Students can practice TS Class 10 Maths Solutions Chapter 8 Similar Triangles Ex 8.1 to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 8 Similar Triangles Exercise 8.1

Question 1.

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

Prove that ∆PQR is an isosceles triangle. (AS_{2})

Solution:

In ∆PQR. ST divides PQ and PR in the same ratio.

\(\frac{\mathrm{PS}}{\mathrm{SQ}}\) = \(\frac{\mathrm{PT}}{\mathrm{TR}}\) (given)

∴ ST // QR

It is also given

that ∠PST = ∠PRQ …………… (1)

Since ST // QR and PQ is a transversal.

∠PST = ∠PQR (Corresponding angles) ………….. (2)

From (1) and (2) we have

∠PRQ = ∠PQR

In ∆PQR, ∠PRQ = ∠PQR

∴ PQ = PR

∴ ∆PQR is an isosceles triangle.

Question 2.

In the given figure, LM // CB and LN // CD, prove that \(\frac{\mathrm{AM}}{\mathrm{AB}}\) = \(\frac{\mathrm{AN}}{\mathrm{AD}}\). (AS_{2})

Solution:

In the figure, it is given that LM // CB and LN // CD.

In ∆ABC, ML // BC.

(By basic Proportionality Theorem,)

\(\frac{\mathrm{AM}}{\mathrm{AB}}\) = \(\frac{\mathrm{AL}}{\mathrm{AC}}\) ……….. (1)

In ∆ADC, LN // CD

(By Basic Proportionality Theorem,)

\(\frac{\mathrm{AN}}{\mathrm{AD}}\) = \(\frac{\mathrm{AL}}{\mathrm{AC}}\) ……….. (2)

From (1) and (2) we have

\(\frac{\mathrm{AM}}{\mathrm{AB}}\) = \(\frac{\mathrm{AN}}{\mathrm{AD}}\)

Alter :

In ∆ABC, ML // BC, AC is a transversal.

∠ALM = ∠ACB (Corresponding angles)

ML // BC, AB is a transversal.

∠AML = ∠ABC (Coressponding angles)

Now in ∆^{les} AML and ABC,

∠ALM = ∠ACB;

∠AML = ∠ABC;

∠A is common.

∴ ∆AML ~ ∆ABC

∴ Their corresponding sides are proportional.

(i.e.,)\(\frac{\mathrm{AM}}{\mathrm{AB}}\) = \(\frac{\mathrm{AL}}{\mathrm{AC}}\) ……….. (1)

Similarity in ∆ADC, LN // CD.

So we can prove that

\(\frac{\mathrm{AN}}{\mathrm{AD}}\) = \(\frac{\mathrm{AL}}{\mathrm{AC}}\) ……….. (2)

Hence from (1) & (2), we get

\(\frac{\mathrm{AM}}{\mathrm{AB}}\) = \(\frac{\mathrm{AN}}{\mathrm{AD}}\)

Question 3.

In the given figure, DE // AC and DF // AE prove that \(\frac{\mathrm{BF}}{\mathrm{FE}}\) = \(\frac{\mathrm{BE}}{\mathrm{EC}}\). (AS_{2})

Solution:

In ∆ABC, DE // AC.

(∴ By Basic Proportionality Theorem,)

In ∆ABE, DF // AE

(∴ By Basic Proportionality Theorem,)

Hence proved.

Question 4.

Prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side (Using basic proportionality theorem). (AS_{2})

Solution:

In ∆ABC, P is the mid-point of AB.

PQ is drawn parallel to BC, intersecting AC in Q.

By Basic Proportionality theorem,

Since P is the mid-point of AB, we have

AP = PB

∴ \(\frac{\mathrm{AQ}}{\mathrm{QC}}\) = \(\frac{\mathrm{AP}}{\mathrm{PB}}\)

⇒ \(\frac{\mathrm{AQ}}{\mathrm{QC}}\) \(\frac{\mathrm{AP}}{\mathrm{AP}}\) (∵ PB = AP)

⇒ \(\frac{\mathrm{AQ}}{\mathrm{QC}}\) = \(\frac{1}{1}\)

⇒ AQ = QC

Since AQ = QC, Q bisects AC.

Question 5.

Prove that a line joining the midpoints of any two sides of a triangle is parallel to the third side. (Using converse of basic proportionality theorem). (AS_{2})

Solution:

In ∆ABC, D and E are the mid-points of the sides AB and AC respectively.

In ∆ABC, DE divides the sides.

AB and AC in the same ratio.

∴ DE||BC (Basic Proportionality Theorem)

Question 6.

In the given figure, DE || OQ and DF || OR. Show that EF || QR. (AS_{2})

Solution:

In the given figure,

DE || OQ and DF || OR

In ∆PQO, DE || OQ

∴ \(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{\mathrm{PD}}{\mathrm{DO}}\) (Basic Proportionality Theorem) ………….. (1)

Similarly in ∆PRO, DF || OR.

∴ \(\frac{\mathrm{PF}}{\mathrm{FR}}\) = \(\frac{\mathrm{PD}}{\mathrm{DO}}\) (Basic Proportionality Theorem) ………….. (2)

From (1) & (2), we have

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

Now, in ∆PQR, we get that \(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{\mathrm{PF}}{\mathrm{FR}}\)

∴ EF || QR (By converse of basic proportionality theorem).

Question 7.

In the adjacent figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR. (AS_{2})

Solution:

In ∆ OPQ, A and B are the points on

OP and OQ such that AB || PQ.

So, in ∆ OPQ, AB || PQ.

∴ \(\frac{\mathrm{OA}}{\mathrm{AP}}\) = \(\frac{\mathrm{OB}}{\mathrm{BQ}}\) (Basic proportionality theorem) …………… (1)

In ∆ OPR, A and C are the points on OP and OR such that AC || PR.

So, in ∆ OPR, AC || PR.

∴ \(\frac{\mathrm{OA}}{\mathrm{AP}}\) = \(\frac{\mathrm{OC}}{\mathrm{CR}}\) (Basic proportionality theorem) …………… (2)

From (1) & (2), we get \(\frac{\mathrm{OB}}{\mathrm{BQ}}\) = \(\frac{\mathrm{OC}}{\mathrm{CR}}\)

Now, in ∆ OQR, we get that \(\frac{\mathrm{OB}}{\mathrm{BQ}}\) = \(\frac{\mathrm{OC}}{\mathrm{CR}}\)

Therefore, BC || QR (Converse of the Basic proportionality theorem).

Question 8.

ABCD is a trapezium in which AB ||DC and its diagonals intersect each other at point ‘O’. Show that \(\frac{\mathrm{AO}}{\mathrm{BO}}\) = \(\frac{\mathrm{CO}}{\mathrm{DO}}\) (AS_{2})

Solution:

ABCD is a trapezium. AB || DC.

The diagonals AC and BD intersect in ‘O’.

Draw a line through ‘O’ parallel to AB, meeting AD in E and BC in F.

Question 9.

Draw a line segment of length 7.2 cm and divide it in the ratio 5:3. Measure the two parts. (AS_{3}, AS_{5})

Solution:

Construction :

Draw a line segment AB = 7.2 cm. Make an angle BAX. Mark A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7} on AX such that AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5} = A_{5}A_{6} = A_{6}A_{7} = A_{7}A_{8}. Mark D on AX such that AD = 5 equal parts and DC = 3 equal parts. Join BC. From D, draw DE || BC. E divides AB in the ratio 5:3. The parts are measured. AE = 4.2 cm on and EB = 2.7 cm.