## AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.1

### 10th Class Maths 8th Lesson Similar Triangles Ex 8.1 Textbook Questions and Answers

Question 1.

In △PQR, ST is a line such that \(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\) and also ∠PST = ∠PRQ.

Prove that △PQR is an isosceles triangle.

Answer:

Given : In △PQR,

\(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\) and ∠PST= ∠PRQ.

R.T.P: △PQR is isosceles.

Proof: \(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\)

Hence, ST || QR (Converse of Basic proportionality theorem)

∠PST = ∠PQR …….. (1)

(Corresponding angles for the lines ST || QR)

Also, ∠PST = ∠PRQ ……… (2) given

From (1) and (2),

∠PQR = ∠PRQ

i.e., PR = PQ

[∵ In a triangle sides opposite to equal angles are equal]

Hence, APQR is an isosceles triangle.

Question 2.

In the given figure, LM || CM and LN || CD. Prove that \(\frac{AM}{AB}\) = \(\frac{AN}{AD}\).

Answer:

Given : LM || CB and LN || CD In △ABC, LM || BC (given) Hence,

Adding ‘1’ on both sides.

From (1) and (2)

∴ \(\frac{AM}{AB}\) = \(\frac{AN}{AD}\).

Question 3.

In the given figure, DE || AC and DF || AE. Prove that \(\frac{BF}{FE}\) = \(\frac{BE}{AC}\).

Answer:

In △ABC, DE || AC

Hence \(\frac{BE}{EC}\) = \(\frac{BD}{DA}\) ………. (1)

[∵ A line drawn parallel to one side of a triangle divides the other two sides in the same ratio – Basic proportionality theorem]

Also in △ABE, DF || AE

Hence \(\frac{BF}{FE}\) = \(\frac{BD}{DA}\) ………. (2)

From (1) and (2), \(\frac{BF}{FE}\) = \(\frac{BE}{AC}\) Hence proved.

Question 4.

Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side (Using Basic proportionality theorem).

Answer:

Given: In △ABC; D is the mid-point of AB.

A line ‘l’ through D, parallel to BC, meeting AC at E.

R.T.P: E is the midpoint of AC.

Proof:

DE || BC (Given)

then

\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)(From Basic Proportional theorem)

Also given ‘D’ is mid point of AB.

Then AD = DB.

⇒ \(\frac{AD}{DB}\) = \(\frac{DB}{DB}\) = \(\frac{AE}{EC}\) = 1

⇒ AE = EC

∴ ‘E’ is mid point of AC

∴ The line bisects the third side \(\overline{\mathrm{AC}}\).

Hence proved.

Question 5.

Prove that a line joining the mid points of any two sides of a triangle is parallel to the third side. (Using converse of Basic proportionality theorem)

Answer:

Given: △ABC, D is the midpoint of AB and E is the midpoint of AC.

R.T.P : DE || BC.

Proof:

Since D is the midpoint of AB, we have AD = DB ⇒ \(\frac{AD}{DB}\) = 1 ……. (1)

also ‘E’ is the midpoint of AC, we have AE = EC ⇒ \(\frac{AE}{EC}\) = 1 ……. (2)

From (1) and (2)

\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)

If a line divides any two sides of a triangle in the same ratio then it is parallel to the third side.

∴ DE || BC by Basic proportionality theorem.

Hence proved.

Question 6.

In the given figure, DE || OQ and DF || OR. Show that EF || QR.

Answer:

Given: △PQR, DE || OQ; DF || OR

R.T.P: EF || QR

Proof:

In △POQ;

\(\frac{PE}{EQ}\) = \(\frac{PD}{DO}\) ……. (1)

[∵ ED || QO, Basic proportionality theorem]

In △POR; \(\frac{PF}{FR}\) = \(\frac{PD}{DO}\) ……. (2) [∵ DF || OR, Basic Proportionality Theorem]

From (1) and (2),

\(\frac{PE}{EQ}\) = \(\frac{PF}{FR}\)

Thus the line \(\overline{\mathrm{EF}}\) divides the two sides PQ and PR of △PQR in the same ratio.

Hence, EF || QR. [∵ Converse of Basic proportionality theorem]

Question 7.

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

Answer:

Given:

In △PQR, AB || PQ; AC || PR

R.T.P : BC || QR

Proof: In △POQ; AB || PQ

\(\frac{OA}{AP}\) = \(\frac{OB}{BQ}\) ……… (1)

(∵ Basic Proportional theorem)

and in △OPR, Proof: In △POQ; AB || PQ

\(\frac{OA}{AP}\) = \(\frac{OC}{CR}\) ……… (2)

From (1) and (2), we can write

\(\frac{OB}{BQ}\) = \(\frac{OC}{CR}\)

Then consider above condition in △OQR then from (3) it is clear.

∴ BC || QR [∵ from converse of Basic Proportionality Theorem]

Hence proved.

Question 8.

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at point ‘O’. Show that\(\frac{AO}{BO}\) = \(\frac{CO}{DO}\).

Answer:

Given: In trapezium □ ABCD, AB || CD. Diagonals AC, BD intersect at O.

R.T.P: \(\frac{AO}{BO}\) = \(\frac{CO}{DO}\)

Construction:

Draw a line EF passing through the point ‘O’ and parallel to CD and AB.

Proof: In △ACD, EO || CD

∴ \(\frac{AO}{CO}\) = \(\frac{AE}{DE}\) …….. (1)

[∵ line drawn parallel to one side of a triangle divides other two sides in the same ratio by Basic proportionality theorem]

In △ABD, EO || AB

Hence, \(\frac{DE}{AE}\) = \(\frac{DO}{BO}\)

[∵ Basic proportionality theorem]

\(\frac{BO}{DO}\) = \(\frac{AE}{ED}\) …….. (2) [∵ Invertendo]

From (1) and (2),

\(\frac{AO}{CO}\) = \(\frac{BO}{DO}\)

⇒ \(\frac{AO}{BO}\) = \(\frac{CO}{DO}\) [∵ Alternendo]

Question 9.

Draw a line segment of length 7.2 cm and divide it in the ratio 5 : 3. Measure the two parts.

Answer:

Steps of construction:

- Draw a line segment \(\overline{\mathrm{AB}}\) of length 7.2 cm.
- Construct an acute angle ∠BAX at A.
- Mark off 5 + 3 = 8 equal parts (A
_{1}, A_{2}, …., A_{8}) on \(\stackrel{\leftrightarrow}{\mathrm{AX}}\) with same radius. - Join A
_{8}and B. - Draw a line parallel to \(\stackrel{\leftrightarrow}{\mathrm{A}_{8} \mathrm{~B}}\) at A
_{5}meeting AB at C. - Now the point C divides AB in the ratio 5:3.
- Measure AC and BC. AC = 4.5 cm and BC = 2.7 cm.