5% of salary = Rs 2000 5% × x = 2000 x x = 2000
x = 40,000 Therefore, the total salary of menu is Rs 40,000 4. 25% of the total mangoes which are rotten is 1,250. Find the total number of Mangoes in the basket. Also, find the number of good mangoes. Sol:
Given, 25% of rotten mangoes = 1,250 Let x be the total mangoes 25% of x = 1,250 × x = 1250
x = 5000 Therefore, total mangoes in the basket is 5000Rs
5. The marks obtained by Rani in her twelfth standard exams are tabulated Below. Express these marks as Percentages. Marks (out of 100) Subjects (i) English (ii) Tamil (iii) Mathematics (iv) Physics (v) Chemistry (vi) Biology
I = 100×90 I = 9000 Rs 8. The marked Price of a toy is ` 1,200. The shopkeeper gave a discount of 15%. What is the selling Price of the toy? Sol:
M.P = 1200 Rs Discount = 15% S.P =? Discount =
= 15% of 1200
= 180 Rs S.P
= M.P – discount
= 1200 – 180 = 1020 Rs
9. In an interview for a computer firm 1,500 applicants were interviewed. If 12% of them were selected, how many applicants were selected? Also find the Number of applicants who were not selected. Sol:
Total number of applicants = 1500 Percentage of selected applicants = 12% 1500 × 12% = 1500 ×
= 180 Not selected candidates = 1500 – 180 = 1320 10. An alloy consists of 30% copper and 40% zinc and the remaining is nickel. Find the amount of nickel in 20 kilograms of the alloy.
Given, copper = 30%, zinc = 40% Nickel = [100 – (30 + 40)] % = 100 – 70 = 30% Amount of nickel in 20Kg of o f the alloy 20 ×
= 2 × 3
=6 Therefore, the amount of Nickel in 20Kg of the alloy = 6Kg
11. Pandian and Thamarai contested for the election to the Panchayat committee from their village. Pandian secured 11,484 votes which was 44% of the total votes. Thamarai secured 36% of the votes. Calculate (i) the number of votes cast in the village and (ii) the number of voters who did not vote for both the contestants. Sol:
Given, Pandiyan’s 44% of the total vote = 11484 Let x be the total votes 44% of x = 11484 × x = 11484 × 100 X=
X=26100 Total votes=26100 Total People=(44%+36%+20%) =100% Not voted for both =20% of total votes =
5. The cost Price of 16 note books is equal to the selling Price of 12 note books. Find the gain Percent.
Let S.P of each note book be x S.P of 12 note books =12x S.P of 16 note books =16x Given C.P of 16 note books=S.P of 12 note books Cost Price of 16 note books=12x S.P of 16 notebooks =16x Here S.P>C.P Profit= S.P-C.P =16x-14x =4x Profit%= ×100 = =33 % 6. A man sold two articles at ` 375 each. On the first article, he gains 25% and on the other, he loses 25%. How much does he gain or lose in the whole transaction? Also, find the gain or loss Percent in the whole transaction. S.P=first article =375 Rs Gain=25% C.P of the first article= × 375 × 375 = =300 Rs
of second article =375 RS Loss=25% C.P of the second article article = × 375 × 375 = S.P
=800 Rs S.P of two articles =375 ×2=750 Here C.P > S.P so there
is a loss
Loss=C.P-S.P =800-750 =50 He lost Rs 50 in the whole transaction Loss %= × 100 =6.25 %
7. Anbarasan purchased a house for ` 17,75,000 and spent ` 1,25,000 on its interior decoration. He sold the house to make a Profit of 20%. Find the S.P. of the house.. house Purchase Price of the house=1775000 Amount spent for interior decoration=1775000+125000 =1900000 Profit=20% = × 19000 1900000 00 =380000 S.P of the house=1900000+380000 house=1900000+380000 =2280000
8. After spending Rupees sixty thousand for remodelling a house, Amla sold a house at a Profit of 20%. If the selling Price was Rupees forty two lakhs, how much did she spend to buy the house? S.P of house =4200000 Profit=20% S.P of the house = × 42000 4200000 00 × 420000 4200000 0 = =3500000 Amount spent for remodeling the house =60000
Purchase Price of the house =3500000-60000 =3440000
9. Jaikumar bought a Plot of land in the outskirts of the city for ` 21,00,000. He built a wall around it for which he spent ` 1,45,000. And then he wants to sell it at ` 25,00,000 by making an advertisement in the newspaper which costs him ` 5,000. Now, find his Profit Percent.
Amount spent to build the wall=145000 Rs Amount spent for advertisement =5000 C.P of the Plot =2100000+14500000+5000 C.P=2250000 S.P=2500000 Profit=S.P-C.P =2500000-2250000 =250000 × 100 Profit%= =11 %
10. A man sold two varieties of his dog for ` 3,605 each. On one he made a gain of 15% and on the other a loss of 9%. Find his overall gain or loss. [ Hint: Hint: Find C.P. of each] S.P of first of first dog=3605% Gain=15% C.P of the first dog = × 3605 3605 × 3605 3605 = =3134.78 S.P of second dog=3605 Rs Loss=9% C.P 0f second dog = =
× 3605 3605
× 3605 3605
=3961.54 Rs C.P of the two dogs =3134.78+3961.54 =7096.32 S.P=of the two dogs =3605+3605 =7210 Rs Here S.P>C.P So there is a gain Gain=S.P-C.P =7210-7096.32 =113.68Rs
Exercise 1.3 1. Choose the correct answer: (i) The discount is always on the _______.
(iii) ______ = Marked Price – Discount. (A) Cost Price (B) Selling Price (C) List Price (D) Market Price Ans:Selling Price (iv) The tax added to the value of the Product is called ______ Tax. (A) Sales Tax (B) VAT (C) Excise Tax (D) Service Tax Ans:VAT (v) If the S.P. of an article is ` 240 and the discount given on it is ` 28, then the M.P. is _______. (A) ` 212 (B) ` 228 (C) ` 268 (D) ` 258 Ans:268 Rs 2. The Price marked on a book is ` 450. The shopkeeper gives 20% discount on it a in book exhibition. What is the Selling Price?
Marked Price of the book=450Rs Discount=20% = × 450 =90 S.P of the book=450-90 =360 Rs
3. A television set was sold for ` 5,760 after giving successive discounts of 10% and 20% respectively. What was the Marked Price?
Let the marked of the television be x=100Rs First discount =10%= × 100 =10Rs S.P after
the first discount =100-10
Second discount =20%= × 100
=18 Rs So S.P after second discount =90-18 =72 When S.P=72 Rs,
MATHEMATICS 4. Sekar bought a computer for ` 38,000 and a Printer for ` 8,000. If the rate of sales tax is 7% for these items, find the Price he has to Pay to buy these two items. Price of computer =38000 =38000 RS Price of Printer =8000 Rs So the Price of computer and Printer=38000+8000 =46000Rs Sales tax=7% × 4600 46000 0 = =3220 Rs So the Price of both including sales tax=46000+3220 =49220 Rs
5. The selling Price with VAT, on a cooking range is ` 19,610. If the VAT is 6%, what is the original Price of the cooking range?
Let the original Price of the cooking range =100rs VAT=6% So S.P of the cooking range with VAT=100+6 =106 S.P=106Rs ,original Price =100 =100 RS If S.P=19610 RS then Original Price = × 1961 19610 0 =18500 RS The original Price of the cooking range =18500 Rs
6. Richard got a discount of 10% on the suit he bought. The marked Price was ` 5,000 for the suit. If he had to Pay sales tax of 10% on the Price at which he bought, how much did he Pay? Marked Price of the suit =5000 Rs Discount=10% = × 5000 5000 =500 Rs S.P of the suit =5000-500 S.P=4500Rs Sales =10%
= × 4500 4500
=450 Rs So the Purchase Price of the suit =4500+450 =4950 RS
7. The sales tax on a refrigerator at the rate of 9% is ` 1,170. Find the actual sale Price. Let the S.P of the refrigerator r efrigerator be x Sales tax =9% = × But sales tax=1170 Rs
× =1170 × 100 x= =>
8. A trader marks his goods 40% above the cost Price. He sells them at a discount of 5%. What is his loss or gain Percentage? Let C.P be 100Rs Marked Price =40% above C.P =100+40 =40 Rs Discount =5% × 140 = =7 S.P of the article =140-7 =133 C.P=100 Rs S.P=133 Profit =133-100 =33 RS
× 100 .
9. A T.V. with marked Price ` 11,500 is sold at 10% discount. Due to festival season, the shop keeper allows a further discount of 5%. Find the net selling Price of the T.V. M.P of TV =11500Rs First discount=10%= × 1150 11500 0 =1150 Rs S.P after the first discount =11500-1150 =10350 Second discount =5% =
=517.50 So S.P of the T.V after second discount =10350-517.50 =9832.50 The net selling Price of the T.V=9832.50
10. A Person Pays ` 2,800 for a cooler listed at ` 3,500. Find the discount Percent offered.
Marked Price of cooler=3500 RS S.P of cooler=2800 RS Discount =M.P-S.P =3500-2800 700 Rs Discount%= × 100 =20%
11. DeePa Purchased 15 shirts at the rate of ` 1,200 each and sold them at a Profit of 5%. If the customer has to Pay sales tax at the rate of 4%, how much will one shirt cost to the customer?
C.P of 1 shirt =1200Rs Profit=5% = × 1200 1200 =60 Rs So S.P of 1 shirt =1200+60 =1260 Rs Sales tax=4% =
× 1260 1260
So the amount Paid to buy one shirt =1260+50.40 =1310.40 RS 12. Find the discount, discount Percent, selling Price and the marked Price. Sl. No Items M. P Rate of Discount Amount of Discount S. P (i) Saree` 2,300 20% (ii) Pen set ` 140 ` 105 (iii) Dining table 20% ` 16,000 (iv) Washing Machine ` 14,500 ` 13,775 (v) Crockery set ` 3,224 12½%
=81000/8 A=1012 Amount=Rs.10125 C.I=A-P =Rs.10125-Rs.8000 =Rs.2125 3. Maria invested ` 80,000 in a business. She would be Paid interest at 5% Per annum compounded annually. Find (i) the amount standing to her credit at the end of second year and (ii) the interest for the third year.
P=Rs80,000 N= 2 years R=5% A=80000(1 + =80000(1 +
=80000X × = 80000X ×
=200X441 A=88200 Amount at the end of second year=Rs.88200 ii. So the interest for third year
A=27783 Rs C.I=A-P =27783-24000 =3783 Rs 5. Find the amount that Dravid would receive if he invests ` 8,192 for 18 months at 12½% Per annum, the interest being compounded half - yearly. P=8192 Rs N=1 Years r=12 %
8. Vicky borrowed ` 26,400 from a bank to buy a scooter at the rate of 15% P.a. compounded yearly. What amount will he Pay at the end of 2 years and 4 months to clear the loan?
P=26400 Rs r=15% n=2 Years 4 months =2(4\12) years =2 (1\3)years A= 26400((1 + ) (1 + ) =26400× × × ××× = ××× = =36659.70rs Vicky has to Pay Rs 36659.70 to clear the loan. 9. Arif took a loan of ` 80,000 from a bank. If the rate of interest is 10% P. a., find the difference in amounts he would be Paying after 1½ years if the interest is (i) compounded annually and (ii) compounded half - yearly.
When interested compounded annually P=80000 Rs r=10%
=800× 1 1 × 2 1 A=92400 Rs ii) when interest is is compounded half yearly P=80000 r=10% n=1 years A=80000(1+ )3 =80000(1+ )3 =80000( )3 =
80000( × × )
9261 =10× 9261 =92610 Rs The amount between the amount is Rs 92610-92400 =210Rs 10. Find the difference between simple interest and compound interest on ` 2,400 at 2 years at 5% Per annum compounded annually. The difference between S.I and C.I For 2 years is P( )2 =2400( )2 =2400( ) × =6 The difference between C.I and S.I is Rs 6
11. Find the difference between simple interest and compound interest on ` 6,400
for 2 years at 6 ¼% P. a. compounded annually. The difference between between S.I and C.I for for 2 Years is is P( )2 =6400( )2 =6400( 2 )
× ) ×
12. The difference between C. I. and S. I. for 2 years on a sum of money lent at 5% P.a. is ` 5. Find the sum of money lent.
Let the Principle be P N=2years R=5% The difference between between S.I and C.I for for 2 Years is is P( )2 2 =P( ) =5 Rs P(
× ) ×
P=2000 Rs 13. Sujatha borrows ` 12,500 at 12% Per annum for 3 years at simple interest and Radhika borrows the same amount for the same Period at 10% Per annum compounded annually. Who Pays more interest and by how much?
To find the strength after 2 years n=2 Now A= P(1+ )n 2 = 2000(1+ ) = 2000(1+ )2 2 =2000( ) =2000 × × × =5 × 441 =2205 So the number 0f students students after 2 year is 2205
2. A car which costs ` 3,50,000 depreciates by 10% every year. What will be the worth of the car after three years?
Present strength of the car P=3,50,000 Rate of deprecation r=10% Number of year =3 year So the value of the car after 3 years Now A= P(1- )n 3 = 350000(1- ) 3 = 350000( ) 2 =2000( ) ×× =350000 × ×× =350 ×81×9 =255150
3. A motorcycle was bought at ` 50,000. The value depreciated at the rate of 8% Per annum. Find the value after one year.
Value of motor cycle p=50000 Rate of depreciation r=8% To find value of the motorcycle after 1 year Now A= P(1- )n = 50000(1- ) = 50000( ) = 500 × 92 =46000
4. In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% Per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000. The initial amount of bacteria P=506000
Percentage of increase r=2.5% Per hour The amount of bacteria at the end of 2 hours Now A= P(1+ )n 2 = 506000(1+ ) 2 = 506000( ) =506000 × × × =5060 × =531616.25
5. From a village People started migrating to nearby cities due to unemployment Problem. The Population of the village two years ago was 6,000. The migration is taking Place at the rate of 5% Per annum. Find the Present Population.
Population of village P=6000 Rs Percentage of migration r=5% So the Present Population A= P(1- )n = 6000(1- )2 =6000 × × × A=5415
6. The Present value of an oil engine is ` 14,580. What was the worth of the engine 3 years before if the value depreciates at the rate of 10% every year?
Let the worth of the engine before 3 years be P Percentage of depreciation r=10% Per annum So Present value =A n ) 14580= P(1- )3 3 14580= P( ) ×× P=14580 × ×× =×
7. The Population of a village increases by 9% every year which is due to the job opportunities available in that vill village. age. If the Present Population of the village is 11,881, what was the Population two years ago?
Let P be the Population of the village 2 years years ago Percentage of increase r=9% So the Present Population A= P(1+ )n 11881= P(1+ )2 11881= 2 P( ) ×× P=14580 × ×
Exercise 1.6 1. Ponmani makes a fixed deposit of ` 25,000 in a bank for 2 years. If the rate of interest is 4% Per annum, find the maturity value.
=250× 8 Interest= 2000Rs Maturity value=P+I =25000+2000 Maturity value =27000 Rs 2. Deva makes a fixed deposit of ` 75,000 in a bank for 3 years. If the rate of interest is 5% Per annum, find the maturity value.
3. Imran deposits ` 400 Per month in a Post office off ice as R.D. for 2 years. If the rate of interest is 12%, find the amount he will receive at the end of 2 years.
Amount deposited every month P=400 Rs Number of month= 2 years N=2×12=24 month Rate of interest r=12% ( ) Total Deposit made=N= [ ] ×() = [ ] 25 years Interest= ×× =
Total amount due=P n + =9600+1200 =10800 Rs
4. The cost of a microwave oven is ` 6,000. Poorani wants to buy it in 5 instalments. If the company offers it at the rate of 10% P. a. Simple Interest, find the E.M.I. and the total amount Paid by her.
Cost of microwave P=6000 Rs Interest=10% N=5 months = years ×× Interest= × =250 Rs Total amount Paid for the Oven =6000+250 =6250 Rs Number of installment=5 So EMI= =1250 Rs Total amount Paid by Poorani=6250 Rs
5. The cost Price of a refrigerator is ` 16,800. Ranjith wants to buy the refrigerator at 0% finance scheme Paying 3 E.M.I. in advance. A Processing fee of 3% is also collected from Ranjith. Find the E.M.I. E.M. I. and the total amount Paid by him h im for a Period of 24 months.
Cost of refrigerator =16800 Rs N=24 Interest= 0% EMI= =700 Rs 3 EMI is Paid in advance so amount amount Paid Paid in advance=700 ×3=2100 Processing fees=3% = ×16800 =504 Rs Total amount Paid =16800+2100+504 =19404 Rs 6. The cost of a dining table is ` 8,400. Venkat wants to buy it in 10 installments’. If the company offers it for a S.I. of 5% P. a., find the E.M.I. and the total amount Paid by him.
Cost of dining table =8400 Rs Number of installments’=10 So n= years Rate of interest r=5% × Interest =8400× × = 350 Rs =Total amount to be Paid =8400+350 =8750 Rs Number of installments’ =10 E.M.I= =875 Rs
Exercise 1.7 1. Twelve carpenters working 10 hours a day complete a furniture work in 18 days.
How long would it take for 15 carpenters working for 6 hours Per day to complete the same Piece of work?
No of carpenters 12 15
No of hours in a day 10 6
No of days 18 x
Step 1: Consider carpenter and days The multiplying facto factorr is Step 2: Consider no of hour Per day and no of days. The Multiplying factor is ×× X= × =24 days 2. Eighty machines can Produce 4,800 identical mobiles in 6 hours. How many mobiles can one machine Produce in one hour? How many mobiles would 25 machines Produce in 5 hours? No of machines No of Mobiles No of hours 4800 80 6 1 x 1 25 y 5
Step 1: Consider the number of machines and no of mobiles Produced The Multiplying Factor is SteP2: Consider number of hours the machine works and the no of mobiles Produced ×× The Multiplying factor =X= × =10 ii) The Multiplying Multiplying factor is The Multiplying factors is ×× X= × =1250 3. If 14 compositors can compose 70 Pages of a book in 5 hours, how many compositors will compose 100 Pages of this book in 10 hours?
No of compositors and no of Papers of a book The multiplying factors = Step2: The multiplying factor is 5\10 So the number of compositors needed ×× X= × =10 compositors 4. If 2,400 sq.m. of land can be tilled by 12 workers in 10 days, how many workers are needed to till 5,400 sq.m. of land in 18 days? Area of land 2400sqm 5400
No of workers 12 x
No of days 10 18
Step1 : Area of land to be tiled and number of days The multiplying factors = Step 2: The multiplying factor is 10\18 So the number of workers needed ×× X= × =15 workers
5. Working 4 hours daily, Swati can embroid 5 sarees in 18 days. How many days will it take for her to embroid 10 sarees working 6 hours daily?
Step1 : No of hours Per day and number of days The multiplying factors = Step2: The multiplying factor is 10\5
So the number of days needed ×× X= × =24 days
6. A sum of ` 2,500 deposited in a bank gives an interest of ` 100 in 6 months. What will be the interest on ` 3,200 for 9 months at the same rate of interest?
Step1 : Amount deposited and interest received The multiplying factors = Step2: Interest received and number of months The multiplying factor is 9\6 So interested received ×× X= ×
Exercise 1.8 1. A man can complete a work in 4 days, whereas a woman can complete it in only 12 days. If they work together, in how many days, can the work be completed?
Time taken by a man to finish finish the work = 4 days. days. Work done by a man in 1 day = 1/4 Time taken by a woman to finish the work = 12days. Work done by a woman in 1 day = 1/12 Work done by both in 1 day = 1/4 + 1/12 = 4/12= 1/3
Time taken by both to finish the work = 1/3days, days. Hence both can finish the work in 1/3days
2. Two boys can finish a work in 10 days when they work together. The first boy can do it alone in 15 days. Find in how many days will the second boy do it all by himself?
Time taken by (A + B) to finish the work = 1/10days. Time taken by first boy to finish the work 15 days. So the Portion of work done by the second=1/10-1/15 =
so time taken by
the second boy to finish the work=1(1/30)days
3. Three men A, B and C can complete a job in 8, 12 and 16 days respectively. A and B work together for 3 days; then B leaves and C joins. In how many days, can A and C finish the work?
Time taken by A to finish the work = 1/8days. Time taken by B to finish the work= 1/12 days Time taken by C to finish finish the work= 1/16days So the Portion Portion of work done by A and B=1/8+1/12 =
So the Portion Portion of work done by A and Bin Bin 3 days= =3(
4. A tap A can fill a drum in 10 minutes. A second tap B can fill in 20 minutes. A third tap C can empty in 15 minutes. If initially the drum is empty, find when it will be full if all taps are opened op ened together?
Time taken by tap A to fill the drum = 10hours. Work done by tap A in 1 hour = 1/10 Time taken by tap B to fill the drum= 20 hours. Work done by tap B in 1 hour = 1/20 Time taken by tap A to fill the drum = 15hours. Work done by tap A in 1 hour = 1/15 When all the taps opened then the Portion of drum filled in one minute=
5 1 = /60= /12
So the time taken for the taps to fill the drum =1(1/12)=12 minutes 5. A can finish a job in 20 days and B can complete it in 30 days. They work together and finish the job. If ` 600 is Paid as wages, find the share of each.
Time taken by A to finish the work = 1/20days. Time taken by B to finish the work= 1/30 days The ratio of work done by b y A, B in 1 day= da y=1/20: 1/30 The ratio of their wages=3:2 Type
6. A, B and C can do a work in 12, 24 and 8 days respectively. They all work for one day. Then C leaves the group. In how many days will A and B complete the rest of the work?
Time taken by A,B,C to finish the work = 1/12, 1/24, 1/18
So the Portion Portion of work done by A and B and C Together in One Day=1/12+1/24+1/8
1 Remaining Portion=1- /4 =
The Portion of work done done by A,B A,B together is
Time taken for A and B together to complete
of the work= 3
/4×8 = 6 days
7. A tap can fill a tank in 15 minutes. Another tap can empty it in 20 minutes. Initially the tank is empty. If both b oth the taps start functioning, when will the tank become full?
Time taken by tap A to fill the tank = 15 hours. Work done by tap A in 1 min= 1/15 Time taken by tap B to fill the tank = 20 hours. Work done by tap B in 1 min = 1/20 Work done by (A + B) in 1 min = (1/15- 1/20) = 1/60 1 So the time taken to fill the ta nk=1( /60) =60min
Geometry Exercise 2.1 1. Choose the correct answer (i) The Point of concurrency of the medians of a triangle is known as (A) In centre (B) circle centre (C) orthocenter (D) centroid Ans: centroid centroid (ii) The Point of concurrency of the altitudes of a triangle is known as (A) in centre (B) circle centre (C) orthocenter (D) centroid Ans: orthocentre orthocentre (iii) The Point of concurrency of the angle bisectors of a triangle is known as (A) in centre (B) circle centre (C) orthocenter (D) centroid Ans: incentre incentre (iv) The Point of concurrency of the Perpendicular bisectors of a triangle is known as (A) in centre (B) circumcentre (C) orthocenter (D) centroid Ans: circumcent circumcentre re 2. In an isosceles triangle AB = AC and +B = 65c. Which is the shortest side.
Given ∠B=65 °, ∠C=65° Sum of the angles of a triangle is 180 ˚∠ A+∠B+∠C=180˚
∠ A+65˚+65˚=180˚ ∠ A+130=180˚ ∠ A=180˚-130˚ ∠ A=50˚ In ABC ∠ A =50˚,,∠B=∠C=65˚ 43
∠ A is smallest smallest angle. The side opposite to smallest angle is smallest. BC is smallest side.
3. PQR is a triangle right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
In PQR P=90 The hypotenuse is QRR Given PR= 24cm PQ=10cm QR= + 24cm =√ 24 24 + 10 =√ 576 576 + 100 P
= 676 676 10cm QR=26cm 4. Check whether the following can be the sides of a right angledtriangle AB = 25 cm, BC = 24 cm, AC = 7cm. AB=25 cm AB2=252=625
BC2=242=576 AC=7, AC2=7 AC2=72 =49 Now 576+49=625 BC2+AC2=AB2 So the sides AB, BC, AC from the sides of a right angled Triangle. 5. Q and R of a triangle PQR are 25° and 65°. Is DPQR a right angledtriangle?
Moreover PQ is 4cm and PR is 3 cm. Find QR. In PQR ∟Q= 25˚ 25˚ R=65˚ ∟R=65˚ ∟P+∟Q+∟R=180˚ ∟P=180-( ∟P=180-(∟Q+∟R) ∟Q+∟R) 180-(25+65) =180-90 =90 =>PQR is a right triangle Given PQ=4cm PR=3cm ∟P=90˚, QR is the hypotenuse ∟QR= +
=√ 4 + 3 =√ 1 6 + 9 =√ 25 25 =5cm 6. A 15 m long ladder reached a window 12m high from the ground. On Placing it Against a wall at a distance x m. Find x.
Length of ladder AC=15cm Height of window AB= 12m ABC is a right angled triangle t riangle with ∟B=90 2
So AC =AB +BC
Here BC =x, 152=122+x2 X2=152-122 =225-144 X2=81 X=9 m
7. Find the altitude of an equilateral triangle of side 10 cm . A ABC is an equilateral triangle with side 10 cm AD is drawn Perpendicular to BC 10cm 10cm BD=DC BD=DC=10\2 =5cm From the right angled triangle ABD right angledat D. B 5cm 5cm C AB2=AD2+BD2 2 2 2 A 10 =AD +5 2 2 2 AD =10 -5 AD2=100-25 10cm 10cm AD2=75 AD=√ 75 75 =5√ 3 cm B C The length of the altitude is 5√ 3 cm. 8. Are the numbers 12, 5 and 13 form a Pythagorean Triplet? 122=144 52=25 132=169 144+25=169 122+52=132 The numbers 12,5,13 form a Pythagoras Triplet. MATHEMATICS
9. A Painter sets a ladder up to reach the bottom of a Second store window 16 feet above the ground. The base of the ladder is 12 feet from the house. While the Painter mixes the Paint a neighbor’s Dog bumps the ladder which moves the base 2 Feet farther away from the house. How far up side Of the house does the ladder reach?
In ABC ∟B=90° AC is the length of ladder Given AB=16m, BC=12m AC=√ + ==√ 16 16 + 12 =√ 256 + 144 =√ 400 400 =20m Length of ladder is 20m Now AC=20m BC=14m AB2+BC2=AC2 AB2+142=202 AB2+196=400 AB2=400-196 =204 AB=√ 5 1 × 4 =2√ 51 51 ft The ladder reaches upside of the house at 2 √ 51 51 ft
Exercise 2.2 1. Choose the correct answer: (i) The _______ of a circle is the distance from the centre to the circumference. (A) Sector (B) segment (C) diameters (D) radius Ans: Radius (ii) The relation between radius and diameter of a circle is ______ (A) Radius = 2 × diameters (B) radius = diameter + 2 (C) Diameter = radius + 2 (D) diameter = 2 (radius) Ans: diameter diameter = 2 (radius) (radius)
(iii) The longest chord of a circle is (A) radius (B) secant (C) diameter (D) tangent
Ans: ) diameter diameter 2. If the sum of the two diameters is 200 mm, find the radius of the circle in cm.
Sum of 2 diameters=200mm So length of 1 diameters= =100mm
=2xradius Radius= =50mm=5cm
3. Define the circle segment and sector of a circle. Segment of a Circle A chord of a circle divides divides the circular circular region region Into two Parts. Each Part is called as segment of The circle.
Sector of a Circle The circular region enclosed by an arc of a circle and the two radii at its end Points is known as Sector of a circle 4. Define the arc of a circle.
AB is a chord. chord. The chord chord AB divides the Circle into two Parts. The curved Parts ALB and AMB are known as Arcs 5. Define the tangent of a circle and secant of a circle. Secant of a Circle A line Passing Passing through a circle and intersecting intersecting the the circle at two Points Points is called called The secant of the circle Tangent Tangent is a line that touches a circle at exactly one Point, and the Point is Known as Point of contact
Chapter 3: Practical Geometry Exercise 4.1 Draw a concentric circles for the following measurements of radii, find out the width of each circular ring. 1. 4cm and 6cm.
Construction: Draw a rough diagram and mark the measurements With O as centre draw a circle OA = 5.5cm. With O as centre draw a circle OB = 3.5cm. Now the concentric circle is drawn. The width of the circular ring
Construction: Draw a rough diagram and mark the measurements With O as centre draw a circle OA = 5.5cm. With O as centre draw a circle OB = 3.5cm. Now the concentric circle is drawn. The width of the circular ring = OA – OB = 5.5 – 3.5 = 2cm
Construction: Draw a rough diagram and mark the measurements With O as centre draw a circle OA = 6.8cm. With O as centre draw a circle OB = 4.2cm. Now the concentric circle is drawn. The width of the circular ring = OA – OB = 6.8 – 4.2 = 2.6cm
Construction: Draw a rough diagram and mark the measurements With O as centre draw a circle OA = 6.5cm. With O as centre draw a circle OB = 5cm. Now the concentric circle is drawn. The width of the circular ring = OA – OB = 6.5 – 5 = 1.5cm
Construction: Draw a rough diagram and mark the measurements With O as centre draw a circle OA = 8.1cm. With O as centre draw a circle OB = 6.2cm. Now the concentric circle is drawn. The width of the circular ring = OA – OB = 8.1 – 6.2 = 1.9cm
Construction: Draw a rough diagram and mark the measurements With O as centre draw a circle OA = 7cm. With O as centre draw a circle OB = 5.3cm. Now the concentric circle is drawn. The width of the circular ring = OA – OB = 7 – 5.3 = 1.7cm