# CBSE Solutions for Class 10 English

#### GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

D and E are points on the sides AB and AC, respectively, of a ΔABC, such that DEBC. If AD = 3.6 cm, AB = 10 cm and AE = 4.5 cm, find EC and AC.

જવાબ : In  ABC, it is given that DEBC.
Applying Thales' theorem, we get:
AD = 3.6 cm , AB = 10 cm, AE = 4.5 cm​
DB = 10 −- 3.6 = 6.4 cm
or, 3.6/6.4= 4.5EC
or, EC = 6.4×4.5/3.6 or, EC =8 cm

D and E are points on the sides AB and AC, respectively, of a ΔABC, such that DEBC. If AB = 13.3 cm, AC = 11.9 cm and EC = 5.1 cm, find AD.

જવાબ :

In ABC, it is given that DE BC.

Applying Thales' Theorem, we get:

Adding 1 to both sides, we get:

AD/DB +1 = AE/EC+1

AB/DB= AC/EC

13.3/DB = 11.9/5.1

DB = 13.3×5.1/11.9 = 5.7 cm Therefore, AD = AB − DB = 13.5 − 5.7 = 7.6 cm

D and E are points on the sides AB and AC, respectively, of a ΔABC, such that DEBC. If AD/DB=4/7 and AC=6.6 cm, find AE.

જવાબ :

In ABC, it is given that DEBC.

Applying Thales' theorem, we get:

4/7= AE/EC

Adding 1 to both the sides, we get:

11/7= AC/EC

EC = 6.6×7/11 = 4.2 cm Therefore, AE = AC −EC= 6.6−4.2 = 2.4 cm

D and E are points on the sides AB and AC, respectively, of a ΔABC, such that DEBC. If AD/AB=8/15 and EC=3.5 cm, find AE.

જવાબ : In ABC, it is given that DEBC. Applying Thales' theorem, we get:  AD/AB=AE/AC 8/15= AE/AE + EC 8/15 = AE/(AE+3.5) 8AE + 28 = 15AE 7AE = 28 AE = 4 cm

D and E are points on the sides AB and AC respectively of a ΔABc such that DEBC. Find the value of x, when AD=x cm, DB=(x−2)cm,AE=(x+2) cm and EC=(x−1) cm.

જવાબ :

In △ABC, it is given that DE∥BC.

Applying Thales' theorem, we have:

x/x−2=x+2/x−1

x(x−1) = (x−2)(x+2)

x2−x = x2−4

x=4 cm

D and E are points on the sides AB and AC respectively of a ΔABc such that DEBC. Find the value of x, when AD=4 cm, DB=(x−4) cm, AE=8 cmand EC=(3x−19) cm.

જવાબ :

In ABC, it is given that DEBC.

Applying Thales' theorem, we have:

4x−4 = 83x−19

4(3x−19) = 8(x−4)

12x −76 = 8x – 32

4x = 44

x = 11 cm

D and E are points on the sides AB and AC respectively of a ΔABc such that DEBC. Find the value of x, when AD=(7x−4) cm, AE=(5x−2) cm,

DB=(3x+4) cm and EC=3x cm.

જવાબ :

In △ABC, it is given that DE∥BC.

Applying Thales' theorem, we have:

7x−43x+4 = 5x−23x

3x(7x−4) =(5x−2)(3x+4)

21x2 − 12x = 15x2 +14 x−8

6x2−26x+8 = 0

(x−4)(6x−2) = 0

x = 4, 13∵ x≠13     (as if x=13 then AE will become negative)

x =4 cm

and E are points on the sides AB and AC respectively of a ΔABC. In each of the following cases, determine whether DEBC or not. AD=5.7 cm, DB=9.5 cm, BD=4.8 cm and EC=8 cm.AD=5.7 cm, DB=9.5 cm, BD=4.8 cm and EC=8 cm.

જવાબ : We have:
AD/DB = 5.7/9.5 = 0.6 cm
AE/EC= 4.8/8 = 0.6 cm Hence,AD/DB=AE/EC Applying the converse of Thales' theorem, we conclude that DEBC

and E are points on the sides AB and AC respectively of a ΔABC. In each of the following cases, determine whether DEBC or not. AB=11.7 cm, AC=11.2 cm, BD=6.5 cm and AE=4.2 cm.AB=11.7 cm, AC=11.2 cm, BD=6.5 cm and AE=4.2 cm.

જવાબ : We have:
AB = 11.7 cm, DB = 6.5 cm
Therefore,
AD = 11.7 −- 6.5 = 5.2 cm
Similarly,
AC = 11.2 cm, AE = 4.2 cm
Therefore,
EC = 11.2 −- 4.2 = 7 cm

AE/EC = 4.2/7

Thus, AD/DB≠AE/EC Applying the converse of Thales' theorem,
we conclude that DE is not parallel to BC.

and E are points on the sides AB and AC respectively of a ΔABC. In each of the following cases, determine whether DEBC or not. AB=10.8 cm, AD=6.3 cm, AC=9.6 cm and EC=4 cm.AB=10.8 cm, AD=6.3 cm, AC=9.6 cm and EC=4 cm.

જવાબ : We have:
AB = 10.8 cm, AD = 6.3 cm
Therefore,
DB = 10.8 −- 6.3 = 4.5 cm
Similarly,
AC = 9.6 cm, EC = 4 cm
Therefore,
AE = 9.6 −- 4 = 5.6 cm
Now,
AE/EC=5.6/4=7/5
Applying the converse of Thales' theorem,
we conclude that DE∥BC.

and E are points on the sides AB and AC respectively of a ΔABC. In each of the following cases, determine whether DEBC or not. AD=7.2 cm, AE=6.4 cm, AB=12 cm and AC=10 cm.AD=7.2 cm, AE=6.4 cm, AB=12 cm and AC=10 cm.

જવાબ : We have:
AD = 7.2 cm, AB = 12 cm
Therefore,
DB = 12 −- 7.2 =  4.8 cm
Similarly,
AE = 6.4 cm, AC = 10 cm
Therefore,
EC = 10 −- 6.4 = 3.6 cm
Now,
AE/EC = 6.4/3.6= 16/9
Applying the converse of Thales' theorem,
we conclude that DE is not parallel to BC.

In a ΔABC,  AD is the bisector or A. If AB = 6.4 cm, AC = 8 cm and BD = 5.6 cm, find DC.

જવાબ : It is given that AD bisects ∠A.
Applying angle−bisector theorem in △ABC, we get:
BD/DC=AB/AC
⇒5.6/DC=6.4/8
⇒DC = 8×5.6/6.4 = 7 cm

In a ΔABC,  AD is the bisector or A. If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.

જવાબ : It is given that AD bisects ∠A.
Applying angle−bisector theorem in △ABC, we get:
BD/DC =AB/AC
Let BD be x cm.
Therefore, DC = (6−x) cm
⇒x6−x = 10/14
⇒14x = 60−10x
⇒24x = 60
⇒x = 2.5 cm
Thus, BD = 2.5 cm
DC = 6−2.5 = 3.5 cm

In a ΔABC,  AD is the bisector or A. If AB = 5.6 cm, BD = 3.8 cm and BC = 6 cm, find AC.

જવાબ : It is given that AD bisects ∠A.
Applying angle−bisector theorem in △ABC, we get:
BD/DC=AB/AC
BD = 3.2 cm, BC = 6 cm
Therefore, DC = 6−3.2 = 2.8 cm
⇒3.2/2.8=5.6/AC ⇒AC = 5.6×2.8/3.2=4.9 cm

In a ΔABC,  AD is the bisector or A. If AB = 5.6 cm, AC = 4 cm and DC = 3 cm, find BC.

જવાબ : It is given that AD bisects ∠A.
Applying angle−bisector theorem in △ABC, we get:
BD/DC = AB/AC
⇒BD/3 = 5.6/4
⇒BD = 5.6×3/4
⇒BD = 4.2 cm
Hence,  BC = 3 + 4.2 = 7.2 cm

M is a point on the side BC of a parallelogram ABCD. DM when produced meets AB produced at N. Prove that DM/MN=DC/BN જવાબ : Given: ABCD is a parallelogram
To prove: DM/MN=DC/BN
Proof: In △DMC and △NMB
∠DMC =∠NMB      (Vertically opposite angle)
∠DCM =∠NBM       (Alternate angles)
By AAA- similarity
△DMC ~ △NMB
∴DM/MN=DC/BN

M is a point on the side BC of a parallelogram ABCD. DM when produced meets AB produced at N. Prove that DN/DM=AN/DC જવાબ : Given: ABCD is a parallelogram
To prove: DN/DM=AN/DC
Proof: In △DMC and △NMB
∠DMC =∠NMB      (Vertically opposite angle)
∠DCM =∠NBM       (Alternate angles)
By AAA- similarity
△DMC ~ △NMB
∴DM/MN=DC/BN
Now,  MN/DM=BN/DCNow,  MNDM=BNDC
Adding 1 to both sides, we get
MNDM+1=BNDC+1
⇒(MN+DM)/DM=(BN+DC)/DC
[∵ABCD is a parallelogram]
⇒DN/DM=AN/DC

Show that the line segment which joins the midpoints of the oblique sides of a trapezium is parallel to the parallel sides.

જવાબ : Let the trapezium ​be ABCD with E and F as the mid points of AD and BC, respectively.
Produce AD and BC to meet at P. In △PAB, DC ∥ AB.
Applying Thales' theorem, we get:
PD/DA = PC/CB
Now, E and F are the midpoints of AD and BC, respectively.
⇒ PD2/DE = PC2/CF
⇒ PD/DE = PC/CF
Applying the converse of Thales' theorem in ∆PEF,
we get that DC ∥ EF.Hence, EF ∥ AB.
Thus. EF is parallel to both AB and DC.
This completes the proof.

In the adjoining figure, ABCD is a trapezium in which CD  AB and its diagonals intersect at O. If AO = (2x + 1) cm, OC = (5x – 7) cm, DO = (7x − 5) cm and OB = (7x + 1) cm, find the value of x.

જવાબ : In trapezium ABCD, AB∥CDAB∥CD and the diagonals AC and BD intersect at O.
Therefore,  AO/OC=BO/OD
⇒2x+1/5x−7=7x+1/7x−5
⇒(5x − 7)(7x + 1) = (7x − 5)(2x + 1)
⇒35x2 + 5x − 49x − 7 = 14x2 − 10x + 7x – 5
⇒21x2 − 41x − 2 = 0
⇒21x2 − 42x +x− 2 = 0
⇒21x(x − 2) + 1(x − 2) = 0
⇒(x − 2)(21x+1) = 0
⇒x = 2,−121∵ x ≠ −121∴ x = 2

In ABC, M and N are points on AB and AC respectively such that BM = CN. If B = C then show that MNBC.

જવાબ : In ABC, B = C
AB = AC (Sides opposite to equal angle are equal)
Subtracting BM from both sides, we get
AB − BM = AC − BM
AB − BM = AC − CN           (BM = CN)
AM = AN
∴∠AMN = ANM (Angles opposite to equal sides are equal)
Now, In ABC,
A + B + C = 180o                     .....(A)
(Angle Sum Property of triangle)
Again In AMN,
A +AMN + ANM = 180o          ......(B)
(Angle Sum Property of triangle)
From (A) and (B), we get
B + C = AMN + ANM
2B  = 2AMN
⇒∠B  = AMN
Since, B  and AMN are corresponding angles.
MNBC

ABC and ∆DBC lie on the same side of BC, as shown in the figure. From a point P on BCPQ  AB and PR  BD are drawn, meeting AC at Q and CD at R, respectively. Prove that QR  AD. જવાબ : In △CAB , PQ ∥ AB.
Applying Thales' theorem, we get:
CPPB = CQQA       ...(A)
Similarly, applying Thales' theorem in △BDC,
where PR ∥ BD we get:
CPPB = CRRD               ...(B)
Hence, from (A) and (B), we have:
CQ/QA = CR/RD
Applying the converse of Thales' theorem, we conclude that QR∥AD in △ADC.
This completes the proof.

In the given figure, side BC of ∆ABC is bisected at D and O is any point on ADBO and CO produced meet AC and AB at E and F, respectively, and AD is produced to X, so that D is the midpoint of OX. Prove that AO : AX = AF : AB and show that EF  BC. જવાબ : Join BX and CX.
It is given that BC is bisected at D.
BD = DC ​
It is also given that OD = OX
The diagonals OX and BC of quadrilateral BOCX  bisect each other.
Therefore, BOCX is a parallelogram.
BO  CX and BX  CO
BX  CF and CX  BE
BX  OF and CX  OE
Applying Thales' theorem in Δ∆ABX, we get:
AOAX = AFAB            ...(A)

Also, in ACX, CX  OE.

Therefore by Thales' theorem, we get: AO/AX = AE/AC           ...(B)
From (A) and (B), we have:
AF/AB = AE/AC
Applying the converse of Thales' theorem in ABC,

EFCB.
This completes the proof.

ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that CQ=14AC.CQ=14AC. If PQ produced meets BC at R, prove that R is the midpoint BC. જવાબ : Join DB.
We know that the diagonals of a parallelogram bisect each other.
Therefore,
CS = ½ AC      ...(A)
Also, it is given that CQ = ¼ AC        ...(B)
Dividing equation (A) by (A), we get:
CQ/CS = ¼AC/ ½ AC
or, CQ = ½CS
Hence, Q is the midpoint of CS.
Therefore, according to midpoint theorem in CSD
PQDS
if PQ DS ,  we can say that QRSB  In CSB, Q is midpoint of CS and QRSB
Applying converse of midpoint theorem ,

we conclude that R is the midpoint of CB.
This completes the proof.

In the adjoining figure, ABC is a triangle in which AB = AC. If D and are points on AB and AC respectively such that AD AE, show that the points B, C, E and D are concyclic. જવાબ : Given:
AD = AE     ...(A)
AB = AC     ...(B)
Subtracting AD from both sides, we get:
⇒ AB −AD = AC - AD
⇒ AB − AD = AC − AE  (Since, AD = AE)
⇒BD = EC    ...(C)
Dividing  equation  (A) by equation (C), we get: AD/DB=AE/EC  Applying the converse of Thales' theorem, DE∥∥BC ⇒∠DEC + ∠ECB  = 180°   (Sum of interior angles on the same side of a transversal line is 180°.)    ⇒∠DEC + ∠CBD =180°  (Since, AB = AC⇒∠B =∠C) Hence, quadrilateral BCED is cyclic. Therefore, B,C,E and D are concyclic points.

In ∆ABC, the bisector of B meets AC at D. A line PQ  AC meets ABBC and BD at P, Q and R respectively.
Show that PR × BQ = QR × BP. જવાબ : In triangle BQP, BR bisects angle B.
Applying angle bisector theorem, we get:
QR/PR = BQ/BP
⇒ BP × QR = BQ × PR
This completes the proof.

Find the pair of similar triangles. State the similarity criterion and write the similarity relation in symbolic form. જવાબ : We have:
∠BAC = ∠PQR = 50°
∠ABC =∠QPR = 60°
∠ACB =∠PRQ=70°
Therefore, by AAA similarity theorem, △ABC ~QPR

Find the pair of similar triangles. State the similarity criterion and write the similarity relation in symbolic form. જવાબ : We have:
AB/DF=3/6=1/2 and BC/DE=4.5/9=1/2
But, ∠ABC ≠∠EDF (Included angles are not equal)
Thus, this does not satisfy SAS similarity theorem.
Hence, the triangles are not similar.

Find the pair of similar triangles. State the similarity criterion and write the similarity relation in symbolic form. જવાબ : We have:
CA/QR=8/6=4/3 and CB/PQ=6/4.5=4/3
⇒CA/QR=CB/PQ
Also, ∠ACB =∠PQR=80°∠ACB =∠PQR=80°
Therefore, by SAS similarity theorem, △ACB ~△RQP.

Find the pair of similar triangles. State the similarity criterion and write the similarity relation in symbolic form. જવાબ : We have
DE/QR=2.5/5=1/2
EF/PQ=2/4=1/2
DF/PR=3/6=1/2
⇒DE/QR=EF/PQ=DF/PR
Therefore, by SSS similarity theorem, △FED~△PQR

Find the pair of similar triangles. State the similarity criterion and write the similarity relation in symbolic form. જવાબ : In △ABC
∠A+∠B+∠C=180°      (Angle Sum Property)
⇒80°+∠B+70°=180°
⇒∠B=30°
∠A=∠M and ∠B=∠N
Therefore, by AA similarity theorem,
△ABC~△MNR

In the given figure ΔODC~ΔOBA , BOC=115°  and CDO=70° . FindDOC જવાબ : It is given that DB is a straight line.
Therefore,
∠DOC + ∠COB = 180°
∠DOC = 180° − 115° = 65°

In the given figure ΔODC~ΔOBA , BOC=115°  and CDO=70° . Find  DCO જવાબ : In △DOC, we have:
∠ODC + ∠DCO + ∠DOC =180°
Therefore,70°+ ∠DCO + 65° = 180°
⇒∠DCO = 180 − 70 − 65 = 45°

In the given figure ΔODC~ΔOBA , BOC=115°  and CDO=70° . Find  OAB જવાબ : It is given that △ODC ~△OBA
Therefore,
∠OAB =∠OCD = 45°

In the given figure ΔODC~ΔOBA , BOC=115°  and CDO=70° . FindOBA. જવાબ : △ODC ~△OBA△ODC ~△OBA
Therefore,
∠OBA =∠ODC= 70°∠OBA =∠ODC= 70°

In the given figure  ΔOAB~ΔOCD.  If AB=8 cm,  BO=6.4 cm,  OC=3.5 and  CD=5 cm, ∆OAB~∆OCD.  If AB=8 cm, BO=6.4 cm, OC=3.5 and CD=5 cm, Find OA
(ii) DO. જવાબ :  Let OA be x cm.
∵ △OAB ~△OCD
∴ OA/OC=AB/CD
⇒x/3.5=8/5
⇒x = 8 × 3.5/5 = 5.6
Hence, OA = 5.6 cm

In the given figure  ΔOAB~ΔOCD.  If AB=8 cm,  BO=6.4 cm,  OC=3.5 and  CD=5 cm, ∆OAB~∆OCD.  If AB=8 cm, BO=6.4 cm, OC=3.5 and CD=5 cm, Find DO. જવાબ : Let OD be y cm
∵∵ △OAB ~△OCD
∴ AB/CD=OB/OD
⇒8/5 = 6.4y
⇒y = 6.4 × 5/8 = 4
Hence, DO = 4 cm

In the given figure, if ADE = B, show that ∆ADE ABC. If AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm and BC = 4.2 cm, find DE. જવાબ : Given:
∠ADE = ∠ABC  and ∠A = ∠A
Let DE be x cm
Therefore, by AA similarity theorem, △ADE~△ABC
⇒3.8/3.6 + 2.1 = x/4.2
⇒x = 3.8 × 4.2/5.7 = 2.8
Hence, DE = 2.8 cm

The perimeters of two similar triangles ABC and PQR are 32 cm and 24 cm respectively. If PQ = 12 cm. find AB.

જવાબ : It is given that triangles ABC and PQR are similar.
Therefore,
Perimeter (△ABC)/Perimeter (△PQR) =AB/PQ
⇒32/24=AB/12
⇒AB=32×12/24=16 cm

The corresponding sides of two similar triangles ABC and DEF are BC = 9.1 cm and EF = 6.5 cm. If the perimeter of ∆ DEF is 25 cm, find the perimeter of ∆ABC.
It is given that △ABC~△DEF.It is given that △ABC~△DEF.
Therefore, their corresponding sides will be proportional.

જવાબ : Also, the ratio of the perimeters of similar triangles is same as the ratio of their corresponding sides.
⇒Perimeter of △ABC/Perimeter of △DEF=BC/EF
Let the perimeter of ∆ABC be x cm.
Therefore,
x/25=9.1/6.5
⇒x=9.1×25/6.5=35
Thus, the perimeter of ∆ABC is 35 cm.

In the given figure, CAB = 90° and AD  BC. Show that ∆ BDA  ∆ BAC. If AC = 75 cm, AB = 1 m and BC = 1.25 m, find AD. જવાબ : In △BDA and △BAC, we have:
∠BDA =∠BAC = 90°
∠DBA = ∠CBA          (Common)
Therefore, by AA similarity theorem,
△BDA~△BAC
⇒AD=0.75/1.25=0.6 m or 60 cm

In the given figure, ABC = 90° and BD  AC. If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC. જવાબ : It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse. In△BDC and △ABC, we have:
∠ABC = ∠BDC = 90° (given)
∠C = ∠C  (common)
By AA similarity theorem,
we get:△BDC~△ABC
AB/BD = BC/DC
⇒5.7/3.8 = BC/5.4
⇒BC = 5.7/3.8 × 5.4 = 8.1
Hence, BC = 8.1 cm

In the given figure, ABC = 90° and BD  AC. If BD = 8 cm, AD = 4 cm, find CD. જવાબ : It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse. In △DBA and △DCB, we have:
∠BDA = ∠CDB
∠DBA = ∠DCB = 90°
Therefore, by AA similarity theorem, we get:
△DBA~△DCB
CD  = 8×8/4  = 16cm

P and Q are points on the sides AB and AC, respectively, of ∆ ABC. If AP = 2 cm, PB = 4 cm, AQ = 3 cm and QC = 6 cm, show that BC = 3PQ.

જવાબ : We have:
AP/AB = 2/6 = 1/3 and AQ/AC = 3/9 = 1/3
⇒AP/AB = AQ/AC
In △APQ and △ABC, we have:
AP/AB = AQ/AC
∠A = ∠A
Therefore, by AA similarity theorem, we get:
△APQ~△ABC
Hence, PQ/BC = AQ/AC = 1/3
⇒PQ/BC = 1/3
⇒BC = 3PQ
This completes the proof.

In the given figure, DB  BCDE  AB and AC  BC.
Prove that BEDE=ACBC.BEDE=ACBC.  જવાબ : In △BED and △ACB, we have: ∠BED = ∠ACB = 90°
∵ ∠B + ∠C = 180°
∴ BD ∥ AC
∠EBD = ∠CAB (Alternate angles)
Therefore, by AA similarity theorem, we get:
△BED~△ACB
⇒ BE/AC = DE/BC
⇒ BE/DE = AC/BC
This completes the proof.

A vertical stick of length 7.5 m casts a shadow 5 m long on the ground and at the same time a tower casts a shadow 24 m long. Find the height of the tower.

જવાબ : Let AB be the vertical stick and BC be its shadow.
Given:
AB = 7.5 m, BC = 5 m Let PQ be the tower and QR be its shadow.
Given:
QR = 24 m
Let the length of PQ be x m.  In △ABC and △PQR, we have:
∠ABC = ∠PQR = 90°
∠ACB = ∠PRQ   (Angular elevation of the Sun at the same time)
Therefore, by AA similarity theorem, we get:
△ABC~△PQR
⇒AB/BC=PQ/QR
⇒7.5/5=x/24=36 m
Therefore, PQ = 36 m
Hence, the height of the tower is 36 m.

In an isosceles ∆ ABC, the base AB is produced both ways to P and Q, such that AP × BQ = AC2.
Prove that ∆ ACP
∆ BCQ. જવાબ : It should be △APC~△BCQ instead of ∆ACP ∼ ∆BCQ
It is given that △ABC is an isosceles triangle.
Therefore,
CA = CB
⇒ ∠CAB = ∠CBA
⇒ 180° − ∠CAB = 180° − ∠CBA
⇒ ∠CAP = ∠CBQ
Also,
AP × BQ = AC2
⇒ AP/AC = AC/BQ (∵ AC = BC) Thus, by SAS similarity theorem, we get:
△APC~△BCQ
This completes the proof.

In the given figure, 1=2 and ACBD=CBCE.1=2 and ACBD=CBCE.
Prove that ∆ ACB  ∆ DCE. જવાબ : We have:
AC/BD = CB/CE
⇒AC/CB = BD/CE
⇒AC/CB = CD/CE  (Since, BD = DC as ∠1 = ∠2)
Also, ∠1 = ∠2i.e, ∠DBC = ∠ACB
Therefore, by SAS similarity theorem, we get:△ACB~△DCE

ABCD is a quadrilateral in which AD = BC. If P, Q, R , S be the mid points of AB, AC, CD and BD respectively, show that PQRS is a rhombus. જવાબ : In △ABC, P and Q are mid points of AB and AC respectively.
So, PQ|| BC, and PQ=1/2 BC                                             .....(A) Similarly, In △ADC, QR=1/2AD=1/2BC                          .....(B)
Now, In △BCD, SR=1/2BC                                                .....(C) Similarly, In △ABD, PS=1/2AD =1/2 BC                           .....(D)
Using (A), (B), (C) and (D)
PQ = QR = SR = PS
Since, all sides are equal
Hence, PQRS is a rhombus.

In a circle, two chords AB and CD intersect at a point inside the circle. Prove that
(a)PAC~PDB જવાબ : AB and CD are two chords
To Prove:
△PAC~△PDB  Proof: In △PAC and △PDB△PAC and △PDB
∠APC=∠DPB  (Vertically Opposite angles)
∠CAP=∠BDP  (Angles in the same segment are equal)
By AA similarity-criterion
△PAC~△PDB
When two triangles are similar, then the ratios of the lengths of their corresponding sides are porportional.
∴PA/PD=PC/PB
⇒PA.PB=PC.PD

ABCD is a parallelogram and E is a point on BC. If the diagonal BD intersects AE at F, prove that AF × FB = EF × FD. જવાબ : We have:
∠AFD = ∠EFB    (Vertically Opposite angles)
∵ DA ∥ BC
∴ ∠DAF =∠BEF      (Alternate angles)
△DAF~△BEF          (AA similarity theorem)
⇒ AF/EF = FD/FB
or, AF × FB = FD × EF
This completes the proof.

A 13 m long ladder reaches a window of a building 12 m above the ground. Determine the distance of the foot of the ladder from the building.

જવાબ :

Let AB and AC be the ladder and height of the building.
It is given that:
AB = 13 m and AC = 12 m
We need to find the distance of the foot of the ladder from the building, i.e, BC.
In right-angled triangle ABC, we have: AB2 = AC2 + BC2

BC = √ (132 − 122) = √ (169 – 144) = √25 = 5m
Hence, the distance of the foot of the ladder from the building is 5 m

### There are No Content Availble For this Chapter 1    ∠BAC A    ∠ADE 2    ∠ABC B    ∠AED 3    ∠ACB C    ∠CAD 4    ∠CED D    180ᵒ - ∠BCE

જવાબ :

1-C, 2-A, 3-B, 4-D

ABCD is a parallelogram. 1    ∠MCD A    ∠BMN 2    ∠CDM B    ∠MBN 3    ∠CMD C    ∠ADC 4    ∠ABC D    ∠MNB

જવાબ :

1-B, 2-D, 3-A, 4-C

AB ӀӀ EF ӀӀ CD , ∠B = ∠A = 60ᵒ 1    ∠PDC A    60ᵒ 2    ∠DCF B    ∠DEF 3    AP C    ∠BFE 4    ∠P D    BP

જવાબ :

1-B, 2-C, 3-D, 4-A 1    ∠B A    ∠ANM 2    BM B    180ᵒ - ∠B 3    ∠M C    CN 4    ∠C D    ∠C

જવાબ :

1-D, 2-C, 3-B, 4-A

BC ӀӀ EF 1    ∠EFC A    ∠BOC 2    ∠FEB B    ∠EOC 3    ∠FOE C    ∠EBC 4    ∠BOF D    ∠BCF

જવાબ :

1-D, 2-C, 3-A, 4-B

ABCD is parallelogram. 1    ∠B A    ∠ACB 2    ∠A B    ∠ACD 3    ∠CAD C    ∠C 4    ∠CAB D    ∠D

જવાબ :

1-D, 2-C, 3-A, 4-B

AC ӀӀ PQ 1    ∠ABD A    ∠DRQ 2    ∠BAD B    ∠BQR 3    ∠BCD C    ∠BPR 4    ∠PRB D    ∠CBD

જવાબ :

1-D, 2-C, 3-B, 4-A

AB ӀӀ CD 1    ∠AOD A    45ᵒ 2    ∠ABD B    115ᵒ 3    ∠COD C    70ᵒ 4    ∠OCD D    65ᵒ

જવાબ :

1-B, 2-C, 3-D, 4-A 1    ∠CDA A    ∠B 2    ∠A + ∠B + ∠C B    90ᵒ 3    ∠ADC C    ∠C 4    ∠BAD D    180ᵒ

જવાબ :

1-B, 2-D, 3-A, 4-C 1    ∠ABD A    ∠CBD 2    ∠ACB B    ∠ABC 3    ∠DEB C    ∠BAC 4    ∠D D    90ᵒ

જવાબ :

1-C, 2-A, 3-B, 4-D

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Math
Chapter 06 : Triangles

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