CBSE Class 10

RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17A
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17B
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself are helpful to complete your math homework.

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RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17A
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17B
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Correct option: (d)

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17A
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17B
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself

Question 1.
Solution:
Let the length of plot be x meters
Its perimeter = 2 [length + breadth]
= 2(x + 16) = (2x + 32) meters

Length of the rectangle is 21. 5 meter
Area of the rectangular plot = length × breadth
= (16 × 21.5)m²
= 344 m²
The length = 21.5 m and the area = 344 m²

Question 2.
Solution:
The length of a rectangular park is twice its breadth
and its perimeter is 840 m.
Find the area of the park.
Let the breadth of a rectangular park = x m
Then, length of a rectangular park = 2x m
Perimeter of a rectangular park = 840 m

∴ Area of a rectangular park = length x breadth = 140x 280 = 39200 m2

Question 3.
Solution:
Let ABCD be the rectangle in which AB = 12 cm and AC = 37 m

Thus, length = 35 cm and breadth = 12 cm
Area of rectangle = (12 × 35) = 420
Hence, the other side = 35 cm and the area = 420 cm²

Question 4.
Solution:
Let the breadth of the plot be x meter
Area = Length × Breadth = (28 × x) meter
= 28x m²
Breadth of plot is = 16. 5 m
Perimeter of the plot is = 2(length + breadth)

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:
Area of verandah = (36 × 15)m² = 540 m²
Area of stone = (0.6 × 0.5)m² [10 dm = 1 m]
Number of stones required = \(\frac { Area\quad of\quad verandah }{ Area\quad of\quad stone } =\frac { 540 }{ 0.3 } =1800\)
Hence, 1800 stones are required to pave the verandah

Question 9.
Solution:

Question 10.
Solution:
Length of the park = 35 m
Breadth of the park = 18 m

Area of the park = (35 18)m² = 630 m²
Length of the park with grass =(35 – 5) = 30 m
Breadth of the park with grass = (18 – 5) m = 13 m
Area of park with grass = (30 × 13) m² = 390 m²
Area of path without grass = Area of the whole park area of park with grass
= 630 m² – 390 m² = 240 m²
Hence, area of the park to be laid with grass = 240 m²

Question 11.
Solution:
Length of the plot = 125 m
Breadth of the plot = 78 m

Area of plot ABCD = (125 × 78)m² = 9750 m²
Length of the plot including the path= (125 + 3 + 3) m = 131 m
Breadth of the plot including the path = (78 + 3 + 3) m = 84 m
Area of plot PQRS including the path
= (131 × 84) m² = 11004 m²
Area of path = Area of plot PQRS Area of plot ABCD
= (11004 – 9750)m² = 1254 m²
Cost of gravelling = Rs 75 per m²
Cost of gravelling the whole path = Rs. (1254 × 75) = Rs. 94050
Hence, cost of gravelling the path = Rs. 94050.

Question 12.
Solution:
Area of rectangular field including the foot path = (54 × 35)m²
Let the width of the path be x m
Then, area of rectangle plot excluding the path = (54 × 2x) (35 × 2x)
Area of path = (54 × 35) (54 × 2x) (35 × 2x) = 54 ×35) (54 × 2x) (35 × 2x) = 420

Question 13.
Solution:
Let the length and breadth of a rectangular garden be 9x and 5x.

Then, area of garden = (9x 5x)m = 45x² m²
Length of park excluding the path = (9x 7) m
Breadth of the park excluding the path = (5x 7) m
Area of the park excluding the path = (9x 7)(5x 7) m²

Question 14.
Solution:

Question 15.
Solution:
Let the width of the carpet = x meter
Area of floor ABCD = (8 ×5) m²
Area of floor PQRS without border
= (8 × 2x)(5 × 2x)
= 40 × 16 x 10x + 4x²
= 40 × 26x + 4x²
Area of border = Area of floor ABCD Area of floor PQRS

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:

Question 26.
Solution:

Question 27.
Solution:

Question 28.
Solution:
Area of the ||gm = (base height) sq. unit
= (25 × 16.8) cm² = 420 cm²

Question 29.
Solution:

Question 30.
Solution:

Question 31.
Solution:
Area of parallelogram = 2 area of DABC
Opposite sides of parallelogram are equal
AD = BC = 20 cm
And AB = DC = 34 cm
In ∆ABC we have
a = AC = 42 cm
b = AB = 34 cm
c = BC = 20 cm

Question 32.
Solution:

We know that the diagonals of a rhombus, bisect each other at right angles
OA = OC = 15 cm,
And OB = OD = 8 cm
And ∠AOB = 90⁰
∴By Pythagoras theorem, we have

Question 33.
Solution:

Question 34.
Solution:

Question 35.
Solution:

Question 36.
Solution:

Question 37.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17a

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17a.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17A
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17B
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself

Question 1.
Solution:

Question 2.
Solution:
Let a = 42 cm, b = 34 cm and c = 20 cm

(ii)Let base = 42 cm and corresponding height = h cm

Hence, the height corresponding to the longest side = 16 cm

Question 3.
Solution:
Let a = 18 cm, b = 24 cm, c = 30 cm
Then,2s = (18 + 24 + 30) cm = 72 cm
s = 36 cm
(s a) = 18cm, (s b) = 12 cm and (s c) = 6 cm

(ii)Let base = 18 cm and altitude = x cm

Hence, altitude corresponding to the smallest side = 24 cm

Question 4.
Solution:
On dividing 150 m in the ratio 5 : 12 : 13, we get

Question 5.
Solution:
On dividing 540 m in ratio 25 : 17 : 12, we get

Question 6.
Solution:
Let the length of one side be x cm
Then the length of other side = {40 (17 + x)} cm = (23 – x) cm
Hypotenuse = 17 cm
Applying Pythagoras theorem, we get

Question 7.
Solution:
Let the sides containing the right – angle be x cm and (x – 7) cm

∴ perimeter of triangle (15 + 8 + 17) cm = 40 cm

Question 8.
Solution:
Let the sides containing the right angle be x and (x 2) cm

Therefore, perimeter of the triangle = 8 + 6 + 10 = 24 cm

Question 9.
Solution:
Side of an equilateral triangle = a = 10 cm

Question 10.
Solution:
Let each side of the equilateral triangle be a cm

Question 11.
Solution:
Let each side of the equilateral triangle be a cm

Perimeter of equilateral triangle = 3a = (3 12) cm = 36 cm

Question 12.
Solution:
Let each side of the equilateral triangle be a cm

Question 13.
Solution:
Base of right angled triangle = 48 cm

Question 14.
Solution:
Let the hypotenuse of right – angle triangle = 6.5 m
Base = 6 cm

Hence, perpendicular = 2.5 cm and area of the triangle =7.5 cm²

Question 15.
Solution:
The circumcentre of a right – triangle is the midpoint of the hypotenuse

Hypotenuse = 2 × (radius of circumcircle)
= (2 × 8) cm = 16 cm
Base = 16 cm, height = 6 cm
Area of right angled triangle=

Hence, area of the triangle= 48 cm²

Question 16.
Solution:
Let each equal side be a cm in length.
Then,

Hence, hypotenuse = 28.28 cm and perimeter = 68.28 cm

Question 17.
Solution:
Let each equal side be a cm and base = 80 cm

perimeter of triangle = (2a + b) cm
= (2 41 + 80) cm
= (82 + 80) cm = 162 cm
Hence, perimeter of the triangle = 162 cm

Question 18.
Solution:
Let the height be h cm, then a= (h + 2) cm and b = 12 cm

Hence, area of the triangle = 48 cm².

Question 19.
Solution:
Let ∆ABC is a isosceles triangle. Let AC, BC be the equal sides
Then AC = BC = 10cm. Let AB be the base of ∆ABC right angle at C.

Hence, area = 50 cm² and perimeter = 34.14 cm

Question 20.
Solution:

Area of shaded region = Area of ABC – Area of DBC
First we find area of ABC

Area of shaded region = Area of ∆ABC – Area of ∆DBC
= (43.30 – 24) cm² = 19. 30 cm²
Area of shaded region = 19.3 cm²

Hope given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17a are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Practical Based Questions for Class 10 Science Chemistry

Question 1.
A student dipped a strip of pH paper in distilled water taken in a tube. As expected, the pH paper acquired green colour. He then dissolved a pinch of common salt in the same tube. What will be the expected change in colour of the pH paper ?
Answer:
There will not be any change in the colour of the pH paper. It will remain green which indicates that the solution has maintained neutral character. Actually, sodium chloride is a salt prepared from strong base (NaOH) and strong acid (HCl). It is a neutral salt and is not expected to bring any change in the pH of distilled water.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 2.
(a) Five solutions A, B, C, D and E when tested with universal indicator showed pW as 4, 1, 11,7 and 9 respectively. Which solutions is :

1. neutral
2. strongly alkaline
3. strongly acidic
4. weakly acidic acid
5. weakly alkaline.

Arrange the solutions in increasing order of H-ion concentration.
Answer:

1. D
2. C
3. B
4. A
5. E.

The increasing order of H-ion concentration :C<E<D<A<B.

Question 3.
Name the acid and base from which the following salts have been formed :

1. Sodium acetate
2. Ammonium chloride
3. Calcium nitrate
4. Sodium carbonate.

Answer:

1. NaOH and CH₃COOH
2. NH₄OH and HCl
3. Ca(OH)₂ and HNO₃
4. NaOH and H₃CO₃.

Question 4.
When water is added gradually to a white solid X, a hissing sound is heard and a lot of heat is produced forming the product Y. A suspension of Y in water is applied to the walls of a house during white washing.
(a) What could be the solid X ? Write its chemical formula.
(b) What could be the product Y ? Write its chemical formula.
Answer:
(a) The white solid X is quick lime. Its chemical formula is CaO.
(b) When water is added to the white solid X, product Y formed is calcium hydroxide. Its chemical formula is Ca(OH)r

Question 5.
Two drinks P and Q gave acidic and alkaline reactions respectively. One has a pH value of 9 and the other has pH value of 3. Which drink has the pH value of 9 ?
Answer:
The solution Q is of alkaline nature. It has a pH value of 9.

Question 6.
Take five test tubes and label them as A, B, C, D and E. Add 5 mL of five unknown solutions in them. Put a small strip of universal indicator in each of them. Following colours appear in these :
Solution A — Orange;
Solution B — Green;
Solution  C – Red;
Solution D – Blue;
Solution E – Violet.
Predict the nature of the solutions in these from the pH character.
Answer:
Solution-A (Weakly acidic);
Solution-B (Neutral);
Solution-C (Strongly acidic);
Solution-D (Weakly basic);
Solution-E (Strongly basic).

Question 7.
A metal A’ gives a compound ‘B’ (molecular mass 40) when it reacts with water. Compound ‘B’ gives a soluble compound ‘C’ on treatment with aluminium oxide. Identify A’, ‘B’ and ‘C’. Also give the reactions involved.
(CBSE 2013)
Answer:
The available data suggests that the metal A sodium (Na). It reacts with water to form sodium hydroxide ‘B’ with formula NaOH. On reacting with aluminium oxide, the compound ‘B’ forms sodium metaaluminate ‘C.
The chemical equations for the reactions are :

Question 8.
You want to study a decomposition reaction by taking ferrous sulphate crystals in a boiling tube. List two steps you would follow while doing the experiment. (CBSE 2014)
Answer:

1. Take a test tube and dry it completely.
2. Take a small amount of the given sample in the test tube. Hold it with a clamp and heat the tube over a burner. Crystals will first become dirty white and then change to brown.

Question 9.
On keeping iron nails in a blue coloured copper sulphate solution, it is observed that the colour of the solution turns light green after sometime. Give reasons for this colour change. Name the type of reaction.
(CBSE 2014)
Answer:
It is an example of metal displacement reaction. Iron has displaced copper from copper sulphate solution and has changed to iron sulphate which is light green in colour.

Question 10.
While studying the double displacement reaction, the solutions of barium chloride and sodium sulphate are mixed together.
(i) What do you observe as soon as the two solutions are mixed together ?
(ii) What will happen in the above observation made by you after ten minutes ? (CBSE 2015)
Answer:
(i) A white precipitate of barium sulphate is immediately formed.

(ii) The white precipitate will settle down at the bottom of the tube and the solution above the precipitate will become colourless.

Question 11.
You want to perform an experiment to study a double displacement reaction in your school laboratory. Name two aqueous solutions required for the experiment. State the colour change you are likly to observe on mixing the two solutions. (CBSE 2015)
Answer:
(i) A white precipitate of barium sulphate is immediately formed.

(ii) The white precipitate will settle down at the bottom of the tube and the solution above the precipitate will become colourless.

Question 12.
You are given two colourless solutions present in two test tubes. One out of these is ethyl alcohol and the other is acetic acid. Give three tests to identify these.
Answer:
Smell. The tube in which the liquid has a vinegar like or fruity smell is acetic acid while the other with a very little pungent smell is ethyl alcohol.
Litmus Test. Add a strip of blue litmus separately in both the tubes. The tube which turns blue litmus red contains acetic acid while the other which does not change its colour is that of ethyl alcohol.
Sodium hydrogen carbonate test. Add a small amount of solid sodium hydrogen carbonate (NaHCO₃) in both these tubes. The one which gives a brisk effervescence contains acetic acid while the other in which no effervescence is noticed contains ethyl alcohol.

Question 13.
You are provided with two samples of hard water; one containing temporary hardness and the other permanent hardness. Without the help of any chemical, how will you identify the nature of the sample.
Answer:
Boil the two samples separately in beakers for sometime. If precipitate appears, filter it out. Now, add a strip of soap in each. The one which produces lather contained temporary hardness while the other in which no lather is formed, contained permanent hardness.

Question 14.
How will you distinguish between ethane and ethene with the help of a chemical test ?
Answer:
Pass the two gases separately through bromine dissolved in carbon tetrachloride taken in two test tubes. If the yellow colour of bromine gets discharged, the gas is ethene. If the colour remains intact, then the gas is ethane.

Question 15.
An unknown organic liquid does not turn blue litmus red and gives no effervescence with sodium hydrogen carbonate. However, when a dry piece of sodium pellet is added to the liquid, a gas is evolved with brisk effervescence. Identify the liquid.
Answer:
The given liquid is probably an alcohol (e.g., ethyl alcohol) which has no reaction with blue litmus or sodium hydrogen carbonate. But it evolves hydrogen gas on reacting with sodium metal
2C₂H₅OH + 2Na ————> 2C₂H₅ONa + H₂

Question 16.
How will you distinguish between hydrochloric acid and ethanoic acid with the strip of a-universal pH paper ?
Answer:
The strip of universal pH paper will turn red in hydrochloric acid while its colour will change to orange in ethanoic acid. In fact, hydrochloric acid is a stronger acid than acetic acid and they have different pH values.

Question 17.
Give a simple test to distinguish soaps from detergents.
Answer:
Soaps donot give lather with hard water. However, detergents form lather with hard water.

Question 18.
Write the name of apparatus/chemicals required to study the following properties of ethanoic acid in the laboratory.
Nature, odour, solubility and action are sodium hydrogen carbonate. (CBSE 2015)
Answer:
Litmus paper, test tube, test tube, lime water.

Question 19.
A student added sodium hydrogen carbonate solution in ethenoic acid taken in a test tube and the gas evolved was tested with a burning splinter. Write the chemical equation for the evolution of this gas and its effect on burning splinter. (CBSE 2015, Sample Paper 2017)
Answer:
CH₃COOH + NaHCO₃ ———–> CH₃COONa + H₂O + CO₂
Since CO₂ is not a supporter of combustion, the burning splinter will be extinguished.

Question 20.
A student is studying the properties of acetic acid in his school laboratory. List two physical and two chemical properties which he must observe and note in his record book. (CBSE 2016)
Answer:
Complete oxidation: When ethanol is warmed with dilute alkaline solution of potassium permanganate (5% solution) called Baeyer’s reagent, ethanoic acid or acetic acid is formed as the product.
CH₃CH₂OH + 2(O) ————–> CH₃COOH +H₂O
Partial oxidation: Upon oxidation with chromic anhydride (CrO₃) dissolved in acetic acid (CH₃COOH), ethanol forms ethanal, also called acetaldehyde.
CH₃CH₂OH +  (O)  —————– > CH₃CHO + H₂O

Question 21.
A student adds a spoon full of powdered sodium hydrogen carbonate to a flask containing ethanoic acid. List two main observations, he must note in his note book, about the reaction that takes place. Also write chemical equation for the reaction. (CBSE 2016)
Answer:

1. A colourless and odourless gas is evolved accompanied by brisk effervescence.
2. When the gas is bubbled through lime water, it becomes milky.

Question 22.
A gas is liberated immediately with brisk effervescence when you add acetic acid to sodium hydrogen carbonate powder in a test tube. Name the gas and describe a test to confirm the identity of the gas.
(CBSE 2017)
Answer:
The gas evolved is CO₂
CH₃COOH + NaHCO₃ ———–> CH₃COONa + H₂O + CO₂
It turns lime water milky when bubbled in small amount. On bubbling the gas in excess, the milkiness disappears.

Question 23.
If you are asked to report two observations about the following two properties of acetic acid, what would you report (t) Odour (it) Effect on litmus. (CBSE 2017)
Answer:

1. Acetic acid has vinegar smell.
2. On adding a few drops of blue litmus solution to acetic acid, it acquires red colour.

Question 24.
Mention the essential materials (chemicals) to prepare soap in the laboratory. Describe in brief the test of determining the nature (acidic/alkaline) of the reaction mixture of a saponification reaction.
(CBSE 2017)
Answer:
The essential materials are edible oil (from animal or vegetable origin), caustic alkali (NaOH or KOH) and common salt. The reaction mixture in the saponification reaction is of alkaline nature since it contains in it a caustic alkali. The nature can be tested by adding a few drops of red litmus to the mixture. Its colour will change to blue.

Question 25.
What do you observe when you add a few drops of acetic acid to a test tube containing.
(a) phenolphthalein
(b) distilled water
(c) universal indicator
(d) sodium hydrogen carbonate.
(CBSE 2017)
Answer:
(a) Acetic acid will remain colourless in phenolphthalein.
(b) Acetic acid will mix with distilled water to form a clear solution.
(c) A universal indicator will impart orange colour to acetic acid.
(d) A brisk effervescence will be noticed due to the evolution of CO₂ gas.

Hope given Practical Based Questions for Class 10 Science Chemistry helpful to you.

If you have any doubts, please comment below. We try to provide online math tutoring for you.