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Relation Between The Two Sides Of A Triangle And Their Opposite Angles

Class 8th Mathematics West Bengal Board Solution
Lets Work Out 9
  1. Let’s see the isosceles triangles given below and without measuring let’s write which…
  2. Let’s see the isosceles triangles given below and without measuring let’s write which…
  3. Let’s see the isosceles triangles given below and without measuring let’s write which…
  4. let’s see the isosceles triangles below and without measuring let’s write which angles…
  5. let’s see the isosceles triangles below and without measuring let’s write which angles…
  6. Line segments AB and CD bisect each other at O. Let’s prove that AC and BD are…
  7. E and F are twp points on two parallel straight-line AB and CD respectively. O is the…
  8. If we produce the base of an isosceles triangle in both sides then two exterior angles…
  9. Let’s prove that the length of the medians are equal in length in an equilateral…
  10. In trapezium ABCD, AD||BC and ∠ABC=∠BCD, let’s prove that ABCD is an isosceles…
  11. AB is the hypotenuse of the isosceles right triangle AB. AD is the bisector of ∠BAC and…
  12. Two isosceles triangles ABC and DBC whose AB = AC, DB = DC and they are situated on the…
  13. Two line segments PQ and RS intersect each other at X in such a way that XP=XR and…
Lets Work Out 9
Question 1.

Let’s see the isosceles triangles given below and without measuring let’s write which two sides are equal in length in each triangle.
Answer:

It is given that ∆ABC is isosceles and

∠A=70°


∠B=40°


And ∠C=70°


∴ ∠A=∠C


We know angles opposite to the equal sides are equal.


So, BA=BC
Question 2.

Let’s see the isosceles triangles given below and without measuring let’s write which two sides are equal in length in each triangle.
Answer:

It is given that ∆PQR is isosceles and

∠P=45°


∠R=45°


∴ ∠P=∠R


We know angles opposite to the equal sides are equal.


So, QP=QR
Question 3.

Let’s see the isosceles triangles given below and without measuring let’s write which two sides are equal in length in each triangle.
Answer:

It is given that ∆XYZ is isosceles and

∠X=35°


∠Z=35°


And ∠Y=110°


∴ ∠X=∠Z


We know angles opposite to the equal sides are equal.


So, XY=YZ
Question 4.

let’s see the isosceles triangles below and without measuring let’s write which angles are equal in measurement in each triangle.
Answer:

It is given that ∆ABC is isosceles and

AB=5cm


BC=5cm


∴BA=BC


We know angles opposite to the equal sides are equal.


So, ∠A=∠C
Question 5.

let’s see the isosceles triangles below and without measuring let’s write which angles are equal in measurement in each triangle.
Answer:

It is given that ∆PQR is isosceles and

PQ=8cm


PR=8cm


And QR=7cm


∴PQ=PR


We know angles opposite to the equal sides are equal.


So, ∠Q=∠R
Question 6.

Line segments AB and CD bisect each other at O. Let’s prove that AC and BD are parallel. Let’s write what kind of quadrilateral is ABCD.
Answer:


Given here two line segments AB and CD bisects each other at O.


∴ OA=OB and OC=OD


Now, in ∆AOC and ∆BOD,


∠AOC=∠BOD (vertically opposite angles)


OC=OD (given)


OA=OB (given)


∴ ∆AOC ≅ ∆BOD (by SAS rule)


∠ACO=∠ODB (by cpct )


∠CAO=∠OBD (by cpct )


But these are also the alternate interior angles.


So, AC||BD


Similarly, ∆AOD ≅ ∆BOC


And AD||BC


Hence ACBD is a parallelogram.
Question 7.

E and F are twp points on two parallel straight-line AB and CD respectively. O is the midpoint of line segment EF. We draw a straight line passing through O which intersect AB and CD at P and Q respectively. Let’s prove that O bisects the line segment PQ.
Answer:


E and F are two points on straight lines AB and CD respectively.


Also, O is the midpoint of EF and a straight line is drawn passing through O which intersect AB and CD at P and Q respectively.


OE=OF (O is the midpoint)


AB||CD


∠PEO=∠QFO (alternate interior angles) …………… (1)


∠POE=∠QOF (vertically opposite angles) …………. (2)


Now, in ΔPOE and ΔQOF,


OE=OF (O is the midpoint)


∠PEO=∠QFO (from (1))


∠POE=∠QOF (from (2))


∴ ΔPOE ≅ ΔQOF (by ASA rule)


So, OP=OQ (by cpct)


Hence, O bisects PQ
Question 8.

If we produce the base of an isosceles triangle in both sides then two exterior angles formed. Let’s prove that they are equal in measurement.
Answer:


Let an isosceles ΔEFG in which the side FG is extended on both sides to form two exterior angles.


EF=FG


So, ∠EFG=∠EGF…………(1)


As we know that sum of two interior angles is equal to the exterior angle of a triangle.


∠EFI=∠EFG+∠E…….(2)


∠EGJ=∠EGF+∠E…….(3)


Now, Subtracting equation (2) from (3)


∠EGJ-∠EFI=∠EGF+∠E-∠EFG-∠E


⇒ ∠EGJ-∠EFI=∠EGF -∠EFG=0 (from (1))


⇒ ∠EGJ-∠EFI=0


⇒ ∠EGJ=∠EFI


Proved.
Question 9.

Let’s prove that the length of the medians are equal in length in an equilateral triangle.
Answer:


Let an equilateral ΔABC with three medians AE, BF and CG


So, E, F and G are the midpoints of BC , AC and AB


∴ BE=EC, AF=FC and AG=GB


AB=BC=AC


⇒ � AB= � BC= � AC


⇒ AG=BE=AF


Now, in ΔAEC and ΔBFA,


EC=AF


AC=BC (given)


∠ACB=∠BAC (each 60° )


∴ ΔAEC ≅ ΔBFA (by SAS rule)


So, AE=BF (by cpct)………(1)


Similarly, ΔCGB≅ ΔBFA (by SAS rule)


So, CG=BF (by cpct)……….(2)


And, ΔCGB≅ ΔAEC


So, CG=AE…………..(3)


From (1),(2) and (3),


AE=BF=CG


Hence, proved.
Question 10.

In trapezium ABCD, AD||BC and ∠ABC=∠BCD, let’s prove that ABCD is an isosceles trapezium.
Answer:


Given a trapezium ABCD in which AD||BC and ∠ABC=∠BCD


Draw AEꞱBC and DFꞱBC


∠ABC=∠BCD (given)…………..(1)


∠AEB=∠DFC=90°………………(2)


Now, in ΔAEB,


∠BAE+∠AEB+∠ABE=180° (sum of all the angles of a Δ)……….(3)


In ΔDFC,


∠FDC+∠DFC+∠DCF=180° (sum of all the angles of a Δ)……….(4)


Subtracting equation (4) from (3)


∠BAE+∠AEB+∠ABE-∠FDC-∠DFC-∠DCF=180°-180°


⇒ ∠BAE-∠FDC+∠AEB-∠DFC+∠ABE-∠DCF=0


⇒ ∠BAE-∠FDC=0 (from (1) and (2))


⇒ ∠BAE=∠FDC………………..(5)


Now, in ΔBAE and ΔCDF,


AE=DF (by construction)


∠BAE=∠FDC (from (5))


∠AEB=∠DFC (each 90° )


∴ ΔBAE ≅ ΔCDF (by ASA rule)


So, AB=DC (by cpct)


Hence, the trapezium is isosceles as AB=CD.
Question 11.

AB is the hypotenuse of the isosceles right triangle AB. AD is the bisector of ∠BAC and AD intersects BC at D. Let’s prove AC + CD = AB.
Answer:


Let BC = AC = a and CD = b
In a right-angled triangle BCA,


By Pythagoras theorem,


AB2 = BC2 + AC2


AB2 = a2 + a2
AB = a√2
Given AD = b, we get
DB = BC – CD or DB = a – b
We have to prove that AC + CD = AB


or (a + b) = a√2.
By the angle bisector theorem, we get














C + CD = AB [we know that AC = a, CD = b and AB = a√2]
Hence proved.
Question 12.

Two isosceles triangles ABC and DBC whose AB = AC, DB = DC and they are situated on the opposite side of BC. Let’s prove that AD bisects BC perpendicularly.
Answer:


Given that ΔABC and ΔDBC are isosceles triangles.


Here, AB=AC and DB=DC


Also, ∠ABC=∠ACB


And ∠DBC=∠DCB


Now, in ΔABD and ΔACD,


AD=AD (common)


AB=AC (given)


BD=DC (given)


∴ ΔABD ≅ ΔACD (by SSS rule)


∠BAD=∠CAD (by cpct)


Again, in ΔABO and ΔACO,


AO=AO (common)


∠BAO=∠CAO (as ∠BAD=∠CAD)


AB=AC (given)


∴ ΔABD ≅ ΔACD (by SAS rule)


So, BO=OC (by cpct)


And ∠AOB=∠AOC(by cpct)


Also, ∠AOB+∠AOC=180° (straight angle)


⇒ ∠AOB+∠AOB=180°


⇒ 2∠AOB=180°


⇒ ∠AOB=90°


So, ∠AOC=∠AOB=90°


Hence, AD bisects BC perpendicularly.
Question 13.

Two line segments PQ and RS intersect each other at X in such a way that XP=XR and ∠PSX =∠ROX. Let’s prove thatPXS ROX.
Answer:


Given that two Two line segments PQ and RS intersect each other at X in such a way that XP=XR and ∠PSX =∠RQX


Now, ∠PSX=∠RQX and ∠RXQ=∠PXS (vertically opposite angles)……………(1)


In ΔRXQ,


∠RQX+∠RXQ+∠XRQ=180° ……………(2)


And in ΔPXS,


∠PXS+∠PSX+∠XPS=180° ………………(3)


Subtracting equation (2) from (3),


∠RQX+∠RXQ+∠XRQ-∠PXS-∠PSX-∠XPS=180° -180°


⇒ ∠RQX-∠PXS +∠RXQ-∠PSX +∠XRQ -∠XPS=0


⇒ ∠XRQ -∠XPS=0 (from (1))


⇒ ∠XRQ =∠XPS ……………..(4)


Now, in ΔPSX and ΔRQX,


XR=XP (given)


∠XRQ =∠XPS (from (4))


∠RXQ=∠PXS (vertically opposite angles)


∴ ΔPSX ≅ ΔRQX (by ASA rule)