Chapter 11 Constructions Exercise 11.1
Question 1: Construct an angle of 90° at the initial point of a given ray and justify the construction.
Answer:
Construction steps:
1. Draw a ray OA.
2. With O as centre and any radius, draw a semicircle that cuts ray OA at B.
3. With B as centre and same radius, draw an arc to cut the semicircle at C.
4. With C as centre and same radius, draw an arc to cut the semicircle at D.
5. With C and D as centres and same radius, draw two arcs which intersect each other at X.
6. Finally, join XO which makes an angle of 90°.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjZAIJ6q2FUz0JINlDdnjYaz7HhsPowCSJz_ccVfp4nLlgzxHUv6WRtxaqJ-B__zC9E2D1TSiPCggMvhsKAV4s9pUwdTPxdFSN5ZZ3RWorD50u0f8Mf0wVWOgtto64iS8fh50aaL_Lc_YiB/s16000/123.1.png)
To justify that ∠XOA = 90°, we have to join OC and OD.
We have, ∠COA = ∠DOC = 60°. From the fourth and fifth step of this construction, we know that OX is the bisector of ∠DOC.
∴ ∠XOC = 1/2 of ∠DOC
= ∠XOC = 1/2 x 60° = 30°
Also, ∠XOA = ∠COA + ∠XOC
= 60° + 30° = 90°
Hence justified.
Question 2: Construct an angle of 45° at the initial point of a given ray and justify the construction.
Answer:
Construction steps:
1. Draw a ray OA.
2. With O as centre and any radius, draw a semicircle that cuts OA at B.
3. With B as centre and same radius, draw an arc to cut the semicircle at C.
4. With C as centre and same radius, draw an arc to cut the semicircle at D.
5. With C and D as centres and same radius, draw two arcs which intersect each other at X.
6. Join XO to intersect the semicircle at Y, which makes an angle 90°.
7. Now, with Y and B as centres and radius more than half of YB draw intersecting arcs to meet at E.
8. Join OE. Therefore ∠EOA = 45°.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhOebDHz8YGMWe7IncGHfH_Jp5FdKlHAHf4Gy18OLL86qc2wReQV8ynMkT8JWm5-JF5GFIaMkLNi_Gpdc-4xsKHKDVTFgnetzQlm7cyzWa2B2eLOZmZYBZzu-3nH24hujztyZG9BhFMf2zU/s16000/123.2.png)
From the construction,
∠XOA = 90°
As EO is the perpendicular bisector of ∠XOA. So,
∠EOA = 1/2 (∠XOA)
= ∠EOA = 1/2 × 90° = 45°.
Hence justified.
Question 3: Construct the angles of the following measurements:
i) 30°
ii) 22½°
iii) 15°
Answer:
(i) 30°
Construction steps:
1. Draw a ray OA.
2. Taking O as a centre and any radius, draw an arc that cuts ray OA at B.
3. Now with B as centre and same radius draw an arc on the previous arc to cut it at C. This will be 60°.
4. With B and C as centres and radius more than half of BC, draw 2 arcs which intersect each other at E.
5. Join EO. Thus, ∠EOA = 30°.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhiUW5CEkoe2ra74ViZA6-BlD8n_MFKqP3ybBV266O5xci7WyhbuHgjs7av5_F0VadbZ8C3JQQ3Rkg2peQ0Uuf_nzQUPyFKrFLb8O3JgiFKbo7fdCM4_wWUrd848efWhknzBouaYLqIqyfj/s16000/123.3.png)
Construction steps:
1. Draw a ray OA.
2. With O as centre and any radius, draw a arc that cuts OA at B.
3. With B as centre and same radius, draw an arc to cut the semicircle at C.
4. With C as centre and same radius, draw an arc to cut the semicircle at D.
5. With C and D as centres and same radius, draw two arcs which intersect each other at X.
6. Join XO to intersect the semicircle at Y, which makes an angle of 90°.
7. Now, with Y and B as centres and radius more than half of YB draw intersecting arcs to meet at E.
8. Join OE. Therefore ∠EOA = 45°.
9. Finally, with E and B as centres and radius more than the half of BE draw 2 intersecting arcs to meet at F.
10. Join FO. Therefore ∠FOA = 22½°.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgR7AxRyrcnPOgI9rYOp46x784sdvAjErDTPXkfBVYliNgLPqk1wZ00qzzBXlwWrNtD2_y0oOpsH4AlXKswWyZJynzZ_Mtztk9cez6G4W3SjdIAZTHOM_fS1ztlbLo8uKagKNYACsSGpZc4/s16000/123.4.png)
Construction steps:
1. Draw a ray OA.
2. Taking O as a centre and any radius, draw a semicircle that cuts ray OA at B.
3. Now with B as centre and same radius draw an arc on the semicircle to cut it at C. This will be 60°.
4. With B and C as centres and radius more than half of BC, draw 2 arcs which intersect each other at E.
5. Join EO to intersect the arc at F. Thus, ∠EOA = 30°.
6. With F and B as centres and radius more than the half of FB, draw 2 intersecting arcs to intersect at D. Join DO. Thus, ∠DOA = 15°.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgWAvNj04XaMuI6IS5xgaPy30KZegq5Q4SZJ853XoViWywJW6xZGsiZi173Tm-S-K_M6igKiFhzQ7aLvUOpfinwJR9rAGMh_90XqG-2l0FXyeEuWmNRAykmF3ftyFVACoYCa6fcPWhCkvIF/s16000/123.5.png)
Question 4: Construct the following angles and verify by measuring them by a protractor:
i) 75°
ii) 105°
iii) 135°
Answer:
i) 75°
Construction steps:
1. Draw a ray OA.
2. With O as centre and any radius draw an arc to intersect the ray OA at B.
3. With B as centre and same radius draw an arc to cut the semicircle at C.
4. With C as centre and same radius draw an arc to cut the semicircle at D.
5. With D and C as centre and same radius draw 2 arcs to intersect at X. Join OX. Let this line intersect the arc at Q.
6. With Q and C as centre draw 2 arcs to intersect at Y.
7. Join OY. Thus, ∠YOA = 75°.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhH7PEqcOdDPOhOzy7jJyusEmLJ7ThILIp93RpBz923-GTTAo2tfl3hHwqoNlQq-Yo5NtppnDWt0AaQ6-kt7_HuQSgCRz5vCY4kmNuEiHmlfs1gfta_FL-eh3SqyeH_mw50Coa5Ehh9nRPG/s16000/123.8.png)
Construction steps:
1. Draw a ray OA.
2. With O as centre and any radius draw a semicircle to cut ray OA at B.
3. With B as centre and same radius draw an arc to cut the semicircle at C.
4. With C as centre and same radius draw an arc on the arc to cut at D.
5. With D and C as centre and same radius draw 2 arcs to intersect at X. Join OX. Let this line intersect the semicircle at Q.
6. With Q and D as centre draw 2 arcs to intersect at Y.
7. Join OY. Thus, ∠YOA = 105°.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgxp3cG7wXvXod43DV8uen0rVmqdYNZ4WjCtdyYJp4KaVBANgVXmX-Rb0EZ6XqHk43ofZIw5522BlGDnADeor9s_caKi65ytcO-XB9kr8C9j6O13toEjOGg-pzniqQtgj3YA1Xo2LRPACiW/s16000/123.7.png)
Construction steps:
1. Draw a line A’OA.
2. With O as centre and any radius draw a semicircle to intersect the line A’OA at B and B’.
3. With B as centre and same radius draw an arc to cut the semicircle at C.
4. With C as centre and same radius draw an arc to cut the semicircle at D.
4. With D and C as centre and same radius draw 2 arcs to intersect at X. Join OX. Let this line intersect the semicircle at Q.
5. With Q and B’ as centre draw 2 arcs to intersect at Y.
6. Join OY. Thus, ∠YOA = 135°.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjVF-9G8GrnJ2ISydX3RtINuHJqa4QYICDwrwwdKjj8CId5tPfg7vL1w8KJuaaHb4vTMUdzB4Olmvqrep1fDHryvyXMIY6Zp48PgPe48PC-PCnrPVDVOW-h76OQt87hkJAwgHNwzqara31D/s16000/123.8.png)
Question 5: Construct an equilateral triangle, given its side and justify the construction.
Answer:
Construction steps:
1. Draw a line segment, AB of any units (say 5cm).
2. With A and B as centres, draw two semicircle on the line segment AB to cut it at D and E respectively.
3. With D and E as centres, draw 2 arcs that cuts the semicircle at G and F respectively to form an angle of 60° each.
4. Now, join AG and BF and extend such that they meet each other at the point C.
5. Therefore, ABC is the required triangle.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiW00aCTbBoUIhqw2LrEXBSAqqY7fstWJAe8orU5QZCeNQkhOuV5tCnJ44xN6APD434JOAST6T19kZrAZYAQe2M8LzgLA-Faz56cdy4_eTHewUw_J1bAs_F9Tc5DLsPI7uge5hQADp4u_e6/s16000/123.9.png)
After construction we observe
AB = 5 cm, ∠A = 60° and ∠B = 60°
We know that the sum of the interior angles of a triangle is 180°
∴ ∠A + ∠B + ∠C = 180°
= 60° + 60° + ∠C = 180° (as ∠A and ∠B are 60° each)
= 120° + ∠C = 180°
= ∠C = 60°
We have one side equal. To prove that other 2 sides are also equal,
∴ BC = CA = 5 cm (Sides opposite to equal angles are equal)
AB = BC = CA = 5 cm
∠A = ∠B = ∠C = 60°
Hence, justified.
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