Constructions

ICSE Class 7 Mathematics · Selina Solutions Concise Mathematics Class 7 ICSE · 49 solved questions

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1(i)Exercise 22(A)

Using ruler and compasses, construct an angle of 30°.

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Steps of construction :

  1. Draw a line segment OA of any suitable length.

  2. With O as centre, draw an arc of any suitable radius to cut OA at B.

  3. With B as centre and the same radius, draw another arc to cut the previous arc at C.

  4. Join OC and produce it to any point D. Then, ∠DOA = 60°.

  5. With B as centre and radius more than half of BC, draw an arc.

  6. With C as centre and the same radius as taken in Step 5, draw another arc which meets the previous arc at E.

  7. Join OE.

Using ruler and compasses, construct an angle of 30°. Constructions, Mathematics Solutions ICSE Class 7.

∠EOA = $\frac{1}{2}$ ∠DOA = $\frac{1}{2}$ × 60° = 30°

Hence, ∠EOA is the required angle of 30°.

1(ii)Exercise 22(A)

Using ruler and compasses, construct an angle of 15°.

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Steps of construction :

  1. Draw a line segment OA of any suitable length.

  2. With O as centre, draw an arc of any suitable radius to cut OA at B.

  3. With B as centre and the same radius, draw another arc to cut the previous arc at C.

  4. Join OC and produce it to any point D. Then, ∠DOA = 60°.

  5. Bisect ∠DOA so that OE is the bisector. Then, ∠EOA = 30°.

  6. Let OE cut the first arc at F.

  7. With B and F as centres and radius more than half of BF, draw two arcs which cut each other at G.

  8. Join OG.

Using ruler and compasses, construct an angle of 15°. Constructions, Mathematics Solutions ICSE Class 7.

∠GOA = $\frac{1}{2}$ ∠EOA = $\frac{1}{2}$ × 30° = 15°

Hence, ∠GOA is the required angle of 15°.

1(iii)Exercise 22(A)

Using ruler and compasses, construct an angle of 75°.

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As, 75° = 60° + 15°

Steps of construction :

  1. Draw a line segment OA of any suitable length.

  2. With O as centre, draw an arc of any suitable radius to cut OA at B.

  3. With B as centre and the same radius, draw another arc to cut the previous arc at C.

  4. Again with C as centre and the same radius, draw one more arc to cut the first arc at D.

  5. Join OC and produce it to any point E. Then, ∠AOE = 60°.

  6. With C and D as centres, draw two arcs of equal radii (more than half of CD) to cut each other at F.

  7. Join OF and produce it to any point G. Then, ∠AOG = 90°.

  8. Let OG cut the first arc at P.

  9. With C and P as centres and radius more than half of CP, draw two arcs which cut each other at H.

  10. Join OH. Then, OH is the bisector of ∠EOG.

Using ruler and compasses, construct an angle of 75°. Constructions, Mathematics Solutions ICSE Class 7.

∠EOG = ∠AOG − ∠AOE = 90° − 60° = 30°

∠EOH = $\frac{1}{2}$ ∠EOG = $\frac{1}{2}$ × 30° = 15°

∠AOH = ∠AOE + ∠EOH = 60° + 15° = 75°

Hence, ∠AOH is the required angle of 75°.

1(iv)Exercise 22(A)

Using ruler and compasses, construct an angle of 180°.

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Steps of construction :

  1. Draw a straight line AB of any suitable length.

  2. Mark a point O on AB.

Using ruler and compasses, construct an angle of 180°. Constructions, Mathematics Solutions ICSE Class 7.

Since, AOB is a straight line,

⇒ ∠AOB = 180°

Hence, ∠AOB is the required angle of 180°.

1(v)Exercise 22(A)

Using ruler and compasses, construct an angle of 165°.

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As, 165° = 180° − 15°

Steps of construction :

  1. Draw a straight line AB and mark a point O on it. Then, ∠AOB = 180°.

  2. With O as centre, draw an arc of any suitable radius to cut OB at P.

  3. With P as centre and the same radius, draw another arc to cut the previous arc at Q.

  4. Join OQ and produce it to D. Then, ∠BOD = 60°.

  5. With P and Q as centres and radius more than half of PQ, draw two arcs which cut each other at R.

  6. Join OR and produce it to E. Then, ∠BOE = $\frac{1}{2}$ ∠BOD = 30°.

  7. Let OE cut the first arc at S.

  8. With P and S as centres and radius more than half of PS, draw two arcs which cut each other at T.

  9. Join OT and produce it to C. Then, ∠BOC = $\frac{1}{2}$ ∠BOE = 15°.

Using ruler and compasses, construct an angle of 165°. Constructions, Mathematics Solutions ICSE Class 7.

∠AOC = ∠AOB − ∠BOC = 180° − 15° = 165°

Hence, ∠AOC is the required angle of 165°.

1(vi)Exercise 22(A)

Using ruler and compasses, construct an angle of 22.5°.

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As, 22.5° = $\frac{1}{2}$ × 45°

Steps of construction :

  1. Draw a line segment OA of any suitable length.

  2. With O as centre, draw an arc of any suitable radius to cut OA at B.

  3. With B as centre and the same radius, draw another arc to cut the previous arc at C.

  4. Again with C as centre and the same radius, draw one more arc to cut the first arc at D.

  5. With C and D as centres, draw two arcs of equal radii (more than half of CD) to cut each other at E.

  6. Join OE. Then, ∠AOE = 90°.

  7. Let OE cut the first arc at F.

  8. With B and F as centres and radius more than half of BF, draw two arcs which cut each other at G.

  9. Join OG. Then, ∠AOG = ∠GOE = $\frac{1}{2}$ × 90° = 45°.

  10. Let OG cut the first arc at H.

  11. With B and H as centres and radius more than half of BH, draw two arcs which cut each other at K.

  12. Join OK.

Using ruler and compasses, construct an angle of 22.5°. Constructions, Mathematics Solutions ICSE Class 7.

∠AOK = $\frac{1}{2}$ ∠AOG = $\frac{1}{2}$ × 45° = 22.5°

Hence, ∠AOK is the required angle of 22.5°.

1(vii)Exercise 22(A)

Using ruler and compasses, construct an angle of 37.5°.

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As, 37.5° = $\frac{1}{2}$ × 75°

Steps of construction :

  1. Draw a line segment OA of any suitable length.

  2. With O as centre, draw an arc of any suitable radius to cut OA at B.

  3. With B as centre and the same radius, draw another arc to cut the previous arc at C.

  4. Again with C as centre and the same radius, draw one more arc to cut the first arc at D.

  5. Join OC and produce it to any point E. Then, ∠AOE = 60°.

  6. With C and D as centres, draw two arcs of equal radii (more than half of CD) to cut each other at F.

  7. Join OF and produce it to any point G. Then, ∠AOG = 90°.

  8. Let OG cut the first arc at P.

  9. With C and P as centres and radius more than half of CP, draw two arcs which cut each other at H.

  10. Join OH. Then, ∠EOH = $\frac{1}{2}$ × (90° − 60°) = 15° and ∠AOH = 60° + 15° = 75°.

  11. Let OH cut the first arc at Q.

  12. With B and Q as centres and radius more than half of BQ, draw two arcs which cut each other at K.

  13. Join OK.

Using ruler and compasses, construct an angle of 37.5°. Constructions, Mathematics Solutions ICSE Class 7.

∠AOK = $\frac{1}{2}$ ∠AOH = $\frac{1}{2}$ × 75° = 37.5°

Hence, ∠AOK is the required angle of 37.5°.

1(viii)Exercise 22(A)

Using ruler and compasses, construct an angle of 67.5°.

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As, 67.5° = 45° + 22.5°

Steps of construction :

  1. Draw a line segment OA of any suitable length.

  2. With O as centre, draw an arc of any suitable radius to cut OA at B.

  3. With B as centre and the same radius, draw another arc to cut the previous arc at C.

  4. Again with C as centre and the same radius, draw one more arc to cut the first arc at D.

  5. With C and D as centres, draw two arcs of equal radii (more than half of CD) to cut each other at E.

  6. Join OE. Then, ∠AOE = 90°.

  7. Let OE cut the first arc at F.

  8. With B and F as centres and radius more than half of BF, draw two arcs which cut each other at G.

  9. Join OG. Then, ∠AOG = ∠GOE = $\frac{1}{2}$ × 90° = 45°.

  10. Let OG cut the first arc at H.

  11. With H and F as centres and radius more than half of HF, draw two arcs which cut each other at K.

  12. Join OK. Then, ∠GOK = $\frac{1}{2}$ ∠GOE = $\frac{1}{2}$ × 45° = 22.5°.

Using ruler and compasses, construct an angle of 67.5°. Constructions, Mathematics Solutions ICSE Class 7.

∠AOK = ∠AOG + ∠GOK = 45° + 22.5° = 67.5°

Hence, ∠AOK is the required angle of 67.5°.

2Exercise 22(A)

Draw ∠ABC = 120°. Bisect the angle using ruler and compasses only. Measure each angle so obtained and check whether the angles obtained on bisecting ∠ABC are equal or not.

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Steps of construction :

  1. Draw a line segment BC of any suitable length.

  2. With B as centre, draw an arc to cut BC at P.

  3. With P as centre, draw an arc of the same radius (as taken in Step 2) to cut the first arc at Q.

  4. With Q as centre, draw one more arc of the same radius which cuts the first arc at R.

  5. Join BR and produce it upto any point A. Then, ∠ABC = 120°.

  6. With P as centre and radius more than half of PR, draw an arc.

  7. With R as centre and the same radius as taken in Step 6, draw another arc which meets the previous arc at D.

  8. Join B and D. Then, BD is the bisector of ∠ABC.

Draw ∠ABC = 120°. Bisect the angle using ruler and compasses only. Measure each angle so obtained and check whether the angles obtained on bisecting ∠ABC are equal or not. Constructions, Mathematics Solutions ICSE Class 7.

On measuring, we get :

∠ABD = 60° and ∠DBC = 60°

⇒ ∠ABD = ∠DBC

Hence, the angles obtained on bisecting ∠ABC are equal, each being 60°.

3Exercise 22(A)

Draw a line segment PQ = 6 cm. Mark a point A in PQ so that AP = 2 cm. At point A, construct angle QAR = 60°.

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Given, PQ = 6 cm and AP = 2 cm.

Steps of construction :

  1. Draw a line segment PQ = 6 cm.

  2. From PQ, cut off PA = 2 cm.

  3. With A as centre, draw an arc of any suitable radius to cut AQ at B.

  4. With B as centre and the same radius, draw another arc to cut the previous arc at C.

  5. Join AC and produce it to any point R.

Draw a line segment PQ = 6 cm. Mark a point A in PQ so that AP = 2 cm. At point A, construct angle QAR = 60°. Constructions, Mathematics Solutions ICSE Class 7.

Hence, ∠QAR is the required angle of 60°.

4Exercise 22(A)

Draw a line segment AB = 8 cm. Mark a point P in AB so that AP = 5 cm. At P, construct angle APQ = 30°.

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Given, AB = 8 cm and AP = 5 cm.

Steps of construction :

  1. Draw a line segment AB = 8 cm.

  2. From AB, cut off AP = 5 cm.

  3. With P as centre, draw an arc of any suitable radius to cut PA at C.

  4. With C as centre and the same radius, draw another arc to cut the previous arc at D.

  5. Join PD and produce it to any point E. Then, ∠APE = 60°.

  6. With C and D as centres and radius more than half of CD, draw two arcs which cut each other at F.

  7. Join PF and produce it to any point Q.

Draw a line segment AB = 8 cm. Mark a point P in AB so that AP = 5 cm. At P, construct angle APQ = 30°. Constructions, Mathematics Solutions ICSE Class 7.

∠APQ = $\frac{1}{2}$ ∠APE = $\frac{1}{2}$ × 60° = 30°

Hence, ∠APQ is the required angle of 30°.

5Exercise 22(A)

Construct an angle of 75° and then bisect it.

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Steps of construction :

  1. Draw an angle AOD = 90° at a point O on the line segment OA.

  2. At the same point O, draw an angle AOE = 60°.

  3. Bisect ∠DOE, so that ∠EOC = ∠DOC = 15°.

In the given figure : ∠AOC = ∠AOE + ∠EOC = 60° + 15° = 75°.

  1. With O as center draw an arc intersecting OC at P and OA at Q.

  2. With P and Q as centres and radius more than half of PQ, draw two arcs which cut each other at F.

  3. Join OF.

Construct an angle of 75° and then bisect it. Constructions, Mathematics Solutions ICSE Class 7.

∠AOF = ∠FOC = $\frac{1}{2}$ ∠AOC = $\frac{1}{2}$ × 75° = 37.5°

Hence, ∠AOC is the required angle of 75° and OF is its bisector, dividing it into two angles of 37.5° each.

6Exercise 22(A)

Draw a line segment of length 6.4 cm. Draw its perpendicular bisector.

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Steps of construction :

  1. Draw a line segment AB = 6.4 cm.

  2. Taking any suitable radius, which must be more than half of AB (i.e., more than 3.2 cm), draw arcs with centres at points A and B on both the sides of AB.

  3. Let the arcs drawn intersect each other at points C and D.

  4. Draw a line through the points C and D, which cuts AB at P.

Draw a line segment of length 6.4 cm. Draw its perpendicular bisector. Constructions, Mathematics Solutions ICSE Class 7.

Here, AP = PB = $\frac{1}{2}$ AB = 3.2 cm and ∠APC = 90°.

Hence, CD is the required perpendicular bisector of AB.

7Exercise 22(A)

Draw a line segment AB = 5.8 cm. Mark a point P in AB such that PB = 3.6 cm. At P, draw perpendicular to AB.

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Given, AB = 5.8 cm and PB = 3.6 cm.

Steps of construction :

  1. Draw a line segment AB = 5.8 cm.

  2. From AB, cut off BP = 3.6 cm.

  3. With P as centre, draw an arc of any suitable radius to cut PB at C.

  4. With C as centre and radius as taken in Step 3, draw the arc to cut the previous arc at D.

  5. Again with D as centre and with the same radius, draw one more arc to cut the first arc at E.

  6. With D and E as centres, draw two arcs of equal radii to cut each other at Q.

  7. Join PQ.

Draw a line segment AB = 5.8 cm. Mark a point P in AB such that PB = 3.6 cm. At P, draw perpendicular to AB. Constructions, Mathematics Solutions ICSE Class 7.

Here, ∠QPB = ∠QPA = 90°.

Hence, PQ is the required perpendicular to AB at point P.

8(i)Exercise 22(A)

Draw a line through point P and parallel to AB.

Draw a line through point P and parallel to AB. Constructions, Mathematics Solutions ICSE Class 7.
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Steps of construction :

  1. Mark any point Q on the given line AB and join PQ.

  2. With Q as centre and any suitable radius (less than PQ), draw an arc which cuts QB at C and QP at D.

  3. With P as centre and the same radius, draw an arc on the side of PQ opposite to that of B, which cuts PQ at E.

  4. With E as centre and radius equal to CD, draw an arc which cuts the arc drawn in Step 3 at F.

  5. Join PF and produce it on both the sides to get the line ST.

Draw a line through point P and parallel to AB. Constructions, Mathematics Solutions ICSE Class 7.

Here, ∠QPF = ∠PQB and these are alternate angles.

∴ ST ∥ AB

Hence, line ST is the required line through the given point P and parallel to the given line AB.

8(ii)Exercise 22(A)

Draw a line through point P and parallel to AB.

Draw a line through point P and parallel to AB. Constructions, Mathematics Solutions ICSE Class 7.
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Steps of construction :

  1. Mark any point Q on the given line AB and join PQ.

  2. With Q as centre and any suitable radius (less than PQ), draw an arc which cuts QB at C and QP at D.

  3. With P as centre and the same radius, draw an arc on the side of PQ opposite to that of B, which cuts PQ at E.

  4. With E as centre and radius equal to CD, draw an arc which cuts the arc drawn in Step 3 at F.

  5. Join PF and produce it on both the sides to get the line ST.

Draw a line through point P and parallel to AB. Constructions, Mathematics Solutions ICSE Class 7.

Here, ∠QPF = ∠PQB and these are alternate angles.

∴ ST ∥ AB

Hence, line ST is the required line through the given point P and parallel to the given line AB.

8(iii)Exercise 22(A)

Draw a line through point P and parallel to AB.

Draw a line through point P and parallel to AB. Constructions, Mathematics Solutions ICSE Class 7.
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Steps of construction :

  1. Mark any point Q on the given line AB and join PQ.

  2. With Q as centre and any suitable radius (less than PQ), draw an arc which cuts QB at C and QP at D.

  3. With P as centre and the same radius, draw an arc on the side of PQ opposite to that of B, which cuts PQ at E.

  4. With E as centre and radius equal to CD, draw an arc which cuts the arc drawn in Step 3 at F.

  5. Join PF and produce it on both the sides to get the line ST.

Draw a line through point P and parallel to AB. Constructions, Mathematics Solutions ICSE Class 7.

Here, ∠QPF = ∠PQB and these are alternate angles.

∴ ST ∥ AB

Hence, line ST is the required line through the given point P and parallel to the given line AB.

1(i)Exercise 22(B)

Construct a ΔABC such that AB = 6 cm, BC = 4 cm and CA = 5.5 cm.

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Given, in ΔABC, AB = 6 cm, BC = 4 cm and CA = 5.5 cm.

Steps of construction :

  1. Draw a line segment AB = 6 cm.

  2. With A as centre, draw an arc of radius 5.5 cm and with B as centre draw another arc with radius 4 cm. Let these arcs meet at C.

  3. Join BC and AC.

Construct a ΔABC such that AB = 6 cm, BC = 4 cm and CA = 5.5 cm. Constructions, Mathematics Solutions ICSE Class 7.

Hence, triangle ABC so obtained is the required triangle.

1(ii)Exercise 22(B)

Construct a ΔABC such that CB = 6.5 cm, CA = 4.2 cm and BA = 5.1 cm.

Show SolutionHide Solution↓

Given, in ΔABC, CB = 6.5 cm, CA = 4.2 cm and BA = 5.1 cm.

Steps of construction :

  1. Draw a line segment CB = 6.5 cm.

  2. With C as centre, draw an arc of radius 4.2 cm and with B as centre draw another arc with radius 5.1 cm. Let these arcs meet at A.

  3. Join CA and BA.

Construct a ΔABC such that CB = 6.5 cm, CA = 4.2 cm and BA = 5.1 cm. Constructions, Mathematics Solutions ICSE Class 7.

Hence, triangle ABC so obtained is the required triangle.

1(iii)Exercise 22(B)

Construct a ΔABC such that BC = 4 cm, AC = 5 cm and AB = 3.5 cm.

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Given, in ΔABC, BC = 4 cm, AC = 5 cm and AB = 3.5 cm.

Steps of construction :

  1. Draw a line segment BC = 4 cm.

  2. With C as centre, draw an arc of radius 5 cm and with B as centre draw another arc with radius 3.5 cm. Let these arcs meet at A.

  3. Join AB and AC.

Construct a ΔABC such that BC = 4 cm, AC = 5 cm and AB = 3.5 cm. Constructions, Mathematics Solutions ICSE Class 7.

Hence, triangle ABC so obtained is the required triangle.

2(i)Exercise 22(B)

Construct a ΔABC such that AB = 7 cm, BC = 5 cm and ∠ABC = 60°.

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Given, in ΔABC, AB = 7 cm, BC = 5 cm and ∠ABC = 60°.

Steps of construction :

  1. Draw a line segment BC = 5 cm.

  2. At B, draw an angle PBC = 60°.

  3. With B as centre, draw an arc of 7 cm radius, which cuts PB at A. Join AC.

Construct a ΔABC such that AB = 7 cm, BC = 5 cm and ∠ABC = 60°. Constructions, Mathematics Solutions ICSE Class 7.

Hence, ABC is the required triangle.

2(ii)Exercise 22(B)

Construct a ΔABC such that BC = 6 cm, AC = 5.7 cm and ∠ACB = 75°.

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Given, in ΔABC, BC = 6 cm, AC = 5.7 cm and ∠ACB = 75°.

Steps of construction :

  1. Draw a line segment BC = 6 cm.

  2. At C, draw an angle PCB = 75°.

  3. With C as centre, draw an arc of 5.7 cm radius, which cuts PC at A. Join AB.

Construct a ΔABC such that BC = 6 cm, AC = 5.7 cm and ∠ACB = 75°. Constructions, Mathematics Solutions ICSE Class 7.

Hence, ABC is the required triangle.

2(iii)Exercise 22(B)

Construct a ΔABC such that AB = 6.5 cm, AC = 5.8 cm and ∠A = 45°.

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Given, in ΔABC, AB = 6.5 cm, AC = 5.8 cm and ∠A = 45°.

Steps of construction :

  1. Draw a line segment AB = 6.5 cm.

  2. At A, draw an angle PAB = 45°.

  3. With A as centre, draw an arc of 5.8 cm radius, which cuts PA at C. Join BC.

Construct a ΔABC such that AB = 6.5 cm, AC = 5.8 cm and ∠A = 45°. Constructions, Mathematics Solutions ICSE Class 7.

Hence, ABC is the required triangle.

3(i)Exercise 22(B)

Construct a ΔPQR such that PQ = 6 cm, ∠Q = 60° and ∠P = 45°. Measure ∠R.

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Given, in ΔPQR, PQ = 6 cm, ∠Q = 60° and ∠P = 45°.

Steps of construction :

  1. Draw a line segment PQ = 6 cm.

  2. At P, draw an angle of 45°.

  3. At Q, draw an angle of 60°.

  4. Let the lines of 45° and 60° meet at R.

Construct a ΔPQR such that PQ = 6 cm, ∠Q = 60° and ∠P = 45°. Measure ∠R. Constructions, Mathematics Solutions ICSE Class 7.

Hence, PQR is the required triangle.

On measuring, we get :

∠R = 75°

3(ii)Exercise 22(B)

Construct a ΔPQR such that QR = 4.4 cm, ∠R = 30° and ∠Q = 75°. Measure PQ and PR.

Show SolutionHide Solution↓

Given, in ΔPQR, QR = 4.4 cm, ∠R = 30° and ∠Q = 75°.

Steps of construction :

  1. Draw a line segment QR = 4.4 cm.

  2. At Q, draw an angle of 75°.

  3. At R, draw an angle of 30°.

  4. Let the lines of 75° and 30° meet at P.

Construct a ΔPQR such that QR = 4.4 cm, ∠R = 30° and ∠Q = 75°. Measure PQ and PR. Constructions, Mathematics Solutions ICSE Class 7.

Hence, PQR is the required triangle.

On measuring, we get :

PQ = 2.3 cm and PR = 4.4 cm

3(iii)Exercise 22(B)

Construct a ΔPQR such that PR = 5.8 cm, ∠P = 60° and ∠R = 45°. Measure ∠Q and verify it by calculations.

Show SolutionHide Solution↓

Given, in ΔPQR, PR = 5.8 cm, ∠P = 60° and ∠R = 45°.

Steps of construction :

  1. Draw a line segment PR = 5.8 cm.

  2. At P, draw an angle of 60°.

  3. At R, draw an angle of 45°.

  4. Let the lines of 60° and 45° meet at Q.

Construct a ΔPQR such that PR = 5.8 cm, ∠P = 60° and ∠R = 45°. Measure ∠Q and verify it by calculations. Constructions, Mathematics Solutions ICSE Class 7.

Hence, PQR is the required triangle.

On measuring, we get :

∠Q = 75°

Verification :

In ΔPQR,

∠P + ∠Q + ∠R = 180° [Angle sum property of triangle]

⇒ 60° + ∠Q + 45° = 180°

⇒ 105° + ∠Q = 180°

⇒ ∠Q = 180° − 105° = 75°

Hence, ∠Q = 75°, which verifies the measured value.

4(i)Exercise 22(B)

Construct an isosceles ΔABC such that base BC = 4 cm and base angle = 30°. Measure the other two sides of the triangle.

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In an isosceles triangle, the base angles are equal.

Given, in isosceles ΔABC, base BC = 4 cm and ∠B = ∠C = 30°.

Steps of construction :

  1. Draw a line segment BC = 4 cm.

  2. At B, draw an angle of 30°.

  3. At C also, draw an angle of 30°. Let these 30° lines meet at A.

Construct an isosceles ΔABC such that base BC = 4 cm and base angle = 30°. Measure the other two sides of the triangle. Constructions, Mathematics Solutions ICSE Class 7.

Hence, ABC is the required triangle.

On measuring, we get :

AB = AC = 2.3 cm (approx.)

4(ii)Exercise 22(B)

Construct an isosceles ΔABC such that base AB = 6.2 cm and base angle = 45°. Measure the other two sides of the triangle.

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In an isosceles triangle, the base angles are equal.

Given, in isosceles ΔABC, base AB = 6.2 cm and ∠A = ∠B = 45°.

Steps of construction :

  1. Draw a line segment AB = 6.2 cm.

  2. At A, draw an angle of 45°.

  3. At B also, draw an angle of 45°. Let these 45° lines meet at C.

Construct an isosceles ΔABC such that base AB = 6.2 cm and base angle = 45°. Measure the other two sides of the triangle. Constructions, Mathematics Solutions ICSE Class 7.

Hence, ABC is the required triangle.

On measuring, we get :

AC = BC = 4.4 cm (approx.)

4(iii)Exercise 22(B)

Construct an isosceles ΔABC such that base AC = 5 cm and base angle = 75°. Measure the other two sides of the triangle.

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In an isosceles triangle, the base angles are equal.

Given, in isosceles ΔABC, base AC = 5 cm and ∠A = ∠C = 75°.

Steps of construction :

  1. Draw a line segment AC = 5 cm.

  2. At A, draw an angle of 75°.

  3. At C also, draw an angle of 75°. Let these 75° lines meet at B.

Construct an isosceles ΔABC such that base AC = 5 cm and base angle = 75°. Measure the other two sides of the triangle. Constructions, Mathematics Solutions ICSE Class 7.

Hence, ABC is the required triangle.

On measuring, we get :

AB = BC = 9.6 cm (approx.)

5(i)Exercise 22(B)

Construct an isosceles ΔABC such that AB = AC = 6.5 cm and ∠A = 60°.

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Vertex angle is the angle between two equal sides.

Given, in isosceles ΔABC, AB = AC = 6.5 cm and ∠BAC = 60°.

Steps of construction :

  1. Draw a line segment AB = 6.5 cm.

  2. At A, draw AP so that angle BAP = 60°.

  3. With A as centre draw an arc of 6.5 cm radius, which cuts AP at C. Join C and B.

Construct an isosceles ΔABC such that AB = AC = 6.5 cm and ∠A = 60°. Constructions, Mathematics Solutions ICSE Class 7.

Hence, ABC is the required isosceles triangle.

5(ii)Exercise 22(B)

Construct an isosceles ΔABC such that one of the equal sides = 6 cm and vertex angle = 45°. Measure the base angles.

Show SolutionHide Solution↓

Vertex angle is the angle between two equal sides.

Let the equal sides be AB = AC = 6 cm and vertex angle ∠BAC = 45°.

Steps of construction :

  1. Draw a line segment AB = 6 cm.

  2. At A, draw AP so that angle BAP = 45°.

  3. With A as centre draw an arc of 6 cm radius, which cuts AP at C. Join C and B.

Construct an isosceles ΔABC such that one of the equal sides = 6 cm and vertex angle = 45°. Measure the base angles. Constructions, Mathematics Solutions ICSE Class 7.

Hence, ABC is the required isosceles triangle.

On measuring, we get :

∠B = ∠C = 67.5° = 67° 30′

5(iii)Exercise 22(B)

Construct an isosceles ΔABC such that BC = AB = 5.8 cm and ∠B = 30°. Measure ∠A and ∠C.

Show SolutionHide Solution↓

Vertex angle is the angle between two equal sides.

Given, in isosceles ΔABC, BC = AB = 5.8 cm and ∠B = 30°.

Here, ∠B is the vertex angle.

Steps of construction :

  1. Draw a line segment BC = 5.8 cm.

  2. At B, draw BP so that angle CBP = 30°.

  3. With B as centre draw an arc of 5.8 cm radius, which cuts BP at A. Join A and C.

Construct an isosceles ΔABC such that BC = AB = 5.8 cm and ∠B = 30°. Measure ∠A and ∠C. Constructions, Mathematics Solutions ICSE Class 7.

Hence, ABC is the required isosceles triangle.

On measuring, we get :

∠A = ∠C = 75°

6(i)Exercise 22(B)

Construct an equilateral ΔABC such that AB = 5 cm. Draw the perpendicular bisectors of BC and AC. Let P be the point of intersection of these two bisectors. Measure PA, PB and PC.

Show SolutionHide Solution↓

Given, in equilateral ΔABC, AB = BC = CA = 5 cm.

Steps of construction :

  1. Draw a line AB = 5 cm.

  2. With A as centre, draw an arc of radius 5 cm.

  3. With B as centre, draw another arc of radius 5 cm. Let the two arcs intersect each other at point C. Join AC and BC.

  4. Taking a radius more than half of BC, draw arcs with centres B and C on both the sides of BC. Let these arcs intersect each other at D and E. Draw a line through D and E.

  5. Taking a radius more than half of AC, draw arcs with centres A and C on both the sides of AC. Let these arcs intersect each other at F and G. Draw a line through F and G.

  6. Let DE and FG intersect each other at P. Join PA, PB and PC.

Construct an equilateral ΔABC such that AB = 5 cm. Draw the perpendicular bisectors of BC and AC. Let P be the point of intersection of these two bisectors. Measure PA, PB and PC. Constructions, Mathematics Solutions ICSE Class 7.

Hence, ABC is the required triangle.

On measuring, we get :

PA = PB = PC = 2.9 cm (approx.)

6(ii)Exercise 22(B)

Construct an equilateral ΔABC such that each side is 6 cm.

Show SolutionHide Solution↓

Given, in equilateral ΔABC, AB = BC = CA = 6 cm.

Steps of construction :

  1. Draw a line AB = 6 cm.

  2. With A as centre, draw an arc of radius 6 cm.

  3. With B as centre, draw another arc of radius 6 cm. Let the two arcs intersect each other at point C.

  4. Join AC and BC.

Construct an equilateral ΔABC such that each side is 6 cm. Constructions, Mathematics Solutions ICSE Class 7.

Hence, ABC is the required triangle.

7(i)Exercise 22(B)

Construct a ΔABC such that AB = 6 cm, BC = 4.5 cm and AC = 5.5 cm. Construct the circumcircle of this triangle.

Show SolutionHide Solution↓

Given, in ΔABC, AB = 6 cm, BC = 4.5 cm and AC = 5.5 cm.

Steps of construction :

  1. Draw a line segment AB = 6 cm.

  2. With A as centre, draw an arc of radius 5.5 cm and with B as centre draw another arc with radius 4.5 cm. Let these arcs meet at C.

  3. Join AC and BC.

  4. Draw the perpendicular bisectors of the sides AB and AC. Let these bisectors intersect each other at point O.

  5. Taking O as centre and OA or OB or OC as radius, draw a circle.

Construct a ΔABC such that AB = 6 cm, BC = 4.5 cm and AC = 5.5 cm. Construct the circumcircle of this triangle. Constructions, Mathematics Solutions ICSE Class 7.

Hence, the circle so drawn, passing through the vertices A, B and C, is the required circumcircle of ΔABC.

7(ii)Exercise 22(B)

Construct an isosceles ΔPQR such that PQ = PR = 6.5 cm and ∠PQR = 75°. Using ruler and compasses only construct the circumcircle to this triangle.

Show SolutionHide Solution↓

Given, in isosceles ΔPQR, PQ = PR = 6.5 cm and ∠PQR = 75°.

In an isosceles triangle, the base angles are equal.

⇒ ∠PRQ = ∠PQR = 75°

In ΔPQR,

∠QPR + ∠PQR + ∠PRQ = 180° [Angle sum property of triangle]

⇒ ∠QPR + 75° + 75° = 180°

⇒ ∠QPR = 180° − 150° = 30°

Steps of construction :

  1. Draw a line segment PQ = 6.5 cm.

  2. At P, draw PX so that angle QPX = 30°.

  3. With P as centre draw an arc of 6.5 cm radius, which cuts PX at R. Join QR.

  4. Draw the perpendicular bisectors of the sides PQ and PR. Let these bisectors intersect each other at point O.

  5. Taking O as centre and OP or OQ or OR as radius, draw a circle.

Construct an isosceles ΔPQR such that PQ = PR = 6.5 cm and ∠PQR = 75°. Using ruler and compasses only construct the circumcircle to this triangle. Constructions, Mathematics Solutions ICSE Class 7.

Hence, the circle so drawn, passing through the vertices P, Q and R, is the required circumcircle of ΔPQR.

7(iii)Exercise 22(B)

Construct an equilateral triangle ABC such that its one side = 5.5 cm. Construct the circumcircle to this triangle.

Show SolutionHide Solution↓

Given, in equilateral ΔABC, AB = BC = CA = 5.5 cm.

Steps of construction :

  1. Draw a line AB = 5.5 cm.

  2. With A as centre, draw an arc of radius 5.5 cm.

  3. With B as centre, draw another arc of radius 5.5 cm. Let the two arcs intersect each other at point C. Join AC and BC.

  4. Draw the perpendicular bisectors of the sides AB and AC. Let these bisectors intersect each other at point O.

  5. Taking O as centre and OA or OB or OC as radius, draw a circle.

Construct an equilateral triangle ABC such that its one side = 5.5 cm. Construct the circumcircle to this triangle. Constructions, Mathematics Solutions ICSE Class 7.

Hence, the circle so drawn, passing through the vertices A, B and C, is the required circumcircle of ΔABC.

8(i)Exercise 22(B)

Construct a ΔABC such that AB = 6 cm, BC = 5.6 cm and CA = 6.5 cm. Inscribe a circle to this triangle.

Show SolutionHide Solution↓

Given, in ΔABC, AB = 6 cm, BC = 5.6 cm and CA = 6.5 cm.

Steps of construction :

  1. Draw a line segment BC = 5.6 cm.

  2. With B as centre, draw an arc of radius 6 cm and with C as centre draw another arc with radius 6.5 cm. Let these arcs meet at A.

  3. Join AB and AC.

  4. Bisect ∠ABC and ∠ACB. Let these two bisectors meet at I.

  5. From I, draw a perpendicular on BC. This perpendicular meets BC at point L.

  6. With I as centre and IL as radius, draw a circle.

Construct a ΔABC such that AB = 6 cm, BC = 5.6 cm and CA = 6.5 cm. Inscribe a circle to this triangle. Constructions, Mathematics Solutions ICSE Class 7.

Hence, the circle so drawn, touching the sides BC, AB and AC, is the required incircle of ΔABC.

8(ii)Exercise 22(B)

Construct the isosceles ΔMNP such that base MN = 5.8 cm, base angle MNP = 30°. Construct the incircle to this triangle.

Show SolutionHide Solution↓

Given, in isosceles ΔMNP, base MN = 5.8 cm and base angle ∠MNP = 30°.

In an isosceles triangle, the base angles are equal.

⇒ ∠NMP = ∠MNP = 30°

Steps of construction :

  1. Draw a line segment MN = 5.8 cm.

  2. At M, draw an angle of 30°.

  3. At N also, draw an angle of 30°. Let these 30° lines meet at P.

  4. Bisect ∠PMN and ∠PNM. Let these two bisectors meet at I.

  5. From I, draw a perpendicular on MN. This perpendicular meets MN at point L.

  6. With I as centre and IL as radius, draw a circle.

Construct the isosceles ΔMNP such that base MN = 5.8 cm, base angle MNP = 30°. Construct the incircle to this triangle. Constructions, Mathematics Solutions ICSE Class 7.

Hence, the circle so drawn, touching the sides MN, NP and MP, is the required incircle of ΔMNP.

8(iii)Exercise 22(B)

Construct an equilateral ΔDEF whose one side is 5.5 cm. Construct the incircle to this triangle.

Show SolutionHide Solution↓

Given, in equilateral ΔDEF, DE = EF = FD = 5.5 cm.

Steps of construction :

  1. Draw a line EF = 5.5 cm.

  2. With E as centre, draw an arc of radius 5.5 cm.

  3. With F as centre, draw another arc of radius 5.5 cm. Let the two arcs intersect each other at point D. Join DE and DF.

  4. Bisect ∠DEF and ∠DFE. Let these two bisectors meet at I.

  5. From I, draw a perpendicular on EF. This perpendicular meets EF at point L.

  6. With I as centre and IL as radius, draw a circle.

Construct an equilateral ΔDEF whose one side is 5.5 cm. Construct the incircle to this triangle. Constructions, Mathematics Solutions ICSE Class 7.

Hence, the circle so drawn, touching the sides EF, DE and DF, is the required incircle of ΔDEF.

8(iv)Exercise 22(B)

Construct the ΔPQR such that PQ = 6 cm, ∠QPR = 45° and angle PQR = 60°. Locate its incentre and then draw its incircle.

Show SolutionHide Solution↓

Given, in ΔPQR, PQ = 6 cm, ∠QPR = 45° and ∠PQR = 60°.

Steps of construction :

  1. Draw a line segment PQ = 6 cm.

  2. At P, draw an angle of 45°.

  3. At Q, draw an angle of 60°. Let the lines of 45° and 60° meet at R.

  4. Bisect ∠QPR and ∠PQR. Let these two bisectors meet at I. Then, I is the incentre of ΔPQR.

  5. From I, draw a perpendicular on PQ. This perpendicular meets PQ at point L.

  6. With I as centre and IL as radius, draw a circle.

Construct the ΔPQR such that PQ = 6 cm, ∠QPR = 45° and angle PQR = 60°. Locate its incentre and then draw its incircle. Constructions, Mathematics Solutions ICSE Class 7.

Hence, I is the incentre and the circle so drawn, touching the sides PQ, QR and PR, is the required incircle of ΔPQR.

11 Mark · Multiple Choice Questions

Angle AOB = 120°, is divided into four equal parts. The measure of angle ROA is:

Angle AOB = 120°, is divided into four equal parts. The measure of angle ROA is:. Constructions, Mathematics Solutions ICSE Class 7.
Show SolutionHide Solution↓

Since, ∠AOB = 120° is divided into four equal parts by OP, OQ and OR,

⇒ ∠AOP = ∠POQ = ∠QOR = ∠ROB = $\frac{120°}{4}$ = 30°

∠ROA = ∠AOP + ∠POQ + ∠QOR

= 30° + 30° + 30°

= 90°

Hence, option 4 is the correct option.

21 Mark · Multiple Choice Questions

Point P lies on perpendicular bisector of line segment AB. Then:

Point P lies on perpendicular bisector of line segment AB. Then:. Constructions, Mathematics Solutions ICSE Class 7.
Show SolutionHide Solution↓

Every point on the perpendicular bisector of a line segment is equidistant from the end points of the line segment.

Since, P lies on the perpendicular bisector of AB,

⇒ PA = PB

Hence, option 3 is the correct option.

31 Mark · Multiple Choice Questions

Point P is the point of intersection of the bisectors of angles ABC and DCB, then:

Point P is the point of intersection of the bisectors of angles ABC and DCB, then:. Constructions, Mathematics Solutions ICSE Class 7.
Show SolutionHide Solution↓

Every point on the bisector of an angle is equidistant from the arms of the angle.

Since, P lies on the bisector of ∠ABC,

⇒ P is equidistant from AB and BC.

Since, P lies on the bisector of ∠DCB,

⇒ P is equidistant from BC and CD.

∴ P is equidistant from AB, BC and CD.

Hence, option 4 is the correct option.

41 Mark · Multiple Choice Questions

In ΔABC, the perpendicular bisectors of sides AB and BC meet at point P, then:

In ΔABC, the perpendicular bisectors of sides AB and BC meet at point P, then:. Constructions, Mathematics Solutions ICSE Class 7.
Show SolutionHide Solution↓

Every point on the perpendicular bisector of a line segment is equidistant from its end points.

Since, P lies on the perpendicular bisector of AB,

⇒ PA = PB

Since, P lies on the perpendicular bisector of BC,

⇒ PB = PC

∴ PA = PB = PC

Hence, option 3 is the correct option.

51 Mark · Multiple Choice Questions

The number of parallel lines, that can be drawn parallel to a given line from a point outside it, is:

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Through a point outside a given line, one and only one line can be drawn parallel to the given line.

Hence, option 2 is the correct option.

61 Mark · Multiple Choice Questions

An angle that cannot be constructed using ruler and compass only is:

Show SolutionHide Solution↓

Using ruler and compass only, angles such as 60°, 90° and their bisections and combinations can be constructed.

22.5° = $\frac{1}{2}$ × 45°

67.5° = 45° + 22.5°

105° = 90° + 15°

52° cannot be obtained from such combinations, so it cannot be constructed using ruler and compass only.

Hence, option 3 is the correct option.

71 Mark · Multiple Choice Questions

The length of line segment PQ = 10 cm. The number of triangles PAQ that can be drawn with AP = 3 cm and AQ = 6 cm is:

Show SolutionHide Solution↓

A triangle can be constructed only if the sum of the lengths of every pair of two sides is greater than the length of the third side.

AP + AQ = 3 cm + 6 cm = 9 cm

⇒ AP + AQ < PQ (as 9 cm < 10 cm)

∴ No triangle PAQ can be drawn.

Hence, option 3 is the correct option.

81 Mark · Multiple Choice Questions

The length of line segment PQ = 10 cm. The number of triangles APQ (point A is above PQ) that can be drawn with AP = AQ = 10 cm.

Show SolutionHide Solution↓

Given, AP = AQ = PQ = 10 cm.

⇒ ΔAPQ is an equilateral triangle.

With P and Q as centres and radius 10 cm, the two arcs drawn above PQ intersect at only one point A.

∴ Only one triangle APQ can be drawn above PQ.

Hence, option 1 is the correct option.

Press ↓ to reveal a solution and move to the next question. Back to Selina Solutions Concise Mathematics Class 7 ICSE.