Construction

EXERCISE 11.2

Given the base BC, a base angle, say ∠B and the sum AB + AC of the other two sides of a triangle ABC, you are required to construct it. X

Steps of Construction : D

1. Draw the base BC and at the point B make an Y

angle, say XBC equal to the given angle. A

2. Cut a line segment BD equal to AB + AC from

the ray BX.

B C

3. Join DC and make an angle DCY equal to ∠BDC.�4. Let CY intersect BX at A (see Fig. ).�Then, ABC is the required triangle.��Let us see how you get the required triangle.�Base BC and ∠B are drawn as given. � Next in triangle ACD,�∠ACD = ∠ ADC (By construction)�Therefore, AC = AD and then�AB = BD – AD = BD – AC�AB + AC = BD

Alternative method :�Follow the first two steps as above. Then draw�perpendicular bisector PQ of CD to intersect BD at�a point A (see Fig 11.5). Join AC. Then ABC is the Y�required triangle. Note that A lies on the perpendicular D�bisector of CD, therefore AD = AC. �� A P � Q�� B C

Construction 11.5 : To construct a triangle given its base, a base angle and the�difference of the other two sides.��Given the base BC, a base angle, say ∠B and the difference of other two sides�AB – AC or AC – AB, you have to construct the triangle ABC. Clearly there are�following two cases:��Case (i) : Let AB > AC that is AB – AC is given. X� �Steps of Construction : A �1. Draw the base BC and at point B make an angle�say XBC equal to the given angle.� P�2. Cut the line segment BD equal to AB – AC from D�ray BX. � R�3. Join DC and draw the perpendicular � bisector, say PQ of DC. B C� �4. Let it intersect BX at a point A. Join AC� Q�Then ABC is the required triangle. �� � Q

JUSTIFICTION : � We need to prove that AB – AC = BD .� Let perpendicular bisector intersect CD at point R �thus,�AR is the perpendicular bisector of CD.�So, CR = DR ………… …… ( 1 ) X� and ∠ ARC = ∠ ARD = 90⁰ ……….( 2 ) � Now, A� in ∆ ADR and ∆ ACR� AR = AR ( Common )� ∠ ARD = ∠ ARC ( From eq. 2 ) P� DR = CR (From eq. 1 ) D�So, ∆ ADR ≅ ∆ ACR ( SAS rule ) R �So, AC = AD ( CPCT ) ………[3] �Now, B C� BD = AB – AD� BD = AB – AC [ From (3) ]�Thus, our construction is justified. Q

CASE 2 : AC > AB ��Given the base BC , a base angle ∠ B and the difference AC – AB , �We need to construct ∆ ABC .��STEPS OF CONSTRUCTION : ��1. draw base BC .�2. Now, draw ∠ B construction angle B from point B. X � �Let the ray be BX������ B C

�� 3. Open the compass to length AC – AB .��From point B as center , cut an arc on ray BX ( opposite side of BC ) .�Let the arc intersect BX at D .��4. Join CD . X������ � B C�� � D��

5. Now, we will draw perpendicular bisector of CD.� �6. Mark point A where perpendicular bisector intersect BD.��7. Join AC.�� X���Triangle ABC is required triangle .� � A��� B C� � D R

JUSTIFICTION : �We need to prove that AC – AB = BD . � Let perpendicular bisector CD at point R .��Thus ,�AR is the perpendicular bisector of CD� SO , CR = DR ……..[1]� & ∠ ARC = ∠ ARD = 90⁰ ………..[2] X�Now, in ∆ADR & ∆ACR � AR = AR [ Common ] Q � ∠ ARD = ∠ ARC [From eq.2]� DR = CR [From 1 ] A�So , ∆ADR ≅ ∆ACR [SAS rule ] �=> AC = AD [CPCT] …. (3)�Now, B C� BD = AD – AB � BD = AC – AB [ From (3)]�Thus , our construction � is justified. D R

Construction 11.6 : To construct a triangle, given its perimeter and its two base angles.�Given the base angles, say ∠ B and ∠ C and BC + CA + AB, you have to construct�the triangle ABC.� Steps of Construction :�1. Draw a line segment, say XY equal to BC + CA + AB.�2. Make angles LXY equal to ∠B and MYX equal to ∠C.�3. Bisect ∠ LXY and ∠ MYX. Let these bisectors intersect at a point A.��� L M�� A�� �� X Y� �

��� 4. Draw perpendicular bisectors PQ of AX and RS of AY. �� 5. Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC�[see Fig.]. �� Then ABC is the required triangle ��� � P L M R� A���� B C� X Y� Q S

For the justification of the construction, We�observe that, B lies on the perpendicular bisector PQ of AX.�Therefore, XB = AB and similarly, CY = AC.�This gives BC + CA + AB = BC + XB + CY = XY.�Again ∠BAX = ∠AXB (As in ∆ AXB, AB = XB) and�∠ABC = ∠BAX + ∠AXB = 2 ∠AXB = ∠LXY��Similarly, ∠ACB = ∠MYX as required.

1. Construct a triangle ABC in which BC = 7cm, ∠B = 75o X � and AB + AC = 13 cm D �Solution:�Steps of constructions:�(i) Draw base BC of length 7 cm P �(ii) Now, let’s draw B = 75o A Q �Let the ray be BX �(iii) Open the compass to length AB + AC = 13 cm.�From point B as center, cut an arc on ray BX.�Let the arc intersect BX at D B C�(iv) Join CD�(v) Now, we will draw perpendicular bisector of CD�(vi) Mark point A where perpendicular bisector intersects BD�(vii) Join AC

2. Construct a triangle ABC in which BC = 8 cm, ∠B = 45⁰ and AB − AC = 3.5 cm. X �Solution: A�Steps of construction:�(i) Draw base BC of length 8 cm�(ii) Now let’s draw ∠B = 45⁰ D�Let the ray be BX �(iii) Open the compass to length AB − AC = 3.5 cm B . C�Note: Since AB − AC = 3.5 cm is positive So, BD will be above line BC�From point B as center, cut an arc on ray BX.�Let the arc intersect BX at D�(iv) Join CD�(v) Now, we will draw perpendicular bisector of CD�(vi) Mark point A where perpendicular bisector intersects BD�(vii) Join AC� ∆ ABC IS THE REQUIRED TRIANGLE .

3. Construct a triangle PQR in which QR = 6 cm, ∠Q = 60⁰ and PR − PQ = 2 cm.�Solution: X�Steps of construction: (i) Draw base QR of length 6 cm P �(ii) Now, let’s draw ∠Q = 60⁰ P �Let the ray ne QX Q�(iii) Open the compass to D R � length PR − PQ = 2 cm. �Note:�Since PR − PQ = 2 cm, (PQ − PR) is negative�So, QD will be below line QR�From point Q as center, cut an arc on ray QX. (opposite side of QR).�Let the arc intersect QX at D�

�(vi) Join RD�(v) Now, we will draw perpendicular bisector of RD�(vii) Mark Point P where perpendicular bisector intersects RD�(viii) Join PR�∴ ∆PQR is the required triangle

4. Construct a triangle XYZ in which ∠Y = 30⁰, ∠Z = 90⁰ and XY + YZ + ZX =�11 cm.�Solution:�Given ∠Y = 30⁰, ∠90⁰ and XY + YZ + ZX = 11 cm.�Let’s construct ∆XYZ�Steps of construction:�(i) Draw a line segment AB equal to XY + YZ + ZX = 11 cm�(ii) Make angle equal to ∠Y = 30⁰ from the point A�Let the angle be ∠LAB.�(iii) Make angle equal to ∠Z = 90⁰ from the B�Let the angle be ∠MBA�(vi) Bisect ∠LAB and ∠MBA.�Let these bisector intersect at a point X.

� M� � L�� X� ��� A B� Y Z�� (v) Make perpendicular bisector of XA�Let it intersect AB at point Y�(vi) Make perpendicular bisector of XB�Let it intersect AB at the point Z�(vii) Join XY & YZ��∴ ∆XYZ is the required triangle

90

30⁰

5. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and�other side is 18 cm. D�Solution:�Let ∆ABC be the right angle triangle�Where BC = 12 cm�∠B = 90⁰ and�AC + AB = 18 cm A�Steps of construction:�(i) Draw base BC of length 12 cm�(ii) Now, let’s draw ∠B = 90⁰ B C�Let the ray be BX�(iii) Open the compass to length AB + AC = 18 cm.�From point B as center, cut an arc on ray BX

Let the arc intersects BX at D�(iv) Join CD D�(v) Now we will draw perpendicular bisector of CD���� A��� B C��(vi) Mark point A where perpendicular bisector intersects BD�(vii) Join AC��∴ ∆ABC is the required triangle