Category: CBSE

NCERT Exemplar Class 9 Maths Chapter 6 Lines And Angles

NCERT Exemplar Class 9 Maths Chapter  Exercise 6.1

Write the correct answer in each of the following:

Question 1.
In the given figure, if AB || CD || EF, PQ || RS, ∠RQD = 25° and ∠CQP = 60°, then ∠QRS is equal to

(A) 85°
(B) 135°
(C) 145°
(D) 110°
Solution:
(C): Produced PQ to X which intersect AB at Y.
PQ || RS ⇒ PX || RS

∵ AB || CD and PX is transversal
∠CQP + ∠AYX = 180° [Co-exterior angles]
⇒ 60° + ∠AYX = 180°
⇒ ∠AYX = 180° – 60°
⇒ ∠AYX = 120°
∵ PX || RS and AB is transversal
∠YRS = ∠AYX [Corresponding angles]
⇒ ∠YRS = 120° …(i)
and AB || CD and RQ is transversal
∠YRQ = ∠RQD [Alternate interior angles]
⇒ ∠YRQ = 25° …(ii)
Now, ∠SRQ = ∠SRY + ∠YRQ
⇒ ∠SRQ = 120° + 25° [From (i) and (ii)]
⇒ ∠SRQ = 145°

Question 2.
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
(A) an isosceles triangle
(B) an obtuse triangle
(C) an equilateral triangle
(D) a right triangle
Solution:
(D): Let the angles of a ∆ABC be ∠A, ∠B and ∠C.
We have given, ∠A = ∠B + ∠C …(i)
In ∆ABC, ∠A + ∠B + ∠C = 180° …(ii)
[Angle sum property of a triangle]
From (i) and (ii),
∠A + ∠A = 180°
⇒ 2∠A = 180°
⇒ ∠A = 90°
Hence, the triangle is a right triangle.

Question 3.
An exterior angle of a triangle is 105° and its two interior opposite angles are equal. Each of these equal angles is

Solution:
(B) : Let one of the interior opposite angle of a triangle be x.
∵ Exterior angle
= sum of two opposite interior angles
∴ x + x = 105° [ ∵ Exterior angle = 105° and two interior opposite angles are equal]
⇒ 2x = 105° ⇒ x = \(52 \frac{1}{2}\)°.
Thus, each of the required angles of a triangle \(52 \frac{1}{2}\)°.

Question 4.
The angles of a triangle are in the ratio 5 : 3 : 7. The triangle is
(A) an acute angled triangle
(B) an obtuse angled triangle
(C) a right triangle
(D) an isosceles triangle
Solution:
(A): We have given, the ratio of angles of a triangle is 5 : 3 : 7.
Let angles of a triangle be ∠A, ∠B and ∠C, where ∠A = 5x, ∠B = 3x and ∠C = 7x Now in ∆ABC, ∠A + ∠B + ∠C = 180°
[Angle sum property of a triangle]
∴ 5x + 3x + 7x = 180°
⇒ 15x = 180°
So, ∠A = 5 × 12° = 60°, ∠B = 3 × 12° = 36° and ∠C = 7 × 12° = 84°
∵ All angles are less than 90°, hence the triangle is an acute angled triangle.

Question 5.
If one of the angles of a triangle is 130°, then the angle between the bisectors of the other two angles can be
(A) 50°
(B) 65°
(C) 145°
(D) 155°
Solution:
(D) : Let angles of a triangle be ∠A, ∠B and ∠C, where ∠A = 130°

Question 6.
In the given figure, POQ is a line. The value of x is

(A) 20°
(B) 25°
(C) 30°
(D) 35°
Solution:
(A) :

Since, POQ is a line segment.
∴ ∠POQ = 180°
⇒ ∠POA + ∠AOB + ∠Solution:Q = 180°
⇒ 40° + 4x + 3x = 180°
⇒ 7x = 180° – 40° ⇒ 7x = 140° ⇒ x = 20°

Question 7.
In the given figure, if OP || RS, ∠OPQ = 110° and ∠QRS = 130°, then ∠PQR is equal to

(A) 40°
(B) 50°
(C) 60°
(D) 70°
Solution:
(C) : Draw a line EF parallel to RS through point Q

∵ OP || RS [Given]
⇒ EF || RS [Construction]
∴ OP || EF and PQ is a transversal
⇒ ∠OPQ = ∠PQF [Alternate interior angles]
⇒ ∠PQF = 110° [ ∵ ∠OPQ = 110°]
⇒ ∠PQR + ∠RQF = 110° … (i)
Now, RS || EF and RQ is a transversal
⇒ ∠QRS + ∠RQF = 180° [Co-interior angles]
⇒ 130° + ∠RQF = 180°
⇒ ∠RQF = 180° – 130° = 50°
Now from (i), we have
⇒ ∠PQR + 50° = 110°
⇒ ∠PQR = 110° – 50°
⇒ ∠PQR = 60°

Question 8.
Angles of a triangle are in the ratio 2 : 4 : 3. The smallest angle of the triangle is
(A) 60°
(B) 40°
(C) 80°
(D) 20°
Solution:
(B) : We have given, the ratio of angles of a triangle is 2 : 4 : 3.
Let the angles of a triangle be ∠A, ∠B and ∠C, where ∠A = 2x, ∠B = 4x and ∠C = 3x
Now in ∆ ABC, ∠A + ∠B + ∠C = 180°
[Angle sum property of a triangle]
⇒ 2x + 4x + 3x = 180°
⇒ 9x = 180° ⇒ x = 20°
∠A = 2 × 20° = 40°, ∠B = 4 × 20° = 80° and ∠C = 3 × 20° = 60°
Thus, the smallest angle of the triangle is 40°.

NCERT Exemplar Class 9 Maths Chapter  Exercise 6.2

Question 1.
For what value of x + y in the given figure will ABC be a line? Justify your answer.

Solution:
For ABC to be a line, the sum of the two adjacent angles must be 180° i.e., x + y = 180°.

Question 2.
Can a triangle have all angles less than 60°? Give reason for your answer.
Solution:
No, because if all angles will be less than 60°, then their sum will not be equal to 180° and that will not be a triangle.

Question 3.
Can a triangle have two obtuse angles? Give reason for your answer.
Solution:
No, because if two angles will be more than 90°, then the sum of angles will not be equal to 180°.

Question 4.
How many triangles can be drawn having its angles as 45°, 64° and 72°? Given reason for your answer.
Solution:
The sum of given angles = 45° + 64° + 72° = 181° * 180°.
Thus the sum of all three angles is not equal to 180°. So, no triangle can be drawn with the given angles.

Question 5.
How many triangles can be drawn having its angles as 53°, 64° and 63°? Give reason for your answer.
Solution:
Since, the sum of given angles = 53°+ 64° + 63° = 180°
Thus, we see that the sum of all interior angles of a triangle is 180°. So, we can draw many triangles of the given angles with different sides. Hence, infinitely many triangles can be drawn.

Question 6.
In the given figure, find the value of x for which the lines l and m are parallel.

Solution:
We have given, l || m and a transversal line n,
∴ x + 44° = 180° [Co-interior angles]
⇒ x = 180° – 44° ⇒ x = 136°

Question 7.
Two adjacent angles are equal. Is it necessary that each of these angles will lie a right angle? Justify your answer.
Solution:
No, because each of the two adjacent angles will be right angles only if they will form a linear pair.

Question 8.
If one of the angles formed by two intersecting lines is a right angle, what can you say about the other three angles? Give reason for your answer.
Solution:
Let two lines AB and CD intersect each other at a right angle.
Let ∠AOC = 90°
∠AOC + ∠AOD = 180° [Linear pair]
⇒ ∠AOD = 180° – 90° = 90°
Now, ∠COA = ∠DOB = 90°
[Vertically opposite angles]
and ∠AOD = ∠COB = 90°
[Vertically opposite angles]
Hence, each of the other three angles are right angles.

Question 9.
In the given figure, which of the two lines are parallel and why?

Solution:
Left side figure shows the sum of two interior angles = 132° + 48° = 180° because the sum of two interior angles on the same side of a transversal line n is 180°, thus l ||m.
Right side figure shows the sum of two interior angles = 73° + 106° = 179° ≠ 180° because the sum of two interior angles on the same side of a transversal line r is not equal to 180°, thus p is not parallel to q.

Question 10.
Two lines l and m are perpendicular to the same line n. Are l and m perpendicular to each other? Give reason for your answer.
Solution:
No

Given that the lines l and m are perpendicular to the line n.
∴ ∠1 = ∠2 = 90°
This shows that the corresponding angles are equal.
Thus, l || m.

NCERT Exemplar Class 9 Maths Chapter  Exercise 6.3

Question 1.
In the given figure, OD is the bisector of ∠AOC, OE is the bisector of ∠BOC and OD ⊥ OE. Show that the points A, O and B are collinear.

Solution:
Since, OD and OE bisects ∠AOC and ∠BOC, respectively.
∴ ∠AOC = 2∠DOC …(i)
and ∠COB = 2 ∠COE …(ii)
On adding (i) and (ii), we get
∠AOC + ∠COB = 2 ∠DOC + 2 ∠COE
⇒ ∠AOC + ∠COB = 2 (∠DOC + ∠COE)
⇒ ∠AOC + ∠COB = 2 ∠DOE
⇒ ∠AOC + ∠COB = 2 × 90° [∵ OD ⊥ OE]
⇒ ∠AOC + ∠COB = 180°
As, adjacent angles ∠AOC and ∠COB form a linear pair.
∴ AOB is a straight line.
Thus, points A, O and B are collinear.

Question 2.
In the given figure, ∠1 = 60° and ∠6 = 120°. Show that the lines m and n are parallel.

Solution:
We have ∠1 = 60° and ∠6 = 120°
∠1 = ∠3 = 60° [Vertically opposite angles]
Now, ∠3 + ∠6 = 60° + 120°
⇒ ∠3 + ∠6 = 180°
Since the above shows that the sum of two interior angles on same side of a transversal line l is 180°, thus m || n

Question 3.
AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ.

Solution:
Since, l || m and t is a transversal line.
∴ ∠EAB = ∠ABH [Alternate interior angles]

[On dividing Solution:th sides by 2]
⇒ ∠PAB = ∠ABQ
[∵ AP and BQ are the bisectors of ∠EAB and ∠ABH] Since, ∠PAB and ∠ABQ are alternate interior angles formed by lines AP and BQ and transversal AB. Thus, AP || BQ.

Question 4.
If in the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m.

Solution:
Since, AP || BQ and t is a transversal, therefore, ∠PAB = ∠ABQ
[Alternate interior angles]
⇒ 2 ∠PAB = 2 ∠ABQ
[On multiplying both sides by 2]

⇒ ∠CAB = ∠ABF [ ∵ AP and BQ are the bisectors of ∠CAB and ∠ABF]
Since, ∠CAB and ∠ABF are alternate interior angles formed by lines l and m and transversal t. Thus, l || m.

Question 5.
In given figure, BA || ED and BC || EF. Show that ∠ABC = ∠DEF.
[Hint: Produce DE to intersect BC at P (say)].

Solution:
Let us produce DE, which meets BC at P.

Since BA || ED ⇒ BA || DP
∴ ∠ABP = ∠EPC [Corresponding angles]
or ∠ABC = ∠EPC …(i)
Again, BC || EF or PC || EF
∴ ∠DEF = ∠EPC …(ii)
[Corresponding angles] From (i) and (ii), we get
∠ABC = ∠DEF

Question 6.
In the given figure, BA || ED and BC || EF. Show that ∠ABC + ∠DEF = 180°.

Solution:
Let us produce FE, which meets AB at P.


BC || EF ⇒ BC || PF
∴ ∠EPB + ∠PBC = 180° …(i)
[Co-interior angles]
Now, AB || ED and PF is a transversal.
∴ ∠EPB = ∠DEF …(ii)
[Corresponding angles]
From (i) and (ii), we get
∠DEF + ∠PBC = 180°
⇒ ∠ABC + ∠DEF = 180° [ ∵ ∠PBC = ∠ABC]

Question 7.
In the given figure, DE || QR and AP and BP are bisectors of ∠EAB and ∠RBA, respectively. Find ∠APB.

Solution:
DE || QR and AB is a transversal
∴ ∠EAB + ∠RBA = 180° (Co-interior angles)

[On dividing both sides by 2]
∵ AP and BP are the bisectors of ∠EAB and ∠RBA, respectively.

Using these in (i), we have
⇒ ∠BAP + ∠ABP = 90° …(ii)
In ∆APB, ∠BAP + ∠ABP + ∠APB = 180°
[Angle sum property of a triangle]
⇒ 90° + ∠APB = 180° [From (ii)]
⇒ ∠APB = 90°

Question 8.
The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles of the triangle.
Solution:
Let the angles of a triangle be 2x, 3x and 4x. Since sum of all angles of a triangle is 180°.
∴ 2x + 3x + 4x = 180°

∴ The required three angles are 2 × 20° = 40°, 3 × 20° = 60° and 4 × 20° = 80°.

Question 9.
A triangle ABC is right angled at A. L is a point on BC such that AL ⊥ BC. Prove that ∠BAL = ∠ACB.
Solution:
In ∆ABC and ∆ALB,
∠BAC = ∠ALB [Each 90°]…(i)
and ∠ABC = ∠ABL [Common angle]…(ii)

On adding (i) and (ii), we get
∠BAC + ∠ABC = ∠ALB + ∠ABL …(iii)
Again, in ∆ABC,
∠BAC + ∠ACB + ∠ABC = 180°
[Angle sum property of a triangle]
= ∠BAC + ∠ABC = 180° – ∠ACB …(iv)
In ∆ABE,
∠ABL + ∠ALB + ∠BAL = 180°
[Angle sum property of a triangle]
⇒ ∠ABL + ∠ALB = 180° – ∠BAL …(v)
On substituting the values from (iv) and (v) in (iii), we get
180° – ∠ACB = 180° – ∠BAL
⇒ ∠ACB = ∠BAL

Question 10.
Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.
Solution:
Let two lines m and n are parallel and p and q are respectively perpendicular to m and n.
Since, p ⊥ m ⇒ ∠1 = 90° …(i)
Also, q ⊥ m ⇒ ∠2 = 90° …(ii)
[Since, m || n and q ⊥ n ⇒ q ⊥ m]

From (i) and (ii), we have
Z₁ = Z₂ = 90°
As p and q are two lines and m is transversal. Also corresponding angles ∠1 and ∠2 are equal.
Thus, p || q.

NCERT Exemplar Class 9 Maths Chapter  Exercise 6.4

Question 1.
If two lines intersect, prove that the vertically opposite angles are equal.
Solution:
Let the two lines AB and CD intersect at a point O.
Since, ray OA stands on line CD.
∴ ∠AOC + ∠AOD = 180° [Linear pair]…(i)
Since, ray OD stands on line AB.
∠AOD + ∠AOD = 180° [Linear pair] …(ii)

From (i) and (ii), we get
∠AOC + ∠AOD = ∠AOD + ∠BOD
⇒ ∠AOC + ∠BOD
So, vertically opposite angles ∠AOC and ∠BOD are equal.
Also, ray OB stands on line CD.
∴ ∠DOB + ∠BOC = 180° [Linear pair] …(iii)
From (ii) and (iii), we get
∠AOD + ∠BOD = ∠DOB + ∠BOC
⇒ ∠AOD = ∠BOC
Thus, vertically opposite angles ∠AOD and ∠BOC are equal.

Question 2.
Bisectors of interior ∠B and exterior ∠ACD of a ∆ABC intersect at the point T. Prove that
∠BTC= \(\frac{1}{2}\)∠BAC.
Solution:
In ∆ABC, produce BC to D and the bisectors of ∠ABC and ∠ACD meet at point T.

In ∆ABC,
∠ACD = ∠ABC + ∠CAB
[Exterior angle property of a triangle]

Question 3.
A transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angles so formed are parallel.
Solution:
Let AB and CD are two parallel lines and intersected by a transversal GH at P and Q, respectively. Also, let EP and FQ are the bisectors of corresponding angles ∠APG and ∠CQP, respectively.

Since AB || CD
⇒ ∠APG = ∠CQP [Corresponding angles]

[On dividing both sides by 2]
⇒ ∠EPG = ∠FQP
[∵ EP and FQ are the bisectors of ∠APG and ∠CQP, respectively]
As, these are the corresponding angles made by the lines EP and FQ and transversal line GH.
EP || FQ

Question 4.
Prove that through a given point, we can draw only one perpendicular to a given line.
[Hint: Use proof by contradiction].
Solution:
Let a line l and a point P.

Also let m and n are two lines passing through P and perpendicular to l.
In ∆APB,
∠A + ∠P + ∠B = 180°
[Angle sum property of a triangle]
⇒ 90° + ∠P + 90° = 180°
⇒ ∠P = 180°- 180°
⇒ ∠P = 0°
∴ Lines n and m coincide.
Thus, only one perpendicular line can be drawn through a given point on a given line.

Question 5.
Prove that two lines that are respectively perpendicular to two intersecting lines intersect each other.
[Hint: Use proof by contradiction].
Solution:
Let lines l and m are two intersecting lines. Again, let n and p be another two lines which are perpendicular to m and l respectively.

Let us assume that lines n and p are not intersecting, that means they are parallel to each other i.e., n||p …(i)
Since, lines n and p are perpendicular to m and l, respectively.
But from (i), n || p ⇒ l || m, which shows a contradiction.
So, our assumption was wrong.
Thus lines n and p intersect at a point.

Question 6.
Prove that a triangle must have atleast two acute angles.
Solution:
Let ∆ABC is a triangle.
We know that, the sum of all three angles is 180°.
∴ ∠A + ∠B + ∠C = 180° …(i)
Let us consider the following cases.
Case I: When two angles are 90°.
Suppose two angles ∠B = 90° and ∠C = 90°
So from (i), we get
∠A + 90° + 90° = 180°
⇒ ∠A = 180° -180° = 0
Thus, no triangle is possible.

Case II: When two angles are obtuse.
Suppose ∠B and ∠C are obtuse angles.
From (i), we get ∠A = 180° – (∠B + ∠C)
= 180° – (greater than 180°)
[ ∵ ∠B + ∠C = more than 90° + more than 90°]
⇒ ∠A = negative angle, which is not possible,
Thus, no triangle is possible.

Case III: When one angle is 90°.
Suppose ∠B = 90°.
From (i), ∠A + ∠B + ∠C= 180°
⇒ ∠A + ∠C = 180° – 90° = 90°
So, sum of other two angles are 90°, Hence, both angles are acute.

Case IV: When two angles are acute, then sum of two angles is less than 180°, so that the third angle may be acute or obtuse.
Thus, a triangle must have atleast two acute angles.

Question 7.
In the given figure, ∠Q > ∠R, PA is the bisector of ∠QPR and PM ⊥ QR. Prove that ∠APM = \(\frac{1}{2}\)(∠Q – ∠R).

Solution:
Since, PA is the bisector of ∠QPR.
∴ ∠ QPA = ∠ APR
In ∆PQM, ∠PQM + ∠PMQ + ∠QPM = 180°
[Angle sum property of a triangle]
⇒ ∠PQM + 90° + ∠QPM = 180°
[ ∵ PM ⊥ QR
⇒ ∠PMQ = 90°]
⇒ ∠PQM = 90° – ∠QPM …(ii)
In ∆PMR,
∠PMR + ∠PRM + ∠RPM = 180°
[Angle sum property of a triangle]
⇒ 90° + ∠PRM + ∠RPM = 180°
[∵ PM ⊥ QR ⇒ ∠PMR = 90°]
⇒ ∠PRM = 180° – 90° – ∠RPM
⇒ ∠PRM = 90° – ∠RPM …(iii)
On subtracting (iii) from (ii), we get
∠Q – ∠R = (90° – ∠QPM) – (90° – ∠RPM)
[ ∵ ∠PQM = ∠Q and ∠PRM = ∠R]
⇒ ∠Q – ∠R = ∠RPM – ∠QPM
⇒ ∠Q – ∠R = [∠RPA + ∠APM] -[∠QPA – ∠APM]
⇒ ∠Q – ∠R = ∠RPA + ∠APM – ∠QPA + ∠APM
⇒ ∠Q – ∠R = 2∠APM [By using (i)]
∠APM = \(\frac{1}{2}\) (∠Q – ∠R)

NCERT Exemplar Class 9 Maths

NCERT Exemplar Class 10 Maths Chapter 6 Triangles

NCERT Exemplar Class 10 Maths Chapter 6 Exercise 6.1

Choose the correct answer from the given four options:

Question 1.
In the figure, if ∠BAC = 90° and AD ⊥ BC. Then,

(A) BD . CD = BC²
(B) AB . AC = BC²
(C) BD . CD = AD²
(D) AB . AC = AD²
Solution:
(C)
In ∆ABC,
∠B + ∠BAC + ∠C = 180°
⇒ ∠B + 90° + ∠C = 180°
⇒ ∠B = 90° – ∠C
Similarly, In ∆ADC, ∠D AC = 90° – ∠C
In ∆ADB and ∆ADC,
∠D = ∠D = 90°
∠DBA = ∠D AC [each equal to (90° – ∠C)
∴ ∆ADB ~ ∆CDA
[by AA similarity criterion]
∴ \(\frac{B D}{A D}=\frac{A D}{C D}\)
⇒ BD . CD = AD²

Question 2.
The lengths of the diagonals of a rhombus are 16 cm and 12 cm. Then, the length of the side of the rhombus is
(A) 9 cm
(B) 10 cm
(C) 8 cm
(D) 20 cm
Solution:
(B)
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
Given, AC = 16 cm and BD = 12 cm
∴ AO = 8 cm, BO = 6 cm and ∠AOB = 90°

In right angled ∆AOB,
AB² = AO² + OB² [by Pythagoras theorem]
⇒ AB² = 8² + 6² = 64 + 36 = 100
∴ AB = 10 cm

Question 3.
If ∆ABC ~ ∆EDFand ∆ABC is not similar to ∆DEF, then which of the following is not true?
(A) BC . EF = AC . FD
(B) AB . EF = AC . DE
(C) BC . DE = AB . EF
(D) BC . DE = AB . FD
Solution:
(C)
Given, ∆ABC ~ ∆EDF

Hence, option (B) is true.

Question 4.
If in two triangles ABC and PQR, \(\frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}\) then
(A) ∆PQR ~ ∆CAB
(B) ∆PQR ~ ∆ABC
(C) ∆CBA ~ ∆PQR
(D) ∆BCA ~ ∆PQR
Solution:
(A)
Given, in triangles ABC and PQR,
\(\frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}\)
which shows that sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are also equal, so by SSS similarity, triangles are similar i.e., ∆CAB ~ ∆PQR

Question 5.
In the figure, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠ PBA is equal to

(A) 50°
(B) 30°
(C) 60°
(D) 100°
Solution:
(D): In ∆APB and ∆CPD, ∠APB = ∠CPD = 50°
[vertically opposite angles]

∴ ∆APB ~ ∆DPC [by SAS similarity criterion]
∴ ∠A = ∠D = 30° [corresponding angles of similar triangles]
In ∆APB, ∠A + ∠B + ∠APB = 180° [sum of angles of a triangle = 180°]
⇒ 30° + ∠B + 50° = 180°
∴ ∠B = 180° – (50° + 30°) = 100°
i.e., ∠PBA = 100°

Question 6.
If in two triangles DEF and PQR, ∠D = ∠Q and ∠R = ∠E, then which of the following is not true?

Solution:
(B)
Given, in ∆DEF and ∆PQR, ∠D = ∠Q, ∠R = ∠E

∴ ∆DEF ~ ∆QRP            [by AAA similarity criterion]
⇒ ∠F = ∠P
[corresponding angles of similar triangles]
∴\(\frac{D F}{Q P}=\frac{E D}{R Q}=\frac{F E}{P R}\)

Question 7.
In ∆ABC and ∆DEF, ∠B = ∠E, ∠F = ∠C and AB = 3 DE. Then, the two triangles are
(A) congruent but not similar
(B) similar but not congruent
(C) neither congruent nor similar
(D) congruent as well as similar
Solution:
(B)
In ∆ABC and ∆DEF, ∠B = ∠E,
∠F = ∠C and AB = 3DE

We know that, if in two triangles corresponding two angles are same, then they are similar by AA similarity criterion.
Since, AB ≠ DE
Therefore ∆ABC and ∆DEF are not congruent.

Question 8.
It is given that ∆ABC ~ ∆PQR with \(\frac{B C}{Q R}=\frac{1}{3}\) then \(\frac { { ar }(\Delta PRQ) }{ { ar }(\Delta BCA) } \) equal to
(A) 9
(B) 3
(C) \(\frac{1}{3}\)
(D) \(\frac{1}{9}\)
Solution:
(A)
Given, ∆ABC ~ ∆QR and \(\frac{B C}{Q R}=\frac{1}{3}\)
We know that, the ratio of the areas of two similar triangles is equal to square of the ratio of their corresponding sides.

Question 9.
It is given that ∆ABC ~ ∆DFE, ∠A =30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF= 7.5 cm. Then, the following is true:
(A) DE= 12 cm, ∠F= 50°
(B) DE= 12 cm, ∠F= 100°
(C) EF= 12 cm, ∠D = 100°
(D) EF= 12 cm, ∠D = 30°
Solution:
(B)
Given, ∆ABC ~ ∆DFE, then ∠A = ∠D = 30°, ∠C = ∠E = 50°

Hence, DE = 12 cm, ∠F = 100°

Question 10.
If in ∆ABC and ∆DEF, \(\frac{A B}{D E}=\frac{B C}{F D}\), then they will be similar, when
(A) ∠B = ∠E
(B) ∠A = ∠D
(C) ∠B = ∠D
(D) ∠A = ∠F
Solution:
(C)
Given, in ∆ABC and ∆EDF,

So, ∆ABC ~ ∆EDF if ∠B = ∠D [By SAS similarity criterion]

Question 11.
If ∆ABC ~ ∆QRP, \(\frac { { ar }(\Delta ABC) }{ ar(\Delta PQR) } =\frac { 9 }{ 4 } \), AB= 18 cm and BC = 15 cm, then PR is equal to
(A) 10 cm
(B) 12 cm
(C) \(\frac{20}{3}\) cm
(D) 8 cm
Solution:
(A)
Given, ∆ABC ~ ∆QRP, AB = 18 cm and BC = 15 cm

Question 12.
If S is a point on side PQ of a ∆PQR such that PS = QS = RS, then
(A) PR – QR = RS²
(B) QS² + RS² = QR²
(C) PR² + QR² = PQ²
(D) PS² + RS² = PR²
Solution:
(C)
Given, in ∆PQR,
PS = QS = RS …………. (i)
In ∆PSR, PS = RS [from Eq(i)]
⇒ ∠1 = ∠2 ………… (ii)
[Angles opposite to equal sides are equal]

Similarly, in ∆RSQ, RS = SQ
⇒ ∠3 = ∠4 …………. (iii)
[angles opposite to equal sides are equal]
Now, in ∆PQR, sum of angles = 180°
⇒ ∠P + ∠Q + ∠P = 180°
⇒ ∠2 + ∠4 + ∠1 + ∠3 = 180°
⇒ ∠1 + ∠3 + ∠1 + ∠3 = 180°
⇒ 2(∠1 + ∠3) = 180°
⇒ ∠l + ∠3 = \(\frac{180^{\circ}}{2}\) = 90°
∴ ∠R = 90°
In ∆PQR, by Pythagoras theorem,
PR² + QR² = PQ²

NCERT Exemplar Class 10 Maths Chapter 6 Exercise – 6.2

Question 1.
Is the triangle with sides 25 cm, 5 cm and 24 cm a right triangle? Give reasons for your answer.
Solution:
False
Let a = 25 cm, b = 5 cm and c = 24 cm
Now, b² + c² = (5)² + (24)²
= 25 + 576 = 601 ≠ (25)²
Hence, given sides do not make a right triangle because it does not satisfy the property of Pythagoras theorem.

Question 2.
It is given that ∆DEF ~ ∆RPQ. Is it true to say that ∠D = ∠R and ∠F = ∠P? Why?
Solution:
False
We know that, if two triangles are similar, then their corresponding angles are equal.
∴ ∠D = ∠R, ∠E = ∠P and ∠F = Q

Question 3.
A and B are respectively the points on the sides PQ and PR of A PQR such that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm. Is 4B||Q/?? Give reasons for your answer.
Solution:
True
Given, PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm

and \(\frac{P B}{B R}=\frac{4}{6}=\frac{2}{3}\)
From Eqs. (i) and (ii), \(\frac{P A}{A Q}=\frac{P B}{B R}\)
By converse of basic proportionality theorem, AB || QR

Question 4.
In the figure, BD and CE intersect each other at the point P. Is A∆PBC ~ ∆PDE?Why?

Solution:
True
In ∆PBC and ∆PDE,
∠BPC = ∠EPD [vertically opposite angles]

Since, one angle of ∆PBC is equal to one angle of ∆PDE and the sides including these angles are proportional, so both triangles are similar.
Hence, ∆PBC ~ ∆PDE, by SAS similarity criterion.

Question 5.
In ∆PQR and ∆MST, ∠P = 55°, ∠Q = 25°, ∠M = 100° and ∠S = 25°. Is ∆QPR ~ ∆TSM? Why?
Solution:
False
We know that, the sum of three angles of a triangle is 180°.
In ∆PQR, ∠P +∠Q +∠R = 180°
⇒ 55° + 25 ° + ∠R = 180°
⇒∠R = 180° – (55° + 25 °)
= 180° – 80° = 100°
In ∆TSM, ∠T + ∠S + ∠M = 180°
⇒ ∠T + ∠25° + 100° = 180°
⇒ ∠T = 180° – (25° + 100°) = 180° – 125° = 55°

In ∆PQR and ∆TSM,
∠P = ∠T, ∠Q = ∠S and ∠R = ∠M
∴ ∠PQR = ∠TSM
[since, all corresponding angles are equal]
Hence, ∆QPR is not similar to ∆TSM, since correct correspondence is P ↔ T, Q ↔ S and R ↔ M.

Question 6.
Is the following statement true? Why? “Two quadrilaterals are similar, if their corresponding angles are equal”.
Solution:
False
Two quadrilaterals are similar if their corresponding angles are equal and corresponding sides must also be proportional.

Question 7.
Two sides and the perimeter of one triangle are respectively three times the corresponding sides and the perimeter of the other triangle. Are the two triangles similar? Why?
Solution:
True
Here, the corresponding two sides and the perimeters of two triangles are proportional, then the third side of both triangles will also in proportion.

Question 8.
If in two right triangles, one of the acute angles of one triangle is equal to an acute angle of the other triangle, can you say that the two triangles will be similar? Why?
Solution:
True
Let two right angled triangles be ∆ABC and ∆PQR

In which, ∠A = ∠P = 90° and ∠B = ∠Q = acute angle (Given)
Then, by AA similarity criterion, ∆ABC ~ ∆PQR

Question 9.
The ratio of the corresponding altitudes of two similar triangles is \(\frac{3}{5}\).Is it correct to say that ratio of their areas is \(\frac{6}{5}\) ? Why?
Solution:
False
Ratio of corresponding altitudes of two triangles having areas A₁ and A₂ respectively is \(\frac{3}{5}\).
By the property of area of two similar triangles,
\(\Rightarrow\left(\frac{A_{1}}{A_{2}}\right)=\left(\frac{3}{5}\right)^{2} \Rightarrow \frac{9}{25} \neq \frac{6}{5}\)
So, the given statement is not correct.

Question 10.
D is a point on side QR of ∆PQR such that PD ⊥ QR. Will it be correct to say that ∆PQD ~ A∆RPD? Why?
Solution:
False
In ∆PQD and ∆RPD,
PD = PD [common side]
∠PDQ = ∠PDR [each 90°]

Here, no other sides or angles are equal, so we can say that ∆PQD is not similar to ∆RPD. But if ∠P = 90°, then ∠DPQ = ∠PRD
[each equal to 90° – ∠Q and by ASA similarity criterion, ∆PQD ~ ∆RPD]

Question 11.
In the figure, if ∠D = ∠C, then is it true that ∆ADE ~ ∆ACB? Why?

Solution:
True
In ∆ADE and ∆ACB,
∠A = ∠A [common angle]
∠D = ∠C [given]
∴ ∆ADE ~ ∆ACB [by AA similarity criterion]

Question 12.
Is it true to say that if in two triangles, an angle of one triangle is equal to an angle of another triangle and two sides of one triangle are proportional to the two sides of the other triangle, then the triangles are similar? Give reasons for your answer.
Solution:
False
Because, according to SAS similarity criterion, if one angle of a triangle is equal to an angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
Here, one angle and two sides of two triangles are equal but these sides not including equal angle, so given statement is not correct.

NCERT Exemplar Class 10 Maths Chapter 6 Exercise – 6.3

Question 1.
In a ∆PQR, PR² – PQ² = QR² and M is a point on side PR such that QM⊥ PR. Prove that QM² = PM × MR.
Solution:
Given, In ∆PQR,
PR² – PQ² = QR² and QM ⊥ PR
To prove : QM² = PM × MR
Proof : Since, PR² – PQ² = QR²
⇒ PR² = PQ² + QR²

So, ∆PQR is right angled triangle right angle at Q.
In ∆QMR and ∆PMQ, ∠M = ∠M [each 90°]
∠MQR = ∠QPM [each equal to 90° – ∠R]
∴ ∆QMR ~ ∆PMQ [by AA similarity criterion]
Now, using property of area of similar triangles, we get

Question 2.
Find the value of x for which DE || AB is given figure

Solution:
Given, DE || AB
∴ \(\frac{C D}{A D}=\frac{C E}{B E}\) [by basic proportionality theorem]
\(\frac{x+3}{3 x+19}=\frac{x}{3 x+4}\)
⇒ (x + 3)(3x + 4) = x (3x + 19)
⇒ 3x² + 4x + 9x + 12 = 3x² + 19x
⇒ 19x – 13x = 12
⇒ 6x = 12
∴ x = \(\frac{12}{6}\) = 2
Hence, the required value of x is 2.

Question 3.
In the figure, if ∠1 = ∠2and ∆NSQ ≅ ∆MTR, then prove that ∆PTS ~ ∆PRQ.

Solution:
Given ∆NSQ ≅ ∆MTR and ∠1 = ∠2
To prove : ∆PTS ~ ∆PRQ
Proof : Since, ∆NSQ ≅ ∆MTR
So, SQ = TR ………….. (i)
Also, ∠1 = ∠2 ⇒ PT = PS ………… (ii)
[since, sides opposite to equal angles are also equal]
From Eqs.(i) and (ii), \(\frac{P S}{S Q}=\frac{P T}{T R}\)
⇒ ST || QR            [by converse of basic proportionality theorem]
∴ ∠1 = ∠PQR          [Corresponding angles]
and ∠2 = ∠PRQ
In ∆PTS and ∆PRQ,
∠P = ∠P [common angles]
∠1 = ∠PQR
∠2 = ∠PRQ
∴ ∆PTS ~ ∆PRQ              [by AAA similarity criterion]

Question 4.
Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3 RS. Find the ratio of the areas of ∆POQ and ∆ROS.
Solution:
Given PQRS is a trapezium in which PQ || RS and PQ = 3RS

In ∆POQ and ∆ROS,
∠SOR = ∠QOP     [vertically opposite angles]
∠SRP = ∠RPQ     [alternate angles]
∴ ∆POQ ~ ∆ROS       [by AA similarity criterion]
By property of area of similar triangles,

Hence, the required ratio is 9 : 1

Question 5.
In the figure if AB || DC and AC and PQ intersect each other at the point O, prove that OA . CQ = OC . AP.

Solution:
Given AC an PQ intersect each other at point O and AB || DC
To prove : OA . CQ = OC . AP
Proof: In ∆AOP and ∆COQ,
∠AOP = ∠COQ [vertically opposite angles]
∠APO = ∠CQO
[since AB || DC and PQ is transversal, so alternate angles]
∴ ∆AOP ~ ∆COQ          [by AA similarity criterion]
Then, \(\frac{O A}{O C}=\frac{A P}{C Q}\)
[since, corresponding sides are proportional]
⇒ OA . CQ = OC . AP Hence proved.

Question 6.
Find the altitude of an equilateral triangle of side 8 cm.
Solution:
Let ABC be an equilateral triangle of side 8 cm i.e., AB = BC = CA = 8 cm
Draw altitude AD which is perpendicular to BC. Then, D is the mid-point of BC.
∴ BD = CD = \(\frac{1}{2}\) BC = \(\frac{8}{2}\) = 4 cm
Now, AB² = AD² + BD² [by Pythagoras theorem]

⇒ (8)² = AD² + (4)²
⇒ 64 = AD² + 16
⇒ AD² = 64 – 16 = 48
⇒ AD = \(\sqrt{48}\) = \(4 \sqrt{3}\) cm
Hence, altitude of an equilateral triangle is \(4 \sqrt{3}\) cm

Question 7.
If ∆ABC ~ ∆DEF, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, find the perimeter of ∆ABC.
Solution:
Given AB = 4 cm, DE = 6 cm and EF = 9 cm and FD = 12 cm
Also, ∆ABC ~ ∆DEF


Now, perimeter of ∆ABC = AB + BC + AC = 4 + 6 + 8 = 18 cm

Question 8.
In the figure, if DE || BC, find the ratio of ar(∆ADE) and ar(∆ECB).

Solution:
Given, DE || BC, DE = 6 cm and BC = 12 cm
In ∆ABC and ∆ADE,
∠ABC = ∠ADE         [corresponding angle]
and ∠A = ∠A         [common side]
∴ ∆ABC ~ ∆ADE          [by AA similarity criterion]

Let ar(∆ADE) = k, then ar(∆ABC) = 4k
Now, ar(∆ECB) = ar(ABC) – ar(ADE) = 4k – k = 3k
∴ Required ratio = ar(ADE): ar(DECB)
= k : 3k = 1 : 3

Question 9.
ABCD is a trapezium in which AB || DC and P, Q are points on AD and BC, respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD.
Solution:
Given, a trapezium ABCD in which AB || DC. P and Q are points on AD and BC, respectively such that PQ || DC. Thus,
⇒ AP = 42 cm.
Now; AD = AP + PD = 42 + 18 = 60
∴ AD = 60 cm

Question 10.
Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm², find the area of the larger triangle.
Solution:
Given, ratio of corresponding sides of two similar triangles is 2 : 3 or \(\frac{2}{3}\)
Area of smaller triangle = 48 cm²
By the property of area of two similar triangles,
Ratio of area of both triangles = (Ratio of their corresponding sides)²

Question 11.
In a ∆PQR, N is a point on PR, such that QN ⊥ PR. If PN . NR = QN², prove that ∠PQR = 90°
Solution:
Given, ∆PQR, N is a point on PR, such that QN ⊥ PR and PN . NR = QN²
To prove : ∠PQR = 90°
Proof: We have, PN . NR = QNc
⇒ PN . NR = QN . QN

and ∠PNQ = ∠RNQ [each equal to 90° ]
∴ ∆QNP ~ ∆RNQ        [by SAS similarity criterion]
Then, ∆QNP and ∆RNQ are equiangulars.
i.e., ∠PQN = ∠QRN
⇒ ∠RQN-∠QPN
On adding both sides, we get
∠PQN + ∠RQN = ∠QRN + ∠QPN
⇒ ∠PQR = ∠QRN + ∠QPN …………… (ii)
We know that, sum of angles of a triangle is 180°
In ∆PQR, ∠PQR + ∠QPR + ∠QRP = 180°
⇒ ∠PQR + ∠QPN + ∠QRN = 180°
[ ∵∠QPR = ∠QPN and ∠QRP = ∠QRN]
⇒ ∠PQR + ∠PQR = 180° [using Eq. (ii)]
⇒ 2∠PQR = 180°
⇒ ∠PQR = \(\frac{180^{\circ}}{2}\) = 90°
∴ ∠PQR = 90° Hence proved.

Question 12.
Areas of two similar triangles are 36 cm² and 100 cm² . If the length of a side of the larger triangle is 20 cm, find the length of the corresponding side of the smaller triangle.
Solution:
Given, area of smaller triangle = 36 cm²
and area of larger triangle= 100 cm²
Also, length of a side of the larger triangle = 20 cm
Let length of the corresponding side of the smaller triangle = x cm
By property of area of similar triangles,

Question 13.
In the figure, if ∠ACB = ∠CDA, AC = 8 cm and AD = 3 cm, find BD.

Solution:
Given, AC = 8 cm, AD = 3 cm
and ∠ACB = ∠CDA
In ∆ACD and ∆ABC,
∠A = ∠A       [Common angle]
∠ADC = ∠ACB      [Given]
∴ ∆ADC ~ ∆ACB      [By AA similarity criterion]

Question 14.
A 15 metres high tower casts a shadow 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.
Solution:
Let BC = 15 m be the tower and its shadow AB is 24 m. At that time ∠CAB = θ. Again, let EF = h be a telephone pole and its shadow DE = 16 m. At the same time ∠EDF = θ. Here, ∆ABC and ∆DEF both are right angled triangles.

Hence, the height of the point on the wall where the top of the ladder reaches is 8 m.

Question 15.
Foot of a 10 m long ladder leaning against a vertical wall is 6 m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.
Solution:
Let AB be a vertical wall and AC = 10 m is a ladder. The top of the ladder reached to A and distance of ladder from the base of the wall BC is 6 m.

In right angled ∆ABC
AC² = AB² + BC² [by Pythagoras theorem]
⇒ (10)² = AB² + (6)²
⇒ 100 = AB² + 36
⇒ AB² = 100 – 36 = 64
∴ AB = \(\sqrt{64}\) = 8 m
Hence the height of the point on th wall where the top of the ladder reaches is 8 m.

NCERT Exemplar Class 10 Maths Chapter 6 Exercise – 6.4

Question 1.
In the given figure, if ∠A = ∠C, AB = 6 cm, BP = 15 cm, AP = 12 cm and CP = 4 cm, then find the lengths of PD and CD.

Solution:
Given ∠A = ∠C, AB = 6 cm, BP = 15 cm,
AP = 12 cm and CP = 4 cm
In ∆APB and ∆CPD,
∠A = ∠C [given]
∠APB = ∠CPD [vertically opposite angles]
∴ ∆APB ~ ∆CPD [by AA similarity criterion]

Hence, length of PD is 5 cm and length of CD is 2 cm.

Question 2.
It is given that ∆ABC ~ ∆EDF such that AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm. Find the lengths of the remaining sides of the triangles.
Solution:
Given, ∆ABC ~ ∆EDF, so the corresponding sides of ∆ABC and ∆EDF are in the same ratio
i.e., \(\frac{A B}{E D}=\frac{A C}{E F}=\frac{B C}{D F}\) ………………. (i)

Hence, lengths of the remaining sides of the triangles are EF = 16.8 cm and BC = 6.25 cm.

Question 3.
Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.
Solution:
Let a ∆ABC in which a line DE parallel to BC intersects AB at D and AC at E.
To prove : DE divides the two sides in the same ratio.


Now, since, ∆BDE and ∆DEC lie between the same parallel lines DE and BC and on the same base DE
So, ar(∆BDE) = ar(∆DEC) ………….. (iii)
From Eqs. (i), (ii) and (iii),
\(\frac{A D}{D B}=\frac{A E}{E C}\)
Hence proved

Question 4.
In the figure, if PQRS is a parallelogram and AB || PS, then prove that OC || SR.

Solution:
Given PQRS is a parallelogram, so PQ || SR and PS || QR. Also AB || PS.
To prove : OC || SR
Proof : In ∆OPS and ∆OAB, PS || AB
∠POS = ∠AOB    [common angle]
∠OSP = ∠OBA        [corresponding angles]
∴ ∆OPS ~ ∆OAB           [by AA similarity criterion]
Then, \(\frac{P S}{A B}=\frac{O S}{O B}\) ………… (i)
In ∆CQE and ∆CAB, QR || PS || AB
∠QCR = ∠ACB [common angle]
∠CRQ = ∠CBA [corresponding angles]
∴ ∆CQR ~ ∆CAB

On subtracting 1 from both sides, we get OB CB

By converse of basic proportionality theorem, SR || OC.
Hence proved

Question 5.
A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
Solution:
Let AC be the ladder of length 5 m and BC = 4 m be the height of the wall, which ladder is placed. If the foot of the ladder is moved 1.6 m towards the wall i.e, AD = 1.6 m, then the ladder is slide upward i.e., CE = x m.
In right angled ∆ABC,
AC² = AB² + BC²     [by Pythagoras theorem]
⇒ (5)² = (AB)² + (4)²
⇒ AB² = 25 – 16 = 9
⇒ AB = 3m
Now, DB = AB – AD = 3 – 1.6 = 1.4 m

In right angled ∆EBD,
ED² = EB² + BD²       [by Pythagoras theorem]
⇒ (5)² = (EB)² + (1.4)² [ ∵ BD = 1.4 m]
⇒ 25 = (EB)² + 1.96
⇒ (EB)² = 25 – 1.96 = 23.04
⇒ EB = \(\sqrt{23.04}\) = 4.8
Now, EC = EB – BC = 4.8 – 4 = 0.8
Hence, the top of the ladder would slide upwards on the wall at distance is 0.8 m.

Question 6.
For going to a city B from city A, there is a route via city C such that AC ⊥ CB, AC = 2x km and CB = 2 (x + 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway.
Solution:
Given, AC ⊥ CB, AC = 2xkm,CB = 2(x + 7)km and AB = 26 km
On drawing the figure, we get the right angle ∆ACB right angled at C.
Now, In ∆ACB, by Pythagoras theorem,
AB² = AC² + BC²
⇒ (26)² = (2x)² + {2(x + 7)}²
⇒ 676 = 4x² + 4(x² + 49 + 11x)
⇒ 676 = 4x² + 4x² + 196 + 56x
⇒ 676 = 8x² + 56x + 196
⇒ 8x² + 56x – 480 = 0

On dividing by 8, we get x² + 7x – 60 = 0
⇒ x² + 12x-5x-60 = 0
⇒ x(x + 12) – 5(x + 12) = 0
⇒ (x + 12)(x – 5) = 0
∴ x = -12, x = 5
Since, distance cannot be negative.
∴ x = 5 [∵ x ≠ 12]
Now, AC = 2x = 10 km and BC = 2(x + 7) = 2(5 + 7) = 24 km
The distance covered to reach city B from city A via city C = AC + BC = 10 + 24 = 34 km
Distance covered to reach city B from city A after the construction of the highway is
BA = 26 km
Hence, the required saved distance is 34 – 26
i.e., 8 km

Question 7.
A flag pole 18 m high casts a shadow 9.6 m long. Find the distance of the top of the pole from the far end of the shadow.
Solution:
Let BC = 18 m be the flag pole and its shadow be AB = 9.6 m. The distance of the top of the pole, C from the far end i.e., A of the shadow is AC

In right angled ∆ABC
AC² = AB² + BC² [by Pythagoras theorem]
⇒ AC² = (9.6)² + (18)²
⇒ AC² = 92.16 + 324
⇒ AC² = 416.16
∴ AC = \(\sqrt{416.16}\) = 20.4 m
Hence, the required distance is 20.4 m.

Question 8.
A street light bulb is fixed on a pole 6 m above the level of the street. If a woman of height 1.5 m casts a shadow of 3m, find how far she is away from the base of the pole.
Solution:
Let A be the position of the street bulb fixed on a pole AB = 6 m and CD = 1.5 m be the height of a woman and her shadow be ED = 3 m. Let distance between pole and woman be x m.

Here, woman and pole both are standing vertically
So, CD || AB
In ∆CDE and ∆ABE,
∠E = ∠E [common angle]
∠ABE = ∠CDE [each equal to 90°]
∴ ∆CDE ~ ∆ABE [by AA similarity criterion]
Then \(\frac{E D}{E B}=\frac{C D}{A B} \Rightarrow \frac{3}{3+x}=\frac{1.5}{6}\)
⇒ 3 × 6 = 1.5(3 + x)
⇒ 18 = 1.5 × 3 + 1.5x
⇒ 1.5x = 18 – 4.5
∴ x = \(\frac{13.5}{1.5}\) = 9 m
Hence, she is at the distance of 9 m from the base of the pole.

Question 9.
In the figure, ABC is a triangle right angled at B and BD ⊥ AC. If AD = 4 cm, and CD = 5 cm, find BD and AB.

Solution:
Given, ∆ABC in which ∠B = 90° and BD ⊥ AC
Also, AD = 4 cm and CD = 5 cm
In ∆DBA and ∆DCB,
∠ADB = ∠CDB [each equal to 90°]
and ∠BAD = ∠DBC [each equal to 90° – ∠C] ;.
∴ ∆DBA ~ ∆DCB [by AA similarity criterion]

Question 10.
In the figure, PQR is a right triangle right angled at Q and QS ⊥ PR. If PQ = 6 cm and PS = 4 cm, find QS, RS and QR.

Solution:
Given, ∆PQR in which ∠Q = 90°, QS ⊥ PR and PQ = 6 cm, PS = 4 cm
In ∆SQP and ∆SRQ,
∠PSQ = ∠RSQ [each equal to 90°]
∠SPQ = ∠SQR [each equal to 90° – ∠R]
∴ ∆SQP ~ ∆SRQ [By AA similarity criterion]
Then, \(\frac{S Q}{P S}=\frac{S R}{S Q}\)
⇒ SQ² = PS × SR ………….. (i)
In right angled ∆PSQ,
PQ² = PS² + QS² [by Pythagoras theorem]
⇒ (6)² = (4)².+ QS²
⇒ 36 = 16 + QS²
⇒ QS² = 36 – 16 = 20
∴ QS.= \(\sqrt{20}=2 \sqrt{5}\) cm
On putting the value of QS in Eq(i), we get

Question 11.
In ∆PQR, PD ⊥ QR such that D lies on QR . If PQ = a, PR = b, QD = c and DR = d, prove that [a + b)(a – b) = (c + d)(c – d).
Solution:
Given: In ∆PQR, PD ⊥ QR, PQ = a, PR = b,
QD = c and DR = d
To prove : (a + b)(a -b) = (c + d)(c – d)
Proof : In right angled ∆PDQ,
PQ² = PD² + QD² [by Pythagoras theorem]
⇒ a² = PD² + c²
⇒ PD² = a² – c² …………. (i)

In right angled ∆PDR,
PR² = PD² + DR² [by Pythagoras theorem]
⇒ b² = PD² + d²
⇒ PD² = b² – d² ………….. (ii)
From Eqs. (i) and (ii)
a² – c² = b² – d²
⇒ a² – b² = c² – d²
⇒ (a – b)(a + b) = (c – d)(c + d)
Hence proved.

Question 12.
In a quadrilateral ABCD, ∠A + ∠D = 90°. Prove that AC² + BD² = AD² + BC² [Hint: Produce AB and DC to meet at E]
Solution:
Given : Quadrilateral ABCD, in which ∠A + ∠D = 90°
To prove : AC² + BD² = AD² + BC²
Construct: Produce AB and CD to meet at E
Also join AC and BD

Proof: In ∆AED, ∠A + ∠D = 90° [given]
∴ ∠E = 180° – (∠A + ∠D) = 90°
[ ∵ sum of angles of a triangle = 180°]
Then, by Pythagoras theorem,
AD² = AE² + DE²
In ∆BEC, by Pythagoras theorem,
BC² = BE² + EC²
On adding both equations, we get
AD² + BC² = AE² + DE² + BE² + CE² ………… (i)
In ∆AEC, by Pythagoras theorem,
AC² = AE² + CE²
and in ∆BED, by Pythagoras theorem,
BD² = BE² + DE²
On adding both equations, we get
AC² + BD² = AE² + CE² + BE² + DE² ………… (ii)
From Eqs. (i) and (ii)
AC² + BD² = AD² + BC²
Hence proved.

Question 13.
In the given figure, l || m and line segments AB, CD and EF are concurrent at point P. Prove that \(\frac{A E}{B F}=\frac{A C}{B D}=\frac{C E}{F D}\)

Solution:
Given l || m and line segments AB, CD and EF are concurrent at point P

Question 14.
In the figure, PA, QB, RC and SD are all perpendiculars to a line l,AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm. Find PQ,

Solution:
Given, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm
Also, PA, QB, RC and SD are all perpendiculars to line l
∴ PA || QB || RC || SD By Basic proportionality theorem,
PQ : QR : RS = AB : BC : CD = 6 : 9 : 12
Let PQ = 6x, QR = 9x and RS = 12x
Since, length of PS = 36 cm
∴ PQ + QR + RS = 36
⇒ 6x + 9x + 12x = 36
⇒ 27x = 36
∴ x = \(\frac{36}{27}=\frac{4}{3}\)
Now, PQ = 6x = 6 × \(\frac{4}{3}\) = 8 cm
QR = 9x = 9 × \(\frac{4}{3}\)= 12 cm
and RS = 12x = 12 × \(\frac{4}{3}\) = 16 cm

Question 15.
O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through 0, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO = QO.
Solution:
Given ABCD is a trapezium. Diagonals AC and BD are intersect at O.
PQ || AB || DC
To prove : PO = QO

Proof : In ∆ABD and ∆POD, PO || AB [∵ PQ || AB]
∠D = ∠D [common angle]
∠ABD = ∠POD [corresponding angles]
∴ ∆ABD ~ ∆POD[by AA similarity criterion]
Then, \(\frac{O P}{A B}=\frac{P D}{A D}\) …………… (i)
In ∆ABC and ∆OQC, OQ || AB
∠C = ∠C [common angle]
∠B AC = ∠QOC [corresponding angles]
∴ ∆ABC ~ ∆OQC [by AA similarity criterion]

Question 16.
In the figure, line segment DF intersect the side AC of a ∆ABC at the point E such that E is the mid-point of CA and ∠AEF = ∠AFE. Prove that \(\frac{B D}{C D}=\frac{B F}{C E}\)
[Hint:Take point G on AB such that CG || DF]

Solution:
Given ∆ABC, E is the mid-point of CA and ∠AEF = ∠AFE
To prove : \(\frac{B D}{C D}=\frac{B F}{C E}\)
Construction : Take a point G on AB such that CG || DF
Proof : Since, E is the mid-point of CA

∴ CE = AE
In ∆ACG, CG || EF and E is mid-point of CA
So, CE = GF …………… (ii) [by mid-point theorem]
Now, in ∆BCG and ∆BDF, CG || DF

Question 17.
Prove that the area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the triangle.
Solution:
Let ABC be a right triangle, right angled at B and AB = y, BC = x
Three semi-circles are drawn on the sides AB,
BC and AC, respectively with diameters AB,
BC and AC, respectively
Again, let area of circles with diameters AB,
BC and AC are respectively A₁, A₂ and A₃
To prove : A₃ = A₁ + A₂
Proof : In ∆ABC, by Pythagoras theorem,

Question 18.
Prove that the area of the equilateral triangle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the equilateral triangles drawn on the other two sides of the triangle.
Solution:
Lett a right triangle BAC in which ∠A is right angle and AC = y, AB = x
Three equilateral triangles ∆AEC, ∆AFB and ∆CBD are drawn on the three sides of ∆ABC.
Again, let area of triangles made on AC, AB and BC are A₁, A₂ and A₃, respectively.
To prove : A₃ = A₁ + A₂
Proof : In ∆CAB, by Pythagoras theorem,
BC² = AC² + AB²
⇒ BC² = y² + x²
⇒ BC = \(\sqrt{y^{2}+x^{2}}\)

NCERT Exemplar Class 10 Maths