CBSE Class X

Question 1:

State which pairs of triangles in the following figure are similar? Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Answer:

(i) ∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

Therefore, ΔABC ∼ ΔPQR [By AAA similarity criterion]

(ii) ∴ ∆ABC ~ ∆QRP [By SSS similarity criterion]

(iii)The given triangles are not similar as the corresponding sides are not proportional.

(iv)The given triangles are not similar as the corresponding sides are not proportional.

(v)The given triangles are not similar as the corresponding sides are not proportional.

(vi) In ∆DEF,

∠D +∠E +∠F = 180º

(Sum of the measures of the angles of a triangle is 180º.)

70º + 80º +∠F = 180º

∠F = 30º

Similarly, in ∆PQR,

∠P +∠Q +∠R = 180º

(Sum of the measures of the angles of a triangle is 180º.)

∠P + 80º +30º = 180º

∠P = 70º

In ∆DEF and ∆PQR,

∠D = ∠P (Each 70°)

∠E = ∠Q (Each 80°)

∠F = ∠R (Each 30°)

∴ ∆DEF ∼ ∆PQR [By AAA similarity criterion]

Question 10:

The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.

Answer:

Let us consider the following figure for the given question.

Draw a line OE || AB

In ΔABD, OE || AB

By using basic proportionality theorem, we obtain

AE/ED = BO/OD        (1)

However, it is given that

AO/OC = OB/OD      (2)

From equation (1) and (2) , we obtain

AE/ED = AO/OC

⇒ EO || DC [By the converse of basic proportionality theorem]

⇒ AB || OE || DC

⇒ AB || CD

∴ ABCD is a trapezium.

Question 9:

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.

Answer:

Draw a line EF through point O, such that EF ∥ CD

In ΔADC, EO ∥ CD

By using basic proportionality theorem, we obtain

AE/ED = AO/OC   (1)

In ΔABD, OE ∥ AB

So, by using basic proportionality theorem, we obtain

Question 8:

Using Converse of basic proportionality theorem, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Answer:

Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively.

i.e., AP = PB and AQ = QC

It can be observed that

AP/PB = 1/1

and AQ/QC = 1/1

∴ AP/PB = AQ/QC

Hence, by using basic proportionality theorem, we obtain PQ ∥ BC.

Question 7:

Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Answer:

Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such that PQ ∥ BC

By using basic proportionality theorem, we obtain

AQ/QC = AP/PB

AQ/QC = 1/1 (P is the mid – point of AB. ∴ AP = PB)

⇒ AQ = QC

Or, Q is the mid-point of AC.

Question 6:

In the following figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Answer:

In Δ POQ, AB || PQ

∴ OA/AP = OB/BQ (Basic proportionality theorem)   (i)

Question 5:

In the following figure, DE || OQ and DF || OR, show that EF || QR.

Answer:

In Δ POQ, DE || OQ

Question 4:

In the following figure, DE || AC and DF || AE. Prove that

Answer:

In ΔABC, DE || AC

∴ BD/DA = BE/EC    (Basic Proportionality Theorem)     (i)

Question 3:

In the following figure, if LM || CB and LN || CD, prove that

Answer:

In the given figure, LM || CB

By using basic proportionality theorem, we obtain

Question 2:

E and F are points on the sides PQ and PR respectively of a ;PQR. For each of the following cases, state whether EF || QR.

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

(iii)PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm

Answer:

(i)

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm

(ii)

PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm

(iii)

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm

Question 1:

In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

(i)

(ii)

Answer:

(i)

Let EC = x cm

It is given that DE || BC.

By using basic proportionality theorem, we obtain

(ii)

Let AD = x cm

It is given that DE || BC.

By using basic proportionality theorem, we obtain

Question 3:

State whether the following quadrilaterals are similar or not:

Answer:

Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional, i.e. 1:2, but their corresponding angles are not equal.

Question 2:

Give two different examples of pair of

(i) Similar figures      (ii)Non-similar figures

Answer:

(i) Two equilateral triangles with sides 1 cm and 2 cm

 Two squares with sides 1 cm and 2 cm

(ii) Trapezium and square

Triangle and parallelogram

Question 1:

Fill in the blanks using correct word given in the brackets:−

(i) All circles are __________. (congruent, similar)

(ii) All squares are __________. (similar, congruent)

(iii) All __________ triangles are similar. (isosceles, equilateral)

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)

Answer:

(i) Similar

(ii) Similar

(iii) Equilateral

(iv) (a) Equal

(b) Proportional

Question 5:

A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete.

Each step has a rise of 1/4 m and a tread of 1/2 m (See figure) calculate the total volume of concrete required to build the terrace.

Answer:

From the figure, it can be observed that

1st step is 1/2 m wide,

2nd step is 1 m wide,

3rd step is 3/2 m wide.

Therefore, the width of each step is increasing by 1/2 m each time whereas their height 1/4 m and length 50 m remains the same.

Therefore, the widths of these steps are 1/2, 1, 3/2, 2, …..

Volume of concrete in 1st step

= (1/4) × (1/2) × 50 = 25/4

Volume of concrete in 2nd step

= (1/4) × (1) × 50 = 25/2

Volume of concrete in 3rd step

= (1/4) × (3/2) × 50 = 75/4

It can be observed that the volumes of concrete in these steps are in an A.P.

Volume of concrete required to build the terrace is 750 m³.

Question 4:

The houses of a row are number consecutively from 1 to 49. Show that there is a value of x such that the sum of numbers of the houses preceding the house numbered x is equal to the sum of the number of houses following it.

Find this value of x.

[Hint S_{x − 1} = S₄₉ − S_x]

Answer:

The number of houses was 1, 2, 3 … 49

It can be observed that the number of houses are in an A.P. having a as 1 and d also as 1.

Let us assume that the number of x^th house was like this.

We know that,

Sum of n terms in an A.P. 

Sum of number of houses preceding x^th house = S_{x − 1}

_{It is given that these sums are equal to each other.}

However, the house numbers are positive integers.

The value of x will be 35 only.

Therefore, house number 35 is such that the sum of the numbers of houses preceding the house numbered 35 is equal to the sum of the numbers of the houses following it.

Question 3:

A ladder has rungs 25 cm apart. (See figure). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are  m apart, what is the length of the wood required for the rungs?

[Hint: number of rungs = 250/25]

Answer:

It is given that the rungs are 25 cm apart and the top and bottom rungs are  m apart.

∴ Total number of rungs 

Now, as the lengths of the rungs decrease uniformly, they will be in an A.P.

The length of the wood required for the rungs equals the sum of all the terms of this A.P.

First term, a = 45

Last term, l = 25

n = 11

Therefore, the length of the wood required for the rungs is 385 cm.

Question 2:

The sum of the third and the seventh terms of an A.P is 6 and their product is 8. Find the sum of first sixteen terms of the A.P.

Answer:

We know that,

a_n = a + (n − 1) d

a₃ = a + (3 − 1) d

a₃ = a + 2d

Similarly, a₇ = a + 6d

Given that, a₃ + a₇ = 6

(a + 2d) + (a + 6d) = 6

2a + 8d = 6

a + 4d = 3

a = 3 − 4d (i)

Also, it is given that (a₃) × (a₇) = 8

(a + 2d) × (a + 6d) = 8

From equation (i),

(3 – 4d + 2d) × (3 – 4d + 6d) = 8

(3 – 2d) × (3 + 2d) = 8

9 – 4d² = 8

4d² = 9 – 8 = 1

d² = ¼

d = ± ½

d = ½ or – ½

From equation (i)

(When d is ½)

a = 3 – 4d

a = 3 – 4(1/2)

 = 3 – 2 = 1

(When d is −1/2)

Question 1:

Which term of the A.P. 121, 117, 113 … is its first negative term? [Hint: Find n for a_n < 0]

Answer:

Given A.P. is 121, 117, 113 …

a = 121

d = 117 − 121 = −4

a_n = a + (n − 1) d

= 121 + (n − 1) (−4)

= 121 − 4n + 4

= 125 − 4n

We have to find the first negative term of this A.P.

Therefore, a_n < 0

125 − 4n < 0

125 < 4n

n > 125/4

n > 31.25

Therefore, 32^nd term will be the first negative term of this A.P.

Question 20:

In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance
the competitor has to run?

[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 ×(5 + 3)]

Answer:

The distances of potatoes are as follows.

5, 8, 11, 14…

It can be observed that these distances are in A.P.

a = 5

d = 8 − 5 = 3

= 5[10 + 9 × 3]

= 5(10 + 27) = 5(37)

= 185

As every time she has to run back to the bucket, therefore, the total distance that the competitor has to run will be two times of it.

Therefore, total distance that the competitor will run = 2 × 185 = 370 m

Alternatively,

The distances of potatoes from the bucket are 5, 8, 11, 14…

Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept. Therefore,
distances to be run are

10, 16, 22, 28, 34,……….

a = 10

d = 16 − 10 = 6

S₁₀ =?

= 5[20 + 54]

= 5 (74)

= 370

Therefore, the competitor will run a total distance of 370 m.