## RS Aggarwal Class 7 Solutions Chapter 17 Constructions Ex 17B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17B.

**Other Exercises**

- RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17A
- RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17B
- RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17C
- RS Aggarwal Solutions Class 7 Chapter 17 Constructions CCE Test Paper

**Question 1.**

**Solution:**

Steps of construction :

(i) Draw a line segment BC = 3.6cm

(ii) At B, draw an arc of the radius 5cm.

(iii) At C, draw another arc of the radius 5.4cm which intersects the first arc at A

(iv) Join AB and AC

(v) With centre B and C and radius more than half of BC, draw arcs intersecting each other at L and M.

(vi) Join LM which intersects BC at Q and produce it to P.

Then PQ is perpendicular bisector of side BC.

**Question 2.**

**Solution:**

Steps of construction :

(i) Draw a line segment QR = 6cm

(ii) With centre Q and radius 4.4 cm draw an arc.

(iii) With centre R and radius 5.3 cm, draw another arc intersecting the first arc at P.

(iv) Join PQ and PR.

(v) With centre P and a suitable radius, draw an arc meeting PR at E and PQ at F.

(vi) With centres E and F, with same radius, draw two arcs intersecting each other at G.

(vii) Join PG and produce it to meet QR at S. Then PS is the bisector of ∠P.

**Question 3.**

**Solution:**

Steps of construction :

(i) Draw a line segment BC = 6.2 cm.

(ii) With centres B and C radius 6.2 cm, draw arcs intersecting each other at A.

(iii) Join AB and AC.

∆ABC is the required equilateral triangle.

On measuring, each angle is equal to 60°.

**Question 4.**

**Solution:**

Steps of construction :

(i) Draw a line segment BC = 5.3 cm.

(ii) With centre B and C, and radius 4.8 cm, draw arcs intersecting each other at A

(iii) Join AB and AC, Then ∆ABC is the required triangle.

(iv) Now, with centre A and a suitable radius draw an arc intersecting BC at L and M.

(v) Then with centre L and M, draw two arcs intersecting eachother at E.

(vi) Join AE intersecting BC at D. Then AD is perpendicular to BC.

On measuring ∠B and ∠C, each is equal to 55°.

**Question 5.**

**Solution:**

Steps of construction :

(i) Draw a line segment AB = 3.8cm.

(ii) At A draw a ray AX making an angle of 60°.

(iii) Cut off AC = 5 cm from AX

(iv) Join CB.

Then ∆ABC is the required triangle.

**Question 6.**

**Solution:**

Steps of construction :

(i) Draw a line segment BC = 4.3 cm.

(ii) At C, draw a ray CY making an angle equal to 45°.

(iii) With centre C, and radius 6cm, draw an arc intersecting CY at A.

(iv) Join AB.

Then ∆ABC is the required triangle.

**Question 7.**

**Solution:**

(i) Draw a line segment AB = 5.2cm.

(ii) At A, draw a ray AX making an angle equal to 120°.

(iii) From AX cut off AC = 5.2cm.

(iv) Join BC.

Then ∆ABC is the required triangle.

(v) With centre A and some suitable radius draw an arc intersecting BC at L and M.

(vi) With centres L and M, draw two arcs intersecting each other at E.

(vii) Join AE intersecting BC at D Then AD is the perpendicular to BC.

**Question 8.**

**Solution:**

Steps of construction :

(i) Draw a line segment BC = 6.2 cm.

(ii) At B draw a ray BX making an angle of 60°.

(iii) At C draw another ray CY making an angle of 45° which intersect the ray BX at A. .

Then ∆ABC is the required triangle.

**Question 9.**

**Solution:**

Steps of construction :

(i) Draw a line segment BC = 5.8 cm.

(ii) At B draw a ray BX making an angle of 30°

(iii) At C draw another ray CY making an angle of 30° intersecting the BX at A

Then ∆ABC is the required triangle.

On measuring AB and AC, AB = 3.5cm and AC = 3.5cm.

AB = AC

∆ABC is an isosceles triangle.

**Question 10.**

**Solution:**

Steps of construction :

In ∆ABC, ∠A = 45° and ∠C = 75°

But ∠A + ∠B + ∠C = 180°

⇒ 45° + ZB + 75° = 180°

⇒ ∠B = 180° – 45° – 75°

⇒ ∠B = 180° – 120° = 60°

(i) Draw a line segment AB = 7cm.

(ii) At A, draw a ray AX making an angle of 45°.

(iii) At B, draw another ray BY making an angle of 60° which intersects AX at C

Then ∆ABC is the required triangle.

**Question 11.**

**Solution:**

Steps of construction :

(i) Draw a line segment BC = 4.8 cm

(ii) At C, draw a ray CX making an angle of 90°.

(iii) With centre B, an radius 6.3cm draw an arc intersecting CX at A.

(iv) Join AB.

Then ∆ABC is the required triangle.

**Question 12.**

**Solution:**

Steps of construction :

(i) Draw a line segment BC = 3.5cm

(ii) At B, draw a ray BX making an angle of 90°.

(iii) With centre C and radius 6cm draw an arc intersecting BX at A

(iv) Join AC.

Then ∆ABC is the required triangle.

**Question 13.**

**Solution:**

One acute angle = 30°, then

second acute angle will be = 90° – 30° = 60° (Sum of acute angles = 90°)

Steps of construction :

(i) Draw a line segment BC = 6cm.

(ii) At B draw a ray BX making an angle of 30°.

(iii) At C draw another ray CY making an angle of 60° which intersects BX at A

Then ∆ABC is the required triangle.

Hope given RS Aggarwal Solutions Class 7 Chapter 17 Constructions Ex 17B are helpful to complete your math homework.

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