **www.rsmaths99.com**

** Work
done by A+B+C in 1 hr = **** hrs**

** Work
done by (A+B) in 1 hrs = (**** )
hrs = ** =

** Work
done by C alone = **** - **** =**
=

** Hence, C can finish the work in 48
hrs.**

**Q7. ****Time
taken by A+B+ C to finish the work =**** 8hrs **

** Time taken by A to finish the work =**** 20 hrs**

** ****Time
taken by B to finish the work =**** 24 hrs **

** Work done by (A+B) in 1 hr =**

**Work done by A in 1 hr
= ****
www.rsmaths99.com**

** Work
done by B in 1 hr =**

** Work
done by A+B together
in 1 hr = ****+**** = **** = **

** Work
done by C alone in 1 hr = **** - **** = **** = **** = **** =
30 hrs.**

**Q8. ****Time
taken by A to finish the work =16 days.**

** ****Time
taken by B to finish the work****
= 12 days**

** Work done by A in 1
day = **

** Work
done by B in 1 day =**

**Work done by (A+B) in 1
day = **** + **** = **** = **

** Work done by A in 2 day = **** x
2 = **

** Work
left 1 - ** =** **

** If ****
work done in 1 day www.rsmaths99.com**

**“ 1**** “ “
**

**“ ** “ “** **** x **** =
6 days**

** Hence , 6+2
days =8 days.**

**www.rsmaths99.com**

**Q9. ****Time
taken by A to finish the work =14 days.**

** ****Time
taken by B to finish the work****
= 21 days**

** Work
done by A in 1 day = **

** Work
done by B in 1 day =**

** Work
done by (A+B) in 1 day = **** + **** = **** = **

** Work
done by (A+B) in 6 day = **** x
6 = **** .**

** Remaining work (1 - **** ) = **

**Now, B does 1 work in
21 days**

**Hence, B will do **** work in **** x
21= 6 days. **

** Hence, total number of days taken = 6 + 6 = 12 days.**

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**Q10. A
can do **** part of work in 16 days**

** A “
1 work “ 16 x **** =
24 days**

** B
can do **** part of work in 3 days **

** B “
1 work in 3 x **** =
12 days **

** Work
done by (A+B) in 1 day = **** + **** = **** =
**** = ****.**

** Hence, ****
work done in 1 day**

** Then, 1 “
“ “ 1 x 8 = 8 days.**

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**Q11. ****Time
taken by A to finish the work =15 days.**

** ****Time
taken by B to finish the work****
= 12 days**

** ****Time
taken by C to finish the work =20 days**

** Work
done by A in 1 day = **

** Work
done by B in 1 day = **

** Work
done by C in 1 day = **

** Work done by (A+B+C) in 1 day = **** = **** = **** **

** Work done by (A+B+C) in 2 day = **** = **** **

** Remaining work = 1 - **** = **

** Work done by (A+B) in 1 day = **** + **** = **** =
**** **

** If
**** work done in 1 day**

** Then, 1 “ “ “
**** “**

** Hence, ****
“ “ “
**** = 4 days.**

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**Q12.
Work done by A+B in 1 day = **

** Work
done by B+C in 1 day = **

** Work
done by C+A in 1 day = **

** 2(A+B+C)’s 1 day work = **** + **** = **** = **

** Hence, (A+B+C)’s 1 day work = **** x **** = ****
= 16 days.**

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**Q13. Work
done by A+B in 1 day = **

** Work
done by B+C in 1 day = **

** Work
done by C+A in 1 day = **

** 2(A+B+C)’s 1 day work = **** + **** = **** =
**

** Hence, (A+B+C)’s 1 day work = **** x **** = **** days.**

**A’s 1 day’s work = **** = ****
= ****.
Hence, A
alone finish the work in 30 days.**

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**Q14. Time
taken by A to fill the tank = 10 hrs**

** Time
taken by B to fill the tank = 15 hrs**

** Work
done by A in 1 hr =
**

**Work
done by b in 1 hr =
**.

** Work done by (A+B) in 1 hr = **** =
****
= **** . **

** Hence, time taken by (A+B) to
fill the tank = 6 hrs.**

**Q15. Time
taken by A to fill the tank = 5 hrs**

** Time
taken by B to empty the tank = 6 hrs**

** Work
done by A in 1 hr =
**

**Work
done by B in 1 hr = - **

** Work done by (A+B) in 1 hr = **** =
**** **

** Hence, time taken
to fill the tank completely = 30
hrs.**

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**Q16. Hints:
Work done by (A+B +C) in 1 hr = **** + **** = **** = **** .**

** Hence, it will take = **** hrs = 2hrs and**** x
60 = 2 hrs 40 min.**

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**Q17. Work
done by A in 1 min = **

** Work
done by B in 1 min =
**

**Work
done by C in 1 min = - **

**Work
done by (A+B +C) in 1 min = **** - **** = **** =
**** **

**Hence,
time taken to
fill the tank completely = 20 minutes.**

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**Q18. Time
taken by A to fill the cistern = 9 hrs**

**Work
done by** **A in 1 hr
= **

** Work done by the leak in 1 hr = ****- **** =
**** =90 hrs.**

**Q19. Work
done by A in 1 hr =
**

**Work
done by B in 1 hr = **

** Work done by (A+B) in 1 hr = **** =
**** **

**
**

** Work done by (A+B) in 2 hr = **** x 2 = **** **

**Remaining part = 1 - **** = **** **

**Now, ****
part is filled by B in 1 hr www.rsmaths99.com**

** Then, 1 “ “ B ‘
8hrs**

** Hence, ****
“ “ B
“ 8 x **** =
**** = 2 **** hrs
= 2 hrs **** x
60 = 3 hrs 20 min.**

** **

** -----------------------X-----------------------------**

__Curve__*-*** A figure traced out on a plane surface
with the help of a sharp pencil without lifting it, is called curve. It may be
open curve and closed curve.**

__Simple
closed curve__**:-**** A closed curve which does not intersect itself is called a
simple closed curve.**

** ****www.rsmaths99.com**

__Polygons____:-__** A simple closed curve made up of only line segments is called a
polygon.**

**Each straight line in a polygon is called
its side.**

__Regular
polygon__**:-**** A polygon having all sides equal and all angles equal is called
a regular polygon. Eg. Equilateral triangle and square.**

__Irregular
polygon__**:-**** A polygon which are not regular are called irregular polygons. Eg. A rectangle and rhombus.**

**For
a regular polygon of n sides, we have**

**[i] each exterior angle = **** **

**[ii]
each interior angle = 180 ^{0} – (each exterior
angle).**

**In
a convex polygon of n sides, we have**

**[i] sum of all exterior angles = 4
right Ls**

**[ii] sum of all interior angles = (2n - 4) right ***L***s**

**Number of diagonals in a polygon of n sides = **** **

** **

**Q3. Prove that the sum of all the angles a
quadrilateral is 360 ^{o}.**

** Let ABCD be a quadrilateral.**

** Join A and C.**

** Now, we know that sum of the angles of a triangle is 180 ^{0}.**

** In ****ABC L 2+ L4 + LB = 180^{0} -----(1)**

** In ****ADC L1**

** **

** Adding (1) and (2)
****www.rsmaths99.com**

** L 2+ L4 + LB + L1 + L3 + LD = 360^{0}**

** Or,
( L1 + L2 + L3 + L4 ) + LB
+ LD = 360^{0}**

** **

** Or, LA + LC
+ LB + LD = 360^{0}**

** Hence, the sum of all the angles of a quadrilateral is
360 ^{o}.**

** **

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**Q4. Let the measure of the forth angle be x ^{o }. **

** We know that the sum of the angle of the quadrilateral =
360 ^{o}.**

** **** 76
+ 54+ 108+ x ^{o} = 360^{o}. **

**Q5. Let the angle of the quadrilateral be (3x) ^{o},(5x)^{o}, (7x)^{o}, (9x)^{o}.**

** We know that the sum of the angle of the quadrilateral =
360 ^{o}.**

** **** 3x
+ 5x+ 7x + 9x = 360 ^{o}. x = **

** Now,
3 x 15 = 45 ^{o}, 5 x 15 = 75^{o}, 7 x 15 = 105^{o}, 15
x 9 = 135^{o}.**

**Q6. 75 ^{o} x 3 + y = 360^{o}. => y = 360 – 225 = 135^{o}.**

**Q7. Let the three equal angle be x = (3x) ^{o}. **

** **** (3x) ^{o} + 120^{o} = 360^{o}. **

**Q8. Let the two equal
angle be x = (2x) ^{o}.**

** **** 2x +
85 + 75 = 360 => 2x = 360 – 160 =
200 ****x = 100 ^{o} .**

**Given ABCD is a parallelogram and LA = 110^{o}. since the sum of any two adjacent angle of a parallelogram
is 180^{o}. we have**

** **

** LA
+ LB = 180^{0}. =>
110 + LB
= 180^{0} => LB
= 70^{o} **

** Also, LB + LC = 180^{O} (Since, LB + LC are adjacent angle ).**

ð **70 + LC = 180 =>
LC = 110^{o}.**

**Also, LC + LD = 180^{O} (Since, LC + LD
are adjacent angle ).**

ð **110 + LD = 180 =>
LD = 70^{O}.**

**‘OR’**

*L***A + LB = 180^{0}. =>
110 + LB
= 180^{0} => LB
= 70^{o}**

**If LA = 110^{o}
and LB = 70^{o}
then in parallelogram opposite angles are equal .**

** LC = 110^{o} and
LD = 70^{O}.**

** Q2. Let
the adjacent angle be = LA & LB . then, LA
+ LB = 180^{0}.**

** Since, both are equal **** 2 L**

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**Q3. Let two adjacent angle are LA &
L**

** Then, 4x + 5x = 180 => 9x = 180 => x = 20**

** Since,
LA = 4x **

** Since,** *L***B = 5x **** ** *L***B = 5 x 20 = 100 ^{o}.**

** ** **Then,
****in parallelogram opposite angles are equal .**

** **** LA = LC **

** And, **** LB = LD **

**Q4. Let L**

**We know, the sum of two adjacent angles of a
parallelogram is 180 ^{o}.**

** **** 3x – 4 + 3x + 16 = 180 =>
6x +12 = 180 => x = 28.**

** Then, 1 ^{st} adjacent angle L**

** And, 2 ^{nd} “ “ L**

** Now, ****in parallelogram opposite angles are equal .**

** **** LA = LC **

** And, **** LB = LD **

**Q5. Let ABCD is a ||grams. If LA**

** Since,**** in parallelogram opposite angles are equal .**

** L**

**Now, the sum of two adjacent angles of a parallelogram is 180 ^{o}.**

** LA + LB
= 180^{o} => 65 + LB
= 180^{o} => LB = 115^{o} .**

** Now, ****in parallelogram opposite angles are equal .**

** **** LC = LD = 65^{0} .**

**Q6. Let ABCD is a ||gm. Given, AB = DC = 5x
& BC = AD = 3x.**

** We know, perimeter = sum of all sides**

** 64 = 5x + 3x +5x + 3x = 16x. ****x = 4 cm.**

** Side AB = 5 x 4 = 20 cm. BC = 3 x 4 = 12 cm.**

**Q7. Let ABCD be a ||gm. and AB = x. ATQ, if AB = x then, BC = 10 +x.**

** Now, AB = x ****
CD = x (since, in ||gm opp.
Sides are equal).**

** And, BC = 10 + x **** AD = 10 + x . **

** Now, we know,**

** Perimeter =
sum of all sides**

** **** x + 10 + x + x + 10 +x = 140**

** ****
x = 30 cm. Hence, AB = 30 cm & BC = 40 cm.**

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**Q8. In ****BMC and ****
DNA**

** ***L ***DNA = LBMC
= 90^{0}**

^{ }*L***BMC = ***L***DAN (alternate angles)**

** BC = DA (Opposite sides)**

** By AAS congruency criteria**

** *** ***BMC*** *** ****DNA***
*

* ***Yes, it is true that BM is equal to
DN **

** (by
corresponding parts of congruent triangles BMC and DNA)*** *

** **

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