AP 10th Class Maths Important Questions Chapter 8 Similar Triangles

These AP 10th Class Maths Chapter Wise Important Questions Chapter 8 Similar Triangles will help students prepare well for the exams.

AP State Syllabus 10th Class Maths 8th Lesson Important Questions and Answers Similar Triangles

Question 1.
Solution:
In a square and rectangle, the corresponding angles are equal. But the corresponding sides are not proportional.

∴ A square and a rectangle are not similar

Question 2.
In ΔABC LM||BC and $$\frac{\mathbf{A L}}{\mathbf{L B}}=\frac{2}{3}$$, AM = 5 cm, find AC.
Solution:
In ΔABC, LM // BC

Given AM = 5 cm and AM : MC = 2:3
$$\frac{\mathrm{AM}}{\mathrm{MC}}=\frac{2}{3} \Rightarrow \frac{5}{\mathrm{MC}}=\frac{2}{3}$$
⇒ MC = $$\frac{5 \times 3}{2}=\frac{15}{2}$$ = 7.5
∴ AC = AM + MC = 5 + 7.5 = 12.5 cm

Question 3.

In the figure, ∠BAC = ∠CED, then verify whether the value of ‘x’ is 3 or not.
Solution:
Given: In ΔABC and ΔECD
∠A = ∠E
∠ACB = ∠ECD [∵ Vertically opposite]
∴ LB ¿D [∵ Angle sum property]
∴ ΔABC ~ ΔEDC
⇒ $$\frac{\mathrm{AB}}{\mathrm{ED}}=\frac{\mathrm{BC}}{\mathrm{DC}}=\frac{\mathrm{AC}}{\mathrm{EC}} \Rightarrow \frac{36}{12}=\frac{9}{\mathrm{x}}$$
⇒ 36x= 108 ⇒ x= $$\frac{108}{36}$$ = 3

Question 4.
In ΔABC, DE || BC and AC = 5.6 cm, AE = 2.1 cm, then find AD : DB.
Solution:

In ΔABC DE || BC,
AE = 2.1 cm
EC = AC – AE
= 5.6 – 2.1 = 3.5 cm
As per B.P.T, AD : DB = AE : EC
= 2.1 : 3.5 = 3:5

Question 5.
In the given figure, $$\overline{\mathbf{A B}} \| \overline{\mathbf{Q R}}$$ and PA = 2 cm, AQ = 3 cm, then find the ratio of areas of ΔPQR and ΔPAB.

Solution:
Given AB || QR; PA = 2, AQ = 3
then, APQR ~ APAB
∴ PQ = 2 + 3 = 5
∴ Ratio of area of similar triangles is equal to square of ratio of their corre-sponding sides.

Question 6.
Give two different examples of pair of i) Similar figures, ii) Non – similar figures. Mar. ’17
By keeping in view of the definition of similar figures.
Solution:
By keeping in view of the definition of similar figures.
(i) Similar figures :

For any two correct examples for similar figures.
(ii) Non similar figures :

For any two correct examples for non similar figures.

Question 7.
Define A.A. and S.S.S. rules for similarity of two triangles in your own words.
Solution:
A.A. similarity : If two angles of a triangle are equal to two corresponding angles of another triangle, then the two triangles are similar.
S.S.S. similarity : If in two triangles, sides of one triangle are proportional to the sides of other triangle, then their corresponding angles are equal and hence the two triangles are similar.

Question 8.
The hypotenuse of a right triangle is 6m more than twice the shortest side. If the third side is 2m less than hypotenuse, find the sides of the triangle.
Solution:
Let the shortest side = ‘x’m
then its hypotenuse = (2x + 6) m
Third side = 2x + 6 – 2 = (2x + 4) m then from Pythagoras theorem, we have

x2 + (2x + 4)2 = (2x + 6)2
⇒ x2 + 4x2 + 16x + 16 = 4x2 + 24x + 36
⇒ x2 + 16x – 24x + 16 – 36 = 0
⇒ x2 – 8x – 20 = 0
⇒ x2 – 1 Ox + 2x – 20 = 0
⇒ x(x- 10) + 2 (x- 10) = 0
⇒ (x – 10) (x + 2) = 0
⇒ x = 10 or x = -2
but ‘x’ being the length cannot be nega¬tive.
So x ≠ -2, we consider x = 10 only then the
shortest side = x = 10m
third side = 2x + 4 = 24 m
hypotenuse = 2x + 6 = 26m

Question 9.
Draw a line segment of length 7.2 cm. and divide it in the ratio 5 : 3 using compasses and ruler.
Solution:

Question 10.
Construct a triangle of sides 4.2 cm, 5.1 cm and 6 cm. Then construct a tri¬angle similar to it, whose sides are 2/3 of corresponding sides of the first tri¬angle.
Solution:

1. Draw a triangle ABC, with sides AB = 4.2cm,BC = 5.1 cm,CA = 6cm.
2. Draw a ray BX making an acute angle with BC on the side opposite to vertex A.
3. Locate 3 points B,, B2, B3 on BX so that BB1 = B1B2 = B2B3.
4. Join B3,C an draw a line through B2 parallel to B3C intersecting BC at C1.
5. Draw a line through .C1 parallel to CA intersect AB at A1.
6.  A1BC1 is required triangle.

Question 11.
Construct a triangle PQR, where QR = 5.5cm,∠Q= 65°andPQ = 6cm. Then draw another triangle, whose sides are $$\frac{2}{3}$$ times of the corresponding sides of ΔPQR.
Solution:
Construction Steps:

1. Draw a triangle ΔPQR with sides QR = 5.5cm, ∠Q = 65°and PQ = 6cm.
2. Draw a ray OX making an acute angle with QR on the side opposite to vertex P
3. Locate 3 points Q1 Q2 Q3 on QX. So that QQ1 = Q1Q2 = Q2Q3
4. Join Q3,R and draw a line through Q2 parallel to Q3R intersecting QR at R.
5. Draw a line through ‘R’ parallel to PR intersect PQ at P.
6. ΔP’QR’ is required triangle.

Question 12.
i) State and prove the “Basic Propor-tionality theorem”.
ii) Using the theorem, find the length of AE, if AD = 1.8 cm, BD .= 5.4 cm, EC = 7.2 cm.

Solution:
If a lines is drawn parallel to one side of a triangle to intersecting the other two sides, in distinct points, then the other two sides are divided in the same ratio.
Given : A triangle ABC in which DE || BC and DE intersects AB in D and AC in E.
To prove : $$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$$

Construction : Join BE, CD and draw EF ⊥ AB, DG ⊥ AC.
Proof : In ∆EAD and ∆EDB, here as EF is perpendicular to AB, therefore, EF is the height for both of triangles EAD and EDB.
Now,
Area of ∆EAD = $$\frac { 1 }{ 2 }$$ (base x height)
= $$\frac { 1 }{ 2 }$$ x AD x EF
Area of ∆EDB = $$\frac { 1 }{ 2 }$$(base x height)
= $$\frac { 1 }{ 2 }$$ x DB x EF

∵ ∆DBE, ∆ECD are on the same base DE and between the same parallels DE || BC, we have
Area of ∆DBE = Area of ∆ECD
Hence, (1) = (2)
i.e., $$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$$ (Q.E.D)

ii) Using the theorem, to find the length of $$\overline{\mathrm{AE}}$$.

AD = 1.8 cm, BD = 5.4 cm, EC = 7.2 cm
From the theorem, $$\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{EC}}$$
⇒ AE = $$\frac{(\mathrm{AD})(\mathrm{EC})}{\mathrm{BD}}=\frac{1.8 \times 7.2}{5.4}$$ = 2.4 cm
∴ AE = 2.4 cm

Question 13.
Draw ΔABC with sides 4.3 cm, 5.2 cm and 6.5 cm and then construct a tri-angle similar to ΔABC, whose sides are $$\frac { 3 }{ 5 }$$th of the corresponding sides.
Solution:
Constructing a similar triangle whose sides are $$\frac { 3 }{ 5 }$$ of corresponding sides.
Steps of construction:
1) Draw a triangle ΔABC with sides AB = 4.3 cm, BC = 5.2 cm and AC = 6.5 cm
2) Draw a ray $$\overline{\mathrm{BX}}$$ making an acute angle with BC on the side opposite to vertex ‘A’.

3) Locate 5 points B1; B2, B3, B4, B5 on the ray $$\overline{\mathrm{BX}}$$ Such that
$$\overline{\mathrm{BB}_{1}}=\overline{\mathrm{B}_{1} \mathrm{~B}_{2}}=\overline{\mathrm{B}_{2} \mathrm{~B}_{3}}=\overline{\mathrm{B}_{3} \mathrm{~B}_{4}}=\overline{\mathrm{B}_{4} \mathrm{~B}_{5}}$$
4) Now join $$\overline{\mathrm{B}_{5} \mathrm{C}}$$ and draw another line through the point ‘B3‘ such that it is parallel to $$\overline{\mathrm{B}_{5} \mathrm{C}}$$ which meets $$\overline{\mathrm{BC}}$$ at the point ‘C’.
5) Now draw another line from the point (C’) to $$\overline{\mathrm{AB}}$$ such that it is par¬allel to $$\overline{\mathrm{AC}}$$, which meets $$\overline{\mathrm{AB}}$$ at the point (A’).
6) Now ΔA’B’C’ is required similar tri-angle.

Question 14.
Draw a triangle ABC in which AB = 4 cm, BC = 6 cm and AC = 4 cm. Draw a triangle similar to ΔABC with its sides $$\frac { 3 }{ 4 }$$ times of the corresponding sides of ΔABC.
Solution:

Question 15.
Divide the line segment AB = 6 cm in the ratio of 3 : 2. Explain the con-struction procedure.
Solution:
Steps to construct:

1) Draw a line segment AB = 6 cm
2) Draw a line AX such that ∠BAX is an acute.
3) Make 5 equal parts on AX. (3 + 2)
4) Now join B with last part A5.
5) Then draw a line from A3 to AB which meets at ‘C’ such that it is parallel to BA5.
6) So ‘C’ is the point which divides AB in the ratio of 3 : 2

Question 16.
In the given ΔABC, E and F are two points on AB and AC respectively. Then verify whether EF || BC or not in the following cases.
Solution:
From the converse of Thales theorem
EF || BC if $$\frac{\mathrm{AE}}{\mathrm{EB}}=\frac{\mathrm{AF}}{\mathrm{FC}}$$

Case I) AE = 3.9, EB = 3,
AF = 3.6, CF = 2.4
Then $$\frac{\mathrm{AE}}{\mathrm{EB}}=\frac{3.9}{3}=\frac{1.3}{1} ; \frac{\mathrm{AF}}{\mathrm{FC}}=\frac{3.6}{2.4}=\frac{3}{2}$$
So $$\frac{A E}{E B} \neq \frac{A F}{F B}$$ then EF ∦ BC
Case II) AE = 4, EB = 4.5;
AF = 8 and CF = 9
So $$\frac{A E}{E B}=\frac{4}{4.5}=\frac{8}{9}$$ and $$\frac{\mathrm{AF}}{\mathrm{FB}}=\frac{8}{9}$$
So then $$\frac{\mathrm{AE}}{\mathrm{EB}}=\frac{\mathrm{AF}}{\mathrm{FB}}$$
hence EF || BC.

Question 17.
In the given figure BC || DE and AD = DB = 3.4 and AC = 14.

So find AE and EC.
Solution:
In the given ΔABC, DE || BC is given.
So from Thales theorem,
$$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$$
In this problem, AD = DB = 3.4 is given.
And also AC = 14 is given.
So $$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$$
⇒ $$\frac{3.4}{3.4}=\frac{\mathrm{AE}}{\mathrm{EC}}$$
⇒ AE = EC
But AC = 14 is given.
AC = AE + EC = 2AE = 14
⇒ AE = EC = 7 cm

Question 18.
In given ΔABC points D and E are mid points of AB and AC and also BC = 6 cm then find DE. .
Solution:
As shown in the figure D, E are mid points of AB and AC.

Then AD = BD, AE = EC
∴ $$\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{EC}}$$ hence DE || BC (From the Converse of Thales theorem)
$$\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{DE}}{\mathrm{BC}}=\frac{\mathrm{AE}}{\mathrm{EC}}$$
⇒ $$\frac{A D}{2 A D}=\frac{D E}{6}$$
⇒ 2DE = 6 and DE = 3 cm
∴ DE = 3 cm

Question 19.
In the given figure, AB || CD || EF given AB = 73cm,DC=ycm,EF 4.5cm, BC = x cm. Calculate the values of x and y.

Solution:
Given that, AB || CD || EF and
AB = 7.5 cm, DC = y cm
According to Thales theorem
Δ𝜏CDE ~ Δ𝜏ABE

From (1) and (2)
$$\frac{7.5-y}{7.5}=\frac{y}{4.5}$$
7.5 y = (7.5) (4.5)-4.5 y
12y = (7.5) (4.5) = 33.75
y = $$\frac{33.75}{12}$$ = 2.8125
From (2)
$$\frac{y}{4.5}=\frac{x}{x+3} ; \frac{2.8}{4.5}=\frac{\dot{x}}{x+3}$$
4.5 x = (2.8) x + 8.4
4.5 x – (2.8) x = 8.4
1.7 x = 8.4
x = $$\frac{8.4}{1.7}$$ = 5 (approx)