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RBSE Class 10 Maths Important Questions Chapter 6 Triangles

April 18, 2022 by Safia Leave a Comment

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—
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 1
(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—
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 2
(A) 1
(B) 2
(C) 3
(D) 4
Answer:
(C) 3

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

Question 4.
The areas of two similar triangles are respectively 25 cm2 and 36 cm2. 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—
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 3
(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°

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

Question 7.
If in △ABC, D is the mid-point of side BC and AB2 + AC2 = x (BD2 + AD2), 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—
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 4
(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
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 5
\(\frac{B D}{D C}\) = \(\frac{A B}{A C}\)
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

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

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.

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

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:
\(\frac{AB}{AC}\) = \(\frac{BD}{DC}\)
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 6
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 cm2, then write the area of △DEF.
Solution:
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 7
= 18 cm2

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:
\(\frac{AQ}{QC}\) = \(\frac{AP}{PB}\)
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 8
= \(\frac{5}{2.5}\) = \(\frac{4}{PB}\)
So PB = \(\frac{4}{2}\) = 2 cm

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

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:
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 9
So AB = 15 cm

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

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—

  1. their corresponding angles are equal.
  2. their corresponding sides are proportional.

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

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.
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 10
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°.
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 11
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°

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

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
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 12
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?
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 13
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)
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 14
or \(\frac{AB}{AD}\) = \(\frac{AC}{AE}\)
So \(\frac{AB}{AD}\) = \(\frac{AC}{AE}\) (Hence Proved)

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

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 \(\frac{AE}{ED}\) = \(\frac{BE}{FC}\)
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 15
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)
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 16
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,
\(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\)
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 17
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)

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

Question 4.
In the given figure, if PQ || RS, prove that △POQ ~ △SOR
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 18
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.
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 19
Solution:
In △ABC and △PQR,
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 20
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°

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

Question 6.
In the given figure, OA. OB = OC . OD. Show that ∠A = ∠C and ∠B = ∠D
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 21
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 \(\frac{\mathbf{B C}^{2}}{\mathrm{AC}^{2}}\) = \(\frac{BD}{AD}\)
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 22
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.]
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 23

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

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
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 24
By Phthagoras Theorem,
AB2 = BC2 + CA2
= (25)2 + (6)2
= 6.25 + 36
AB2 = 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 AB2 + CD2 = BD2 + AC2
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 25
Solution:
From △ADC, we have
AC2 = AD2 + CD2 …(1)
(By Pythagoras Theorem)
From △ADB, we have
AB2 = AD2 + BD2
(By Pythagoras Theorem)
Equation (2) – Equation (1)
AB2 – AC2 = BD2 – CD2
or AB2 + CD2 = BD2 + AC2
(Hence Proved)

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

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
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 26

Question 11.
In the figure, △OPR ~ △OSK, ∠POS = 125° and ∠PRO = 70°.
Find the values of ∠OKS and ∠ROP
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 27
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°

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

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.
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 28
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.
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 29
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 30

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)
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 31
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.

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

Question 3.
In figure, CM and RN are respectively the medians of △ABC and △PQR. If △ABC ~ △PQR, prove that—
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 32
(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) \(\frac{CD}{RS}\) = \(\frac{AB}{PQ}\)
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 33
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),
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 34
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
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 35
Now in △ABM and △PQN,
∠B = ∠Q
(∵ △ABC ~ △PQR)
and ∠M = ∠N (Each is of 90°)
So △ABM ~ △PQN
(AA similarity criterion)
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 36

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

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
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 37
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 38
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 39

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°
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 40
To Prove—AC2 = AB2 + BC2
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
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 41

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

Question 7.
BL and CM are medians of a right triangle ABC and ∠A = 90°, then prove that
4(BL2 + CM2) = 5BC2
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(BE2 + CF2) = 5BC2
Solution:
BL and CM are medians of a AABC in which ∠A = 90°, as shown in the figure.
From △ABC
BC2 = AB2 + AC2 …(i)
(Phythagoras Theorem)
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 42
From △ABL
or BL2 = AL2 + AB2
or BL2 = \(\left(\frac{\mathrm{AC}}{2}\right)^{2}\) + AB2
(∵ L is the mid-point of AC)
or BL2 = \(\frac{\mathrm{AC}^{2}}{4}\) + AB2
or 4BL2 = AC2 + 4AB2 …(ii)
From, △CMA
CM2 = AC2 + AM2
or CM2 = AC2 + \(\left(\frac{\mathrm{AB}}{2}\right)^{2}\)
(∵ M is the mid-point of AB)
or CM2 = AC2 + \(\frac{\mathrm{AB}^{2}}{4}\)
or 4CM2 = 4AC2 + AB2 … (iii)
Adding equations (ii) and (iii), we get
4BL2 + 4CM2 = AC2 + 4AB2 + 4AC2 + AB2
or 4(BL2 + CM2) = 5AC2 + 5AB2
or 4(BL2 + CM2) = 5 (AC2 + AB2)
or 4(BL2 + CM2) = 5BC2
[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—
OB2 + OD2 = OC2 + OA2
Solution:
Given—O is a point inside a rectangle ABCD, OA, OB, OC and OD have been joined.
To Prove :
OB2 + OD2 = OC2 + OA2
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 43
Construction : Through O draw LM || AD which meets DC and AB at Land M respectively.
Proof : We know that
OB2 = OM2 + MB2
(By Pythagoras Theorem)
= OM2 + CL2
and OD2 = OL2 + DL2 = OL2 + AM2
∴ OB2 + OD2 = OM2 + CL2 + OL2 + AM2
= (OM2 + AM2) + (CL2 + OL2)
= OA2 + OC2
So OB2 + OD2 = OC2 + OA2
(Hence Proved)

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

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°
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 44
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
DF2 = DE2 + EF2
⇒ DF2 = AB2 + BC2
[ v DE = AB and EF = BC (by Construction)]
⇒ DF2 = AC2 … (i)
[∵ Given that AB2 + BC2 = AC2]
⇒ 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
OB2 + OD2 = OA22 + OC2
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 45
Solution:
O is a point inside a rectangle ABCD, OA, OB, OC and OD have been joined.
To Prove : OB2 + OD2 = OA2 + OC2
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
OB2 = OP2 + PB2
and OD2 = OR2 + DR2
Adding both
⇒ OB2 + OD2 = OP2 + PB2 + OR2 + DR2
⇒ OB2 + OD2 = OP2 + OR2 + PB2 + DR2 …..(i)
Again, In right angle triangle ORC and OPA, By Pythagoras theorem
OC2 = OR2 + RC2
and OA2 = OP2 + PA2
Adding both
⇒ OC2 + OA2 = OR2 + RC2 + OP2 + PA2
⇒ OC2 + OA2 = OR2 + OP2 + RC2 + PA2
⇒ OC2 + OA2 = OR2 + OP2 + PB2 + DR2
[∵ RC = PB and PA = DR]
⇒ OC2 + OA2 = (OR2 + OP2) + (PB2 + DR2) …..(ii)
From equation (i) and (ii)
OB2 + OD2 = OC2 + OA2
⇒ OB2 + OD2 = OA2 + OC2
(Hence Proved)

RBSE Class 10 Maths Important Questions Chapter 6 Triangles

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.
RBSE Class 10 Maths Important Questions Chapter 6 Triangles 46
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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