Rajasthan Board RBSE Class 10 Maths Important Questions Chapter 6 Triangles Important Questions and Answers.

## RBSE Class 10 Maths Chapter 6 Important Questions Triangles

Objective Type Questions—

Question 1.

Two straight lines which are perpendicular to the same straight line are mutually called—

(A) perpendicular lines

(B) parallel lines

(C) bisected lines

(D) same lines

Answer:

(B) parallel lines

Question 2.

In the given figure, AD is the internal bisector of A. If AB = 6 cm, BD = 4 cm and DC = 3 cm, then the value of AC is—

(A) 3 cm

(B) 4 cm

(C) 4.5 cm

(D)5 cm

Answer:

(C) 4.5 cm

Question 3.

In the adjacent figure if AB || CD then the value of x is—

(A) 1

(B) 2

(C) 3

(D) 4

Answer:

(C) 3

Question 4.

The areas of two similar triangles are respectively 25 cm^{2} and 36 cm^{2}. If the length of a median of the smaller triangle is 10 cm, then the length of the corresponding median of the larger triangle is—

(A) 12 cm

(B) 15 cm

(C) 16cm

(D) 18cm

Answer:

(A) 12 cm

Question 5.

In the given figure QA and PB are perpendicular to AB. If AO = 10 cm, BO = 6 cm and PB = 9 cm. Then the length of AQ is—

(A) 12 cm

(B) 15 cm

(C) 18 cm

(D)21cm

Answer:

(B) 15 cm

Question 6.

△ABC ~ △DEF, if ∠A = 40°, ∠E = 80° then the value of ∠C is—

(A) 70°

(B) 60°

(C) 50°

(D) 40°

Answer:

(B) 60°

Question 7.

If in △ABC, D is the mid-point of side BC and AB^{2} + AC^{2} = x (BD^{2} + AD^{2}), then the value of x will be—

(A) 1

(B) 2

(C) 4

(D) zero

Answer:

(B) 2

Question 8.

In figure ABC is a right isosceles triangle where ∠B = 90°. If BC = 8 cm, then what will be the length of AD? Where D is the mid-point of BC—

(A) 20 cm

(B) 720 cm

(C) 2720 cm

(D) 4720 cm

Answer:

(C) 2720 cm

Very Short Answer Type Questions—

Question 1.

In △ABC, AD is the bisector of ∠BAC. If AB = 4 cm, AC = 6 cm, BD = 2 cm then find the value of BC.

Solution:

By Basic Proportionality Theorem

\(\frac{B D}{D C}\) = \(\frac{A B}{A C}\)

or DC = \(\frac{B D \times A C}{A B}\)

Putting the value

= \(\frac{6 \times 2}{4}\) = 3 cm

∴ BC = BD + DC

= 2 + 3 = 5

Question 2.

How are the angles opposite to equal sides of any triangle?

Answer:

Angles opposite to equal sides of a triangle are equal.

Question 3.

Sides of two similar triangles are in the ratio of 4 : 5 find the ratio of the areas of these triangles.

Answer:

∵ The ratio of the areas of similar triangles is equal to the square of the ratio of corresponding sides, therefore ratio of areas of triangles

= ( 4 : 5)^{2}

= 16 : 25

Question 4.

Write the statement of Baudhayan’s Theorem.

Answer:

Baudhayan Theorem— The area of a square formed by the diagonal of any square is equal to the sum of the squares formed on its both the adjcent sides.

Question 5.

In the given figure AD is the internal bisector of ∠A. If AB = 4 cm, AC = 6 cm then write BD : DC.

Solution:

\(\frac{AB}{AC}\) = \(\frac{BD}{DC}\)

So \(\frac{4}{6}\) = \(\frac{BD}{DC}\)

So BD : DC = 4 : 6

= 2 : 3

Question 6.

If △ABC ~ △DEF, AB = 5 cm, DE = 3 cm and area of △ABC = 50 cm^{2}, then write the area of △DEF.

Solution:

= 18 cm^{2}

Question 7.

In the adjacent figure BC || PQuestion If AP = 4 cm, AQ = 5 cm and QC = 2.5 cm, then find the value of PB.

Solution:

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

= \(\frac{5}{2.5}\) = \(\frac{4}{PB}\)

So PB = \(\frac{4}{2}\) = 2 cm

Question 8.

In the figure of above question if AB = 7 cm, AP = 5 cm and AC = 10.5 cm, then find the value of AQ.

Solution:

\(\frac{AC}{AQ}\) = \(\frac{AB}{AP}\)

= \(\frac{10.5}{AQ}\) = \(\frac{7}{5}\)

So AQ = \(\frac{10.5 \times 5}{7}\) = 7.5 cm

Question 9.

On side BC of △ABC points P and Q are such that BP = 3 cm, BQ = 5 cm and BC = 6 cm, then find the value of ratio

\(\frac{a r(\Delta \mathrm{ABP})}{a r(\Delta \mathrm{APQ})}\)

Solution:

\(\frac{a r(\triangle A B P)}{a r(\triangle A P Q)}\) = \(\frac{(B P)^{2}}{(P Q)^{2}}\) = \(\frac{(3)^{2}}{(2)^{2}}\)

= \(\frac{9}{4}\)

Question 10.

The perimeters of two similar triangles ABC and PQR are 36 cm and 24 cm. If PQ = 10 cm then find AB—

Solution:

So AB = 15 cm

Question 11.

If in two triangles ABC and XYZ \(\frac{\mathbf{A B}}{\mathbf{X Y}}\) = \(\frac{\mathbf{B C}}{\mathbf{Y Z}}\) = \(\frac{\mathbf{C A}}{\mathbf{Z X}}\), then to which angle of triangle XYZ will the value of angle A of △ABC be?

Answer:

Equal to ∠X

Question 12.

If in △ABC and △DEF \(\frac{\mathbf{A B}}{\mathbf{D E}}\) = \(\frac{\mathbf{B C}}{\mathbf{E F}}\) = \(\frac{\mathbf{A C}}{\mathbf{D F}}\), then of what type these triangles will be mutually?

Answer:

Similar triangles.

Question 13.

State any two conditions of two triangles to be similar.

Answer:

Two triangles are similar if—

- their corresponding angles are equal.
- their corresponding sides are proportional.

Question 14.

Write SSS Rule.

Answer:

SSS Rule—If the corresponding sides of two triangles are proportional, then they both are similar.

Question 15.

D and E are points on the sides AB and AC respectively of △ABC such that DE || BC. If AD = 8 cm, AB = 12 cm and AE = 12 cm, then write the measure of CE.

Solution:

By Basic Proportionality Theorem

\(\frac{A D}{B D}\) = \(\frac{A E}{C E}\)

Here AD = 8 cm, BD = (12 – 8) = 4 cm and AE = 12 cm.

Hence putting the values

\(\frac{8}{4}\) = \(\frac{12}{CE}\)

CE = \(\frac{12}{4}\) × 4 = 6 cm

Question 16.

In ABC, D is any point on \(\frac{AB}{AC}\) = \(\frac{BD}{DC}\), such that , ∠B = 70° and ∠C = 50° then write the measure of ∠BAD.

Solution:

Given : △ABC in which D is a point on BC such that \(\frac{AB}{AC}\) = \(\frac{BD}{DC}\) and ∠B = 70° and ∠C = 50°.

To determine: ∠BAD

Proof: Here \(\frac{AB}{AC}\) = \(\frac{BD}{DC}\) (given)

∴ AD is the bisector of ∠BAC.

So ∠BAD = \(\frac{1}{2}\) ∠A = \(\frac{60^{\circ}}{2}\) = 30°

Question 17.

In the given figure ∠ABC = 90° and BD ⊥ AC. If BD = 8 cm and AD = 4 cm, then write the measure of CD.

Solution:

In △ABD and △BDC

∠BDC = ∠BDA (each 90°)

∠DBC = ∠BDC

So △ABD ~ △BDC

∵ \(\frac{AB}{BD}\) = \(\frac{BD}{CD}\)

\(\frac{4}{8}\) = \(\frac{8}{CD}\) or 4CD = 64

So CD = \(\frac{64}{4 }\) = 16 cm

Question 18.

In figure EF || BC, if AE : BE = 4:1 and CF =1.5 cm, then what will be the length of AF?

Solution:

In figure EF || BC

∴ \(\frac{AE}{BE}\) = \(\frac{AF}{CF}\)

⇒ \(\frac{4}{1}\) = \(\frac{AF}{1.5}\)

⇒ AF = 4 × 1.5

= 6.0 cm.

Short Answer Type Questions—

Question 1.

If a line intersects sides AB and AC of a △ABC at D and E respectively and is parallel to side BC, then prove that \(\frac{AD}{AB}\) = \(\frac{AE}{AC}\) (see figure)

Solution:

DE || BC (given)

or \(\frac{AB}{AD}\) = \(\frac{AC}{AE}\)

So \(\frac{AB}{AD}\) = \(\frac{AC}{AE}\) (Hence Proved)

Question 2.

ABCD is a trapezium with AB || DC. E and F are points or non-parallel sides AD and BC respectively such that EF is parallel to AB (see figure). Show that \(\frac{AE}{ED}\) = \(\frac{BE}{FC}\)

Solution:

Join A and C which intersects EF at G. (see figure)

AB || DC and EF || AB (Given)

So, EF || DC (Lines parallel to the same line one parallel to each other)

Now, in △ADC.

EG || DC (As EF || DC)

So, \(\frac{AE}{ED}\) = \(\frac{AG}{GC}\) (Theorem 6.1) ,..(i)

Similarly from △CAB,

\(\frac{CG}{AG}\) = \(\frac{CF}{BF}\)

i.e., \(\frac{AG}{GC}\) = \(\frac{BF}{FC}\) ….(ii)

Therefore From (i) and (ii),

\(\frac{AE}{ED}\) = \(\frac{BF}{FC}\) (Hence Proved)

Question 3.

In figure, \(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\) and ∠PST = ∠PRQuestion Prove that ∠PQR is an isoscales triangle.

Solution:

It is given that,

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

So, ST || QR (Theorem 6.2)

Therefore, ∠PST = ∠PQR

(Corresponding angles)… (i)

Also, it is given that

∠PST = ∠PRQ

So, ∠PRQ = ∠PQR [From (i) and (ii)]

Therefore, PQ = PR (Sides opposite to equal angles)

i.e., PQR is an isosceles triangle. (Hence Proved)

Question 4.

In the given figure, if PQ || RS, prove that △POQ ~ △SOR

Solution:

PQ || RS

So ∠P = ∠S (Alternate angles)

and ∠Q = ∠R (Altemet angles)

Also, ∠POQ = ∠SOR

(Vertically opposite angles)

Therefore, △POQ ~ △SOR

(AAA similarity criterion)

(Hence Proved)

Question 5.

In the given figure find ∠P.

Solution:

In △ABC and △PQR,

So △ABC ~ △RQP (SSS similarity)

Therefore, ∠C = ∠P (Corresponding angles of similar triangles)

But ∠C = 180° – ∠A – ∠B (angle sum property of A)

⇒ ∠C = 180° – 80° – 60° = 40°

So, ∠P = 40°

Question 6.

In the given figure, OA. OB = OC . OD. Show that ∠A = ∠C and ∠B = ∠D

Solution:

OA.OD = OC.OB (given)

So, \(\frac{OA}{OC}\) = \(\frac{OD}{OB}\) …(i)

Also, we have

∠AOD = ∠COB

(Vertically opposite angles) … (2)

Therefore, from equation (1) and (2),

△AOD ~ △COB

(SAS similarity criterion)

So, ∠A = ∠C

∠B = ∠D

(Corresponding angles of similar triangles)

Question 7.

In the given figure ∠ACB = 90° and CD ⊥ AB. Prove that \(\frac{\mathbf{B C}^{2}}{\mathrm{AC}^{2}}\) = \(\frac{BD}{AD}\)

Solution:

△ACD ~ △ABC

[Since we know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.]

Question 8.

A ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall and its top reaches a window 6 m above the ground. Find the length of the ladder.

Solution:

Let AB be the ladder and CA be the wall with the window at A as shown in the figure.

Also, BC = 2.5 m

and CA = 6 m

By Phthagoras Theorem,

AB^{2} = BC^{2} + CA^{2}

= (25)^{2} + (6)^{2}

= 6.25 + 36

AB^{2} = 42.25

∴ AB = \(\sqrt {42.25}\) = 6.5 m

So, AB = 6.5 m

Thus, the length of the ladder is 6.5 m

Question 9.

In the given figure, AD ⊥ BC. Prove that AB^{2} + CD^{2} = BD^{2} + AC^{2}

Solution:

From △ADC, we have

AC^{2} = AD^{2} + CD^{2} …(1)

(By Pythagoras Theorem)

From △ADB, we have

AB^{2} = AD^{2} + BD^{2}

(By Pythagoras Theorem)

Equation (2) – Equation (1)

AB^{2} – AC^{2} = BD^{2} – CD^{2}

or AB^{2} + CD^{2} = BD^{2} + AC^{2}

(Hence Proved)

Question 10.

The length of the shadow of 2 m long student on a plane ground is 1 m. At the same time the length of the shadow of a tower is 5 m, then find the height of the tower.

Solution:

Let the height of the tower be h m. From figure we see that △ABC and △DEC are similar, i.e.,

△ABC ~ △DEC

So, \(\frac{AB}{DE}\) = \(\frac{CB}{CE}\)

⇒ \(\frac{2}{h}\) = \(\frac{1}{5}\)

or h = 5 × 2 = 10 m

Question 11.

In the figure, △OPR ~ △OSK, ∠POS = 125° and ∠PRO = 70°.

Find the values of ∠OKS and ∠ROP

Solution:

According to question

∠POS = 125° and ∠PRO = 70°

According to figure, ROS is a straight line

∴ ∠ROP + ∠POS = 180°

or ∠ROP + 125° = 180°

or ∠ROP = 180° – 125°

= 55°

When ∠ROP = 55° then ∠KOS will also 55° because due to vertically opposite angle

∵ △OPR ~ △OSK

∴ ∠R = ∠S = 70°

Therefore, In △ROP

∠R + ∠O + ∠P = 180°

∴ 70° + 55° + ∠P = 180°

or ∠P = 180° – 70° – 55°

∴ ∠P = 55°

∴ ∠P = ∠K = 55°

So ∠OKS = 55° and ∠ROP = 55°

Long Answer Type Questions—

Question 1.

If a line drawn parallel to one side of any triangle intersects the other two sides in two points, then these points divide those sides in the same ratio. Prove.

Or

If a line is drawn parallel to one side of a triangle to intersect the other two sides in district points, then prove that the other two sides are divided in the same ratio.

Solution:

Given : In △ABC, DE || BC has been drawn and this intersects AB at D and AC at E.

To Prove : \(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)

Construction : Join B to E and C to D and draw EF ⊥ BA and DG ⊥ AC

Proof : △BDE and △CED lie on the same base DE and between the same parallel lines DE and BC

∴ Area of △BDE = Area of △CED … (1)

∵ The common height of △ADE and △BDE is EF.

Question 2.

A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.

Solution:

Let AB devote the lamp-post and CD the girl after walking for 4 seconds away from the lamp-post, (see figure)

According to the figure DE is the length of the shadow of the girl. Let DE be x cm.

Now, BD = 1.2 m × = 4.8 m

Now in △ABE and △CDE

∠B = ∠D (Each is of 90°, because lamp-post as well as the girl are standing vertical to the ground)

and ∠E = ∠E (Same angle)

So, △ABE ~ △CDE

(AA similarity criterion)

Therefore, \(\frac{BE}{DE}\) = \(\frac{AB}{CD}\)

(Corresponding sides of similar triangles)

\(\frac{\mathrm{BD}+\mathrm{DE}}{\mathrm{DE}}\) = \(\)

or \(\frac{4.8+x}{x}\) = \(\frac{3.6}{0.9}\)

(90 cm = \(\frac{90}{100}\) m = 0.9 m)

or, 4.8 + x = 4x

or, 3x = 4.8

x = 1.6

So, the length of the shadow of the girl after walking for 4 seconds is 1,6 m.

Question 3.

In figure, CM and RN are respectively the medians of △ABC and △PQR. If △ABC ~ △PQR, prove that—

(i) △AMC ~ △PNR

(ii) \(\frac{CM}{RN}\) = \(\frac{AB}{QP}\)

(iii) △CMB ~ △RNQ

Or

In figure CD and RS are respectively the medians of △ABC and △PQR. If △ABC ~ △PQR, then prove that

(i) △ADC ~ △PSR

(ii) \(\frac{CD}{RS}\) = \(\frac{AB}{PQ}\)

Solution:

(i) △ABC ~ △PQR (Given)

So \(\frac{AB}{PQ}\) = \(\frac{BC}{QR}\) = \(\frac{CA}{RP}\) …(i)

and ∠A = ∠P, ∠B = ∠Q

and ∠C = ∠R …(ii)

AB = 2AM and PQ = 2PN (Since CM and RN are medians)

So from (i),

or \(\frac{CM}{RN}\) = \(\frac{BC}{QR}\) = \(\frac{BM}{QN}\)

[From (ix) and (x)]

So △CMB ~△RNQ (SSS similarity)

(Hence Proved)

Question 4.

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Solution:

Given— Two triangles ABC and PQR are given such that

△ABC ~ △PQR

Now in △ABM and △PQN,

∠B = ∠Q

(∵ △ABC ~ △PQR)

and ∠M = ∠N (Each is of 90°)

So △ABM ~ △PQN

(AA similarity criterion)

Question 5.

D and E are points on sides AB and AC respectively of ABC such that DE || BC. If line DE divides the triangle ABC into two figures of equal areas, then prove that \(\frac{\mathbf{B D}}{\mathbf{A B}}\) = \(\frac{2-\sqrt{2}}{2}\)

Solution:

Given : △ABC in which DE || BC.

To prove : \(\frac{\mathbf{B D}}{\mathbf{A B}}\) = \(\frac{2-\sqrt{2}}{2}\)

Proof: In △ABC, DE || BC

∴ ∠1 = ∠2 = ∠3

∴ △ABC ~ △ADE

Question 6.

In a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides. Prove.

Solution:

Given—

In a △ABC, ∠B = 90°

To Prove—AC^{2} = AB^{2} + BC^{2}

Construction—Draw BD ⊥ AC

Proof—In △ABC and △ADB,

∵ ∠ABC = ∠ADB (Each 90°)

and ∠A = ∠A (common)

∴ By Angle-Angle similarity corollary

△ABC ~ △ADB

∵ The corresponding sides of similar triangles are proportional

Question 7.

BL and CM are medians of a right triangle ABC and ∠A = 90°, then prove that

4(BL^{2} + CM^{2}) = 5BC^{2}

Or

BE and CF are the medians of a right triangle ABC and angle A of this triangle is a right angle, Prove that

4(BE^{2} + CF^{2}) = 5BC^{2}

Solution:

BL and CM are medians of a AABC in which ∠A = 90°, as shown in the figure.

From △ABC

BC^{2} = AB^{2} + AC^{2} …(i)

(Phythagoras Theorem)

From △ABL

or BL^{2} = AL^{2} + AB^{2}

or BL^{2} = \(\left(\frac{\mathrm{AC}}{2}\right)^{2}\) + AB^{2}

(∵ L is the mid-point of AC)

or BL^{2} = \(\frac{\mathrm{AC}^{2}}{4}\) + AB^{2}

or 4BL^{2} = AC^{2} + 4AB^{2} …(ii)

From, △CMA

CM^{2} = AC^{2} + AM^{2}

or CM^{2} = AC^{2} + \(\left(\frac{\mathrm{AB}}{2}\right)^{2}\)

(∵ M is the mid-point of AB)

or CM^{2} = AC^{2} + \(\frac{\mathrm{AB}^{2}}{4}\)

or 4CM^{2} = 4AC^{2} + AB^{2} … (iii)

Adding equations (ii) and (iii), we get

4BL^{2} + 4CM^{2} = AC^{2} + 4AB^{2} + 4AC^{2} + AB^{2}

or 4(BL^{2} + CM^{2}) = 5AC^{2} + 5AB^{2}

or 4(BL^{2} + CM^{2}) = 5 (AC^{2} + AB^{2})

or 4(BL^{2} + CM^{2}) = 5BC^{2}

[From equation (i)]

(Hence Proved)

Question 8.

O a point lying inside a rectangle ABCD has been joined with the vertices A, B, C and D. Prove that—

OB^{2} + OD^{2} = OC^{2} + OA^{2}

Solution:

Given—O is a point inside a rectangle ABCD, OA, OB, OC and OD have been joined.

To Prove :

OB^{2} + OD^{2} = OC^{2} + OA^{2}

Construction : Through O draw LM || AD which meets DC and AB at Land M respectively.

Proof : We know that

OB^{2} = OM^{2} + MB^{2}

(By Pythagoras Theorem)

= OM^{2} + CL^{2}

and OD^{2} = OL^{2} + DL^{2} = OL^{2} + AM^{2}

∴ OB^{2} + OD^{2} = OM^{2} + CL^{2} + OL^{2} + AM^{2}

= (OM^{2} + AM^{2}) + (CL^{2} + OL^{2})

= OA^{2} + OC^{2}

So OB^{2} + OD^{2} = OC^{2} + OA^{2}

(Hence Proved)

Question 9.

Prove that if in a triangle the square of one side is equal to the sum of the square of the other two sides, then the angle appropriate the front side is a right angle.

Solution:

Given : A triangle ABC in which

AC = AB + BC

To Prove: ∠ABC = 90°

Construction : Construct a triangle DEF such that DE = AB, EF = BC and ∠E = 90°

Proof: To prove that ∠ABC = 90°, we shall have simply to prove that △ABC ~ △DEF

∵ △DEF is a right triangle in which ∠DEF is a right angle

So by Pythagoras Theorem

DF^{2} = DE^{2} + EF^{2}

⇒ DF^{2} = AB^{2} + BC^{2}

[ v DE = AB and EF = BC (by Construction)]

⇒ DF^{2} = AC^{2} … (i)

[∵ Given that AB^{2} + BC^{2} = AC^{2}]

⇒ DF = AC

So in △ABC and △DEF

AB = DE (by Construction)

BC = EF (by Construction)

and AC = DF (From (i)]

Hence by SSS criterion of conference

△ABC ≅ △DEF

⇒ ∠B = ∠E

[∵ Corresponding angle of congruent triangles are the same]

⇒ ∠B = ∠E = 90°

[∵ ∠E = 90° (by Construction)]

So △ABC is a right angle triangle.

(Hence Proved)

Question 10.

O is any point inside rectangle ABCD. Prove that

OB^{2} + OD^{2} = OA^{2}2 + OC^{2}

Solution:

O is a point inside a rectangle ABCD, OA, OB, OC and OD have been joined.

To Prove : OB^{2} + OD^{2} = OA^{2} + OC^{2}

Construction : Through 0 draw PR || BC which meet AB and DC at P and R respectively.

Proof : In right angle triangle OPB and ORD, By pythagoras theorem

OB^{2} = OP^{2} + PB^{2}

and OD^{2} = OR^{2} + DR^{2}

Adding both

⇒ OB^{2} + OD^{2} = OP^{2} + PB^{2} + OR^{2} + DR^{2}

⇒ OB^{2} + OD^{2} = OP^{2} + OR^{2} + PB^{2} + DR^{2} …..(i)

Again, In right angle triangle ORC and OPA, By Pythagoras theorem

OC^{2} = OR^{2} + RC^{2}

and OA^{2} = OP^{2} + PA^{2}

Adding both

⇒ OC^{2} + OA^{2} = OR^{2} + RC^{2} + OP^{2} + PA^{2}

⇒ OC^{2} + OA^{2} = OR^{2} + OP^{2} + RC^{2} + PA^{2}

⇒ OC^{2} + OA^{2} = OR^{2} + OP^{2} + PB^{2} + DR^{2}

[∵ RC = PB and PA = DR]

⇒ OC^{2} + OA^{2} = (OR^{2} + OP^{2}) + (PB^{2} + DR^{2}) …..(ii)

From equation (i) and (ii)

OB^{2} + OD^{2} = OC^{2} + OA^{2}

⇒ OB^{2} + OD^{2} = OA^{2} + OC^{2}

(Hence Proved)

Question 11.

In the given figure, \(\frac{\mathrm{PK}}{\mathrm{KS}}\)= \(\frac{\mathrm{PT}}{\mathrm{TR}}\) and ∠PKT = ∠PRS. Prove that △PSR is an isosceles triangle.

Solution:

According to the question, it is given that

\(\frac{\mathrm{PK}}{\mathrm{KS}}\)= \(\frac{\mathrm{PT}}{\mathrm{TR}}\)

So KT || SR

∴ ∠PKT = ∠PSR

(Corresponding angle) …..(i)

It is also given that

∠PKT = ∠PRS … (ii)

From equation (i) and (ii),

we get ∠PRS = ∠PSR

∴ PS = PR

(sides opposite to equal angles)

Therefore △PSR is an isosceles triangle.

(Hence Proved)

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