Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 6 Similar Triangles Extra Questions and Answers Notes to clear their doubts.
Kerala SCERT Class 9 Maths Chapter 6 Solutions Similar Triangles
Similar Triangles Class 9 Kerala Syllabus Questions and Answers
HSSLive Kerala State Syllabus 9th Standard Maths Chapter 6 Similar Triangles Solutions Questions and Answers
Class 9 Maths Chapter 6 Kerala Syllabus – Angles And Sides
Intext Questions And Answers
Question 1.
Is there any relation between the sides of two triangles with the same angles?
Answer:
Consider two triangles with the same angles but different sides.
Here,
a and p are the lengths of the sides opposite the 80° angle
b and q are the lengths of the sides opposite the 60° angle
c and r are the lengths of the sides opposite the 40° angle.
When we pair the sides of shorter lengths p, q, r and longer lengths a, b, c opposite to the equal angles of the two triangles as (p, a), (q, b) and (r, c) then we get,
\(\frac{p}{a}=\frac{q}{b}=\frac{r}{c}\)
That is,
In two triangles with the same angles, if we pair the sides opposite equal angles, then the shorter lengths are all the same part of the longer one (or the longer lengths are all same times the shorter).
In short,
In triangles with the same angles, the sides, in the order of lengths, are in the same ratio.
Using scale factor, we can explain this as:
For the triangles having the same angle, then smaller triangle with sides p, q, r are the same part of the larger triangle with side a, b, c. This means the change from the sides of the first triangle to the second has the same scale factor. If this scale factor is taken as k, then the relation will be
a = kp, b = kq, c = kr
That is,
In triangles with the same angles, the sides opposite equal angles are scaled by the same factor.
Textual Questions And Answers
Question 1.
One side of a triangle is 8 centimetres and the two angles on it are 60° and 70°. Draw the triangle with lengths of sides 1½ times that of this triangle and with the same angles.
Answer:
Given the triangle with one side 8 cm, two angles on it are 60° and 70°.
To draw a triangle, we need to take the side as 12 cm (ie.,1½ times 8 cm).
So, draw a line of 12 cm long and draw the same angles (60° and 70°) as those in the given triangle.
Since the angles are equal, the other sides will also be in the same ratio as the given triangle.
Question 2.
In a right triangle, the perpendicular from the square corner to the hypotenuse divides it into pieces 2 centImetres and 3 centimetres long:
i. Prove that the two small right triangles formed by this perpendicular have the same angles.
ii. Taking the height of the perpendicular as h, prove that \(\frac{h}{2}=\frac{3}{h}\)
iii. Calculate the lengths of the perpendicular sides of the original large triangle.
iv. Prove that if the length of the perpendicular from the square corner of a right triangle to the hypotenuse is h and it divides the hypotenuse into pieces of length a and b, then h² = ab.
Answer:
Let the right triangle be ∆ABC
Hence proved.
i. Here,∠A=90°
Therefore, ∠B + ∠C = 90°
Let’s take ∠B = x so ∠C = 90° – x
Consider ADC where
∠ADC = 90°
∠DCA = 90° – x (Given)
Therefore, ∠DAC = x
Thus, the angles of ∠ADC are 90° – x, x and 90° ………(i)
Consider ∠ADB where
∠ADB = 90°
∠DBA = x (Given)
Therefore, ∠BAD = 90° – x
Thus, the angles of ∆ADB are 90° – x, x and 90° ……(ii)
∴ from (î) and (ii) the two small right triangles formed by the perpendicular have the same angles.
Hence proved.
ii. Let AD = h
We know that in triangles with same angles, the sides opposite to equal angles are in the same ratio.
Therefore,
\(\frac{A D}{B D}=\frac{C D}{A D}\)
\(\frac{h}{2}=\frac{3}{h}\)
iii. Here we have to find the length of AB and AC.
We have, \(\frac{h}{2}=\frac{3}{h}\) [From (ii)]
Cross multiplying, h² = 2 × 3 = 6
Consider ∆ADB
Using Pythagoras theorem,
h² + 2² = AB²
AB² = 6 + 4=10
AB = \(\sqrt{10}\) cm
Consider ∆ADC
Using Pythagoras theorem,
AC² = h² + 3²²
AC² = 6 + 9 = 15
AC = \(\sqrt{15}\) cm
iv. Let BD = a and CD = b
In triangles with the same angle, the sides opposite equal angles are in the same ratio. So,
\(\frac{A D}{B D}=\frac{C D}{A D}\)
\(\frac{h}{a}=\frac{b}{h}\)
h² = ab
Question 3.
At the two ends of a horizontal line, angles of the same size are drawn and two points on these slanted lines are joined:
i. Prove that the horizontal line (blue) and the slanted line (red) cut each other Into parts in the same ratio.
ii. Prove that the slanted lines (green) at the ends of the horizontal line are also in the same ratio.
iii. Explain how this idea can be used to divide a 6 centimetre long line in the ratio 3: 4.
Answer:
i. ∠A = ∠B (Given)
∠AMC = ∠BMD (Vertical angles)
Therefore, ∠C = ∠D
Therefore,
\(\frac{\mathrm{MC}}{\mathrm{MD}}=\frac{\mathrm{AC}}{\mathrm{BD}}=\frac{\mathrm{AM}}{\mathrm{MB}}\)
Thus,
\(\frac{A M}{M B}=\frac{M C}{M D}\)
Hence proved.
ii. We have
\(\frac{M C}{M D}=\frac{A C}{B D}=\frac{A M}{M B}\)
The slanted lines at the ends of the horizontal line are also in the same ratio.
iii.
∠CAB = ∠ACD
Draw BD to cut AC at O.
AB: CD = OA: OC
⇒ OA: OC = 3: 4
Question 4.
The picture below shows a square sharing one corner with a right triangle and the other three corners on the sides of this triangle:
i. Calculate the length of the sides of the square.
ii. What is the length of the sides of such a square drawn within a triangle of sides 3 centimetres, 4 centimetres, 5 centimetres?
Answer:
i.
All the angles in ∆ABC and ∆APQ are equal. So, sides opposite to equal angles are in the same ratio. Thus,
\(\frac{x+2}{2}=\frac{x+1}{x}\)
x(x + 2) = 2(x + 1)
x² + 2x = 2x + 2
x² = 2
x = √2
Side of the square = √2 cm
ii. Given AB = 4 cm, BC = 3 and AC = 5 cm
All the angles in ∆APQ and ∆ABC are equal.
Therefore, if we take the side length of the square as x,
\(\frac{A P}{A B}=\frac{P Q}{B C}\)
\(\frac{4-x}{4}=\frac{x}{3}\)
x = \(\frac{12}{7}\)
Question 5.
Calculate the area of the largest right triangle in the picture below:
Answer:
AC = 13 cm
Let BM = h cm
Since h² = ab
h² = 9 × 4 = 36 cm
Therefore, h = 6 cm
Area of ∆ABC = \(\frac{1}{2}\) × AC × BM
= \(\frac{1}{2}\) × 13 × 6
= 39 cm²
Question 6.
Two poles of heights 3 metres and 2 metres are erected upright on the ground and ropes are stretched from the top of each to the foot of the other:
i. At what height above the ground do the ropes cross each other?
ii. Prove that this height would be the same whatever the distance between the poles.
iii. Denoting the heights of the poles as a, b and the height of the point of crossing above the ground as h, find the relation between a, b and h.
Answer:
(i) All the angles in triangle ∆ABC and ∆AFE are equal. So,
\(\frac{b}{h}=\frac{x+y}{x}\)
⇒ \(\frac{h}{b}=\frac{x}{x+y}\) ………..(i)
Also, all the angles in triangle ∆ADB and ∆FEB are equal. So,
\(\frac{a}{h}=\frac{x+y}{y}\)
⇒ \(\frac{h}{a}=\frac{y}{x+y}\) ……(ii)
(i) + (ii) ⇒ \(\frac{h}{b}+\frac{h}{a}=\frac{x}{x+y}+\frac{y}{x+y}\)
h\(\left[\frac{1}{b}+\frac{1}{a}\right]\) = 1
\(\frac{1}{a}+\frac{1}{b}=\frac{1}{h}\)
(where a = 3 and b = 2)
\(\frac{1}{3}+\frac{1}{2}=\frac{1}{h}\)
h = \(\frac{6}{5}\) = 1.2 m
(ii) Since the height ‘h’ is dependent only on the height of the poles, change in the distance between the poles does not affect ‘h’.
(iii) From (i)
\(\frac{1}{h}=\frac{1}{a}+\frac{1}{b}\)
Question 7.
In the picture below, AP is the bisector of ∠A of triangle ABC:
i. Prove that angles of the triangles ABP and CPQ are the same.
ii. Calculate \(\frac{B P}{P C}\).
iii. Prove that in any triangle, the bisector of an angle divides the opposite side in the ratio of the sides containing the angle.
Answer:
i. Consider ∆ACQ where,
AC = QC = 3 cm (Given)
∠CAQ = ∠CQA (isosceles triangle)
∠BAP = ∠PAC (AP bisects ∠A)
Consider ∆APB and ∆QPC where,
∠BAP = ∠CQP
∠APB = ∠QPC (Vertically opposite angle)
Therefore, ∠ABP = ∠PCQ (Third angle)
ii. Consider ∆APB and ∆QPC where,
\(\frac{A B}{Q C}=\frac{B P}{P C}=\frac{5}{3}\)
iii.
In ∆ABC, AP is the bisector of ∠A.
Draw a line from P to AB and AC where,
PQ ⊥ AB and AC ⊥ PR
Consider ∆AQP and ∆ARP where,
∠QAP = ∠RAP (AP is the bisector of ∠A)
∠AQP = ∠ARP = 90°
∠QPA = ∠RPA (Third angle)
AP = AP (Common side)
Therefore, ∆AQP = ∆ARP
Therefore, PQ = PR = h
Area of ∆ABP = \(\frac{1}{2}\) × AB × h²
Area of ∆ACP = \(\frac{1}{2}\) × AC × h²
Ratio of areas = \(\frac{1}{2}\) × AB × h : \(\frac{1}{2}\) × AC × h = AB : AC
BP: PC = Area of ∆ABP: Area of ∆APC
= \(\frac{1}{2}\)AB × h : \(\frac{1}{2}\) × AC × h
Thus, BP: PC = AB: AC
This means the bisector of an angle divides the opposite side in the ratio of the sides containing the angle.
Class 9 Maths Kerala Syllabus Chapter 6 Solutions – Sides And Angles
Textual Questions And Answers
Question 1.
Draw the triangle with angles the same as those of the triangle shown below, and sides scaled by 1\(\frac{1}{4}\).
Answer:
4 × 1\(\frac{1}{4}\) = 4 × \(\frac{5}{4}\) = 5cm
8 × 1\(\frac{1}{4}\) = 8 × \(\frac{5}{4}\) = 10cm
- Draw ∆ABC
- Join AC to E and AB to D where length of CE and BD are 1 cm and 2 cm respectively
- Draw ∆ADE
Question 2.
See the picture of the quadrilateral.
i. Draw the quadrilateral with the same angles as this and sides scaled by a factor of 1\(\frac{1}{2}\).
ii. Draw a quadrilateral with angles different from this and sides scaled by a factor of 1\(\frac{1}{2}\).
Answer:
i. 4 × 1\(\frac{1}{4}\) = 4 × \(\frac{3}{2}\) = 6cm
5 × 1\(\frac{1}{2}\) = 5 × \(\frac{3}{2}\) = 7\(\frac{1}{2}\) cm
6 × 1\(\frac{1}{2}\) = 6 × \(\frac{3}{2}\) = 9 cm
- Draw quadrilateral ABCD
- Join AD to E , AC to F and AB to G where the length of DE, CF and BG where 2cm, 2\(\frac{1}{2}\) cm and 3 cm respectively.
ii. Draw quadrilateral ABCD with diagonal AC = 8.5 cm
The change in the length of the diagonal changes the angles.
Question 3.
The area of a triangle is 6 square centimetres. What is the area of the triangle with lengths of sides four times those of this? What about the one with lengths of sides half of this?
Answer:
Area of a triangle = 6 square centimetres
If lengths of sides four times that of the given triangle then.
The ratio of the length of the side = 1: 4 (Given)
Thus, the area of triangle = 1 : 16
That is, the scale factor of area is the square of the scale factor of the sides.
Therefore, the required area of the triangle = 6 × 16 = 96 cm²
If lengths of sides half of that of the given triangle then,
The ratio of the length of the side = 2 : 1 (Given)
Thus, the area of triangle = 4 : 1
That is, the scale factor of area is the square of the scale factor of the sides. Therefore, the required area of the triangle = \(\frac{6}{4}=\frac{3}{2}\) cm²
SCERT Class 9 Maths Chapter 6 Solutions – Third Way
Intext Questions And Answers
If two triangles have two of the sides scaled by the same factor and the included angles equal, then the third sides are also scaled by the same factor and the other two angles are also equal.
Question 1.
Scaling a triangle without knowing its sides and angles.
To scale a triangle, by stretching the sides by one and a half times where its sides and angles are unknown.
First, draw a circumcircle around the triangle. Next, draw a second circle with a radius, one and a half times larger that the original. Now join the centre of the circles to the vertices of the triangle and extend them to meet the larger circle. Join these points to draw a larger triangle.
Similarly, scale a rectangle using the circle by one and a half times that of the original one
Let’s draw a rectangle PQRS and draw a circumcircle for it.
Then draw another circle with a scale factor of one and a half times the radius.
Now join the centre of the circles to the vertices of the rectangle and extend them to meet the larger circle. Join these points to draw the required rectangle ABCD.
Is this method is possible for all quadrilaterals?
Answer:
No
Because we cannot draw a circle through the four vertices of a parallelogram which is not a rectangle.
However, we can use this method to scale any regular polygon.
If two triangles have any one of the following relations, they also have the other two:
- Have the same angles
- Have all sides scaled by the same factor
- Have two sides scaled by the same factor and the included angle equal
Triangles having any of these relations between them are said to be similar
Textual Questions And Answers
Question 1.
Prove that if the perpendicular sides of two right triangles are scaled by the same factor, then the hypotenuses are also scaled by the same factor.
Answer:
Consider ∆ABC and ∆PQR
Given that
\(\frac{A B}{P Q}=\frac{B C}{Q R}\)
Here ∠B = ∠Q = 90°
Since the two sides are scaled by the same factor and the included angles are equal the two triangles are similar.
Therefore, the third side AC and PQ are also scaled by the same factor.
Question 2.
Prove that the hypotenuse and one side of two triangles are scaled by the same factor, then the third sides are also scaled by the same factor.
Answer:
Consider the right triangle ABC and PQR
Given that
\(\frac{A C}{P R}=\frac{B C}{Q R}\)
Let,
\(\frac{A C}{P R}\) = k \(\frac{B C}{Q R}\) = k
AC = k PR
BC = k QR
By Pythagoras theorem
AB = \(\sqrt{A C^2-B C^2}\)
=\(\sqrt{k^2(P R)^2-k^2(Q R)^2}\)
= k\(\sqrt{P R^2-Q R^2}\)
= kPQ
\(\frac{A B}{P Q}\) = k
Therefore, the third sides are also scaled by the same factor.
Question 3.
Draw a triangle and mark a point inside it. Join this point to the vertices and extend each of them by half its original length. Join the end points of these lines to form another triangle:
Prove that the sides of the larger triangle are one and a half times the sides of the original triangle.
Answer:
Consider ∆BDC and ∆QDR where,
BD: QD = 2x : 3x = 2: 3
CD: RD = 2y : 3y = 2:3
Since the 2 sides are scaled by the same factor and one angle is common,then the two triangles are similar.
BC:QR = 2: 3
\(\frac{B C}{Q R}=\frac{2}{3}\)
QR = \(\frac{3}{2}\)BC = 1\(\frac{1}{2}\) BC
Similarly, PR = \(\frac{3}{2}\)AC = 1\(\frac{1}{2}\)AC
PQ = \(\frac{3}{2}\)AB = 1\(\frac{1}{2}\)AB
Therefore, the sides of the larger triangles is one and a half times the sides of the smaller triangle.
Question 4.
The vertices of a quadrilateral are joined to a point inside it and these line are extended by the same scale factor. The ends of these lines are joined to form another quadrilateral:
i. Prove that the sides of the two quadrilaterals also are scaled by the same factor.
ii. Prove that the angles of the two quadrilaterals are the same.
Answer:
i.
Consider ∆AOB and ∆POQ
∠O = ∠O (Common angle)
OP – OA = OQ – OB
Therefore, ∆AOB and ∆POQ are similar triangle.
So, \(\frac{A B}{P Q}=\frac{O B}{O Q}=\frac{O A}{O P}\) ..(1)
Consider ∆BOC and ∆QOR
∠O = ∠O (Common angle)
OR – OC = OQ – OB
Therefore, ABOC and AQOR are similar triangle.
So, \(\frac{B C}{Q R}=\frac{O C}{O R}=\frac{O B}{O Q}\) …(2)
Consider ADOC and ASOR
∠O = ∠O (Common angle)
OR – OC = OS – OD
Therefore, ∆DOC and ∆SOR are similar triangle
So, \(\frac{D C}{S R}=\frac{O D}{O S}=\frac{O C}{O R}\) …(3)
Consider ∆AOD and ∆SOP
∠O = ∠O (Common angle)
OP – OA = OS – OD
Therefore, ∆AOD and ∆SOP are similar triangles
So, \(\frac{A D}{P S}=\frac{O D}{O S}=\frac{O A}{O P}\) …(4)
(1),(2),(3),(4) ⇒
\(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{D C}{S R}=\frac{A D}{P S}\)
Therefore, the sides of large and small quadrilateral can be scaled by the same factor.
ii.
∠A = a + h; ∠P = a + h ⇒ ∠A = ∠P
∠B = b + c; ∠Q = b + c ⇒ ∠B = ∠Q
∠C = d + e; ∠R = d + e ⇒ ∠C = ∠R
∠D = g + f; ∠S = g + f ⇒ ∠D = ∠S
Therefore, the angles of the two quadrilaterals are the same.
Similar Triangles Class 9 Extra Questions and Answers Kerala Syllabus
Question 1.
In the figure ∠Q = 90°, QR = 5 cm , SR = 3 cm, QS is perpendicular to PR.
a) find the length QS
b) find the length PS
Answer:
(a) QS² = 5² – 3² = 4², QS = 4
(b) PS × SR = QS²
PS × 3 = 4²
PS = \(\frac{4 \times 4}{3}=\frac{16}{3}\)
Question 2.
Draw an equilateral triangle of side 6 cm. Draw another triangle, its sides are 1\(\frac{1}{2}\) times of the originals.
Answer:
6 × \(\frac{3}{2}\) = 9 cm
∆DBE is the required triangle.
Question 3.
In the figure PQR and QST are right triangle. Prove that QR × QS = QP × QT
Answer:
∆PQR and ∆STQ are similar
\(\frac{Q R}{Q T}=\frac{P R}{T S}=\frac{P Q}{Q S}\)
QR × QS = PQ × QT
Question 4.
A boy of height 90 cm walking away from the base of a lamp post at the speed of 1.2 m/s. If the lamp is 3.6 m above the ground then find the length of his shadow after 4 second.
Answer:
Speed = 1.2 m/s
After 4 seconds, he covers 4 × 1.2 meters = 4.8 m
\(\frac{\mathrm{AB}}{\mathrm{BE}}=\frac{\mathrm{CD}}{\mathrm{DE}}\)
\(\frac{3.6}{(4.8+x)}=\frac{0.9}{x}\)
3.6x = 4.8 × 0.9 + 0.9x
2.7x = 4.32
x = 1.6 m
Question 5.
In the figure ABCD is a rectangle BC = 24 cm, DP = 10 cm, CD = 15 cm. Find AQ and CQ.
Answer:
CD = 15 cm, DP = 10 cm
PC = 5 cm.
Let x be the length of CQ.
BQ = x + 24
Triangles ∆ABQ and PCQ are similar.
\(\frac{P C}{A B}=\frac{C Q}{B Q}\)
Since ABCD is a rectangle
AB = CD = 15 cm
\(\frac{5}{15}=\frac{x}{x+24}\), solving x = 12 cm
AQ² = AB² + BQ² = 15² + 36² = 1521
AQ = 39 cm
Question 6.
In the figure ∠A = ∠P, ∠B = ∠Q, AB = 5 cm. BC = 4 cm, AC = 2 cm, PR = 6 cm.
(a) What is the length of PQ?
(b) What is the ratio of the permetres of ∆ABC and ∆PQR?
Answer:
(a) PQ = 5 × 3 = 15 cm, QR = 12 cm
(b) \(\frac{\text { Perimeter of } \triangle A B C}{\text { Perimeter of } \triangle P Q R}=\frac{11}{6+15+12}=\frac{11}{33}\) = 1 : 3
Question 7.
In the figure ∠B = ∠D = 90 AB = 15 cm, AD = 5 cm
a) If ∠DAE = 40° find ∠AED, ∠BAC.
b) What is the measure of ∠C?
c) \(\frac{\mathrm{BC}}{\mathrm{DE}}\) = ………… [3, 4, 5]
Answer:
(a) ∠AED = 90 – 40 = 50°
∠BAC = 40°
(b) ∠C = 50°
(c) ∆ADE and ∆ABC are similar
\(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{15}{5}=\frac{3}{1}\)
So,
\(\frac{\mathrm{BC}}{\mathrm{DE}}\) is also \(\frac{3}{1}\) = 3
Question 8.
Draw the triangle with angles the same as those of the triangle shown below, and sides scaled by 1\(\frac{1}{2}\).
Answer:
6 × \(\frac{3}{2}\) = 9 cm
7 × \(\frac{3}{2}\) = 10.5 cm
