Mathematics Semester ii Solutions Solutions for Class 7 Math Chapter 7 Construction Of Triangles are provided here with simple step-by-step explanations. These solutions for Construction Of Triangles are extremely popular among class 7 students for Math Construction Of Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Semester ii Solutions Book of class 7 Math Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Semester ii Solutions Solutions. All Mathematics Semester ii Solutions Solutions for class 7 Math are prepared by experts and are 100% accurate.

#### Question O1:

What is the sum of three angles in a triangle?

#### Answer:

The sum of three angles in a triangle is 180°.

#### Question W1:

Name the sides and angles of the following triangles. #### Answer:

(1) Sides are AB, BC and AC.

Angles are A, B and C.

(2) Sides are PQ, PR and RQ.

Angles are P, Q and R.

(3) Sides are XY, YZ and ZX

Angles are X, Y and Z.

#### Question O2:

What is the measure of each angle in an equilateral triangle?

#### Answer:

The measure of each angle in an equilateral triangle is60°.

#### Question W2:

Name the types of triangles. (Sides are in cm)

 SI. No. AB BC CA ∠A ∠B ∠C Name of the ΔABC 1. 5.4 6.7 5.4 ---- ---- ---- 2. 6 6 6 ---- ---- ---- 3. 6 8 5.5 ---- ---- ---- 4. 7 ---- ---- ---- 30° 120° 5. 5 ---- ---- 30° 90° ---- 6. ---- 8 ---- ---- 50° 60°

#### Answer:

 SI. No. AB BC CA ∠A ∠B ∠C Name of the ΔABC 1. 5.4 cm 6.7 cm 5.4 cm … … … Isosceles triangle 2. 6 cm 6 cm 6 cm … … … Equilateral triangle 3. 6 cm 8 cm 5.5 cm … … … Scalene triangle 4. 7 cm … … … 30° 120° Obtuse triangle 5. 5 cm … … 30° 90° … Right triangle 6. … 8 cm … … 50° 60° Acute triangle

#### Question O3:

In an obtuse angled triangle, what is the type of the two remaining angles?

#### Answer:

In an obtuse angled triangle, the two remaining angles are acute angles.

Reason: Measure of an obtuse angle is greater than 90° and the sum of three angles in a triangle is 180°. So, each of the two remaining angles is less than 90°.

#### Question W3:

Fill in the blanks with suitable answers.

(i) A triangle has ………. Vertices.

(ii) Triangles are grouped into ………. types according to their sides.

(iii) Triangles are grouped into ………. types according to their angles.

(iv) In a scalene triangle three sides are not ……….

(v) If one angle of a triangle is 90°, the triangle is said to be ………..

#### Answer:

(i) A triangle has three vertices.

(ii) Triangles are grouped into three types according to their sides.

(iii) Triangles are grouped into four types according to their angles.

(iv) In scalene triangle three sides are not equal

(v) If one side of a triangle is 90°, the triangle is said to be right angled triangle.

#### Question O1:

Why is it not possible to construct a triangle with sides 3 cm, 4 cm and 8 cm?

#### Answer:

We know that if the sum of any two sides of a triangle is always greater than the third side (greater side), then the triangle can be constructed.

For measures 3 cm, 4 cm and 8 cm, we observe that 3 cm + 4 cm < 8 cm.

So, it is not possible to construct a triangle with the sides 3 cm, 4 cm and 8 cm.

#### Question W1:

Written:

Construct a triangle PQR, in which PQ = QR = RP = 6 cm.

#### Answer:

Construction of ΔPQR, where PQ = QR = RP = 6 cm.

Steps of construction:

(i) Draw PQ = 6 cm.

(ii) Using the compass, with P and Q as centers and radii as 6 cm, draw arcs which cut each other at point R, Here, triangle PQR is the required triangle.

#### Question O2:

The three sides of a triangle are 3 cm, 4 cm and 5 cm. What type of triangle is this?

#### Answer:

Since the given sides 3 cm, 4 cm and 5 cm are different from each other, the triangle can be classified as a scalene triangle.

#### Question W2:

Written:

Construct an equilateral triangle XYZ. Whose sides are 4.5 cm

#### Answer:

Construction of ΔXYZ, where XY = YZ = ZX = 4.5 cm,

Steps of construction:

(i) Draw XY = 4.5 cm.

(ii) Using the compass, with X and Y as centers and radii as 4.5 cm, draw arcs which cut each other at point Z. Here, triangle XYZ is the required triangle.

#### Question W3:

Written:

Construct a triangle ABC in which BC = 5 cm, AB = AC = 4.5 cm. Name the type of the triangle.

#### Answer:

Construction of ΔABC, where BC = 5 cm, AB = AC = 4.5 cm,

Steps of construction:

(i) Draw BC = 5 cm.

(ii) Using the compass, with B and C as centers and radii as 4.5 cm, draw arcs which cut each other at point A. Here, triangle ABC is the required triangle.

This triangle can be classified as an isosceles triangle as AB = AC.

#### Question W4:

Written:

Construct a triangle PQR, in which PQ = 6.7 cm, QR = PR = 5.4 cm.

#### Answer:

Construction of ΔPQR, where PQ = 6.7 cm, QR = PR = 5.4 cm.

Steps of construction:

(i) Draw PQ = 6.7 cm.

(ii) Using the compass, with P and Q as centers and radii as 5.4 cm, draw arcs which cut each other at point R. Here, ΔPQR is the required triangle.

#### Question W5:

Written:

Construct a triangle XYZ, in which XY = 6 cm, YZ = 8 cm and ZX = 5.5 cm.

#### Answer:

Construction of ΔXYZ, where XY = 6 cm, YZ = 8 cm and ZX = 5.5 cm.

Steps of construction:

(i) Draw XY = 6 cm.

(ii) Using the compass, with X and Y as centers and respective radii as 5.5 cm and 8 cm, draw arcs which cut each other at point Z. Here, ΔXYZ is the required triangle.

#### Question W6:

Written:

Construct a triangle ABC in which AB = 6 cm, BC = 10 cm and AC = 8 cm. Write the steps of construction.

#### Answer:

Construction of ΔABC, where AB = 6 cm, BC = 10 cm and AC = 8 cm.

Steps of construction:

(i) Draw AB = 6 cm.

(ii) Using the compass, with A and B as centers and respective radii as 8 cm and 10 cm, draw arcs which cut each other at point C. Here, ΔABC is the required triangle.

#### Question W1:

Construct a triangle EPG, given EP = 8.4 cm, E = 65°, P = 30° Measure the sides PG and EG. Write the steps of construction.

#### Answer:

Construction of ΔEPG, where EP = 8.4 cm, and .

Steps of construction:

(i) Draw EP = 8.4 cm.

(ii) Using a protractor at E and P, construct and .

(iii) Let them meet at G. Thus, triangle EPG is the required triangle.

#### Question O1:

In ΔABC, if A = 50°, B = 60°, then what is the measure of C?

#### Answer:

∠C = 70°.

Explanation: #### Question O2:

In ΔXYZ, if X = Y = 60°, then what is the measure of Z?

#### Answer:

∠Z = 60°.

Explanation: #### Question W2:

Construct the triangle MRB in which RB = 5.2 cm. R = 30°, B = 120° and measure M.

#### Answer:

Construction of ΔMRB, where RB = 5.2 cm, and .

Steps of construction:

(i) Draw RB = 5.2 cm.

(ii) Using a protractor at R and B, construct and .

Let them meet at M. Thus, triangle MRB is the required triangle.

On measuring, it is observed that M = 30°

#### Question W3:

Construct the triangle XYZ, in which XY = 6 cm X = 60°, Y = 30°. Measure the third angle.

#### Answer:

Construction of ΔXYZ, where XY = 6 cm, and .

Steps of construction:

(i) Draw XY = 6 cm.

(ii) Using a protractor at X and Y; construct and respectively.

Let them meet at Z. Thus, triangle XYZ is the required triangle.

On measuring, it is observed that measure of the third angle .

#### Question W4:

Construct the triangle ABC, in which AB = 5.5 cm, A = 45°, B = 85°. Measure the sides AC and BC.

#### Answer:

Construction of ΔABC, where AB = 5.5 cm, and .

Steps of constructions:

(i) Draw AB = 5.5 cm.

(ii) Using a protractor at A and B, construct and .

Let them meet at C. Thus, triangle ABC is the required triangle.

On measuring, it is observed that AC = 7.2 cm and BC = 5.1 cm.

#### Question W5:

Construct triangle CDE, in which CD = 6 cm C = D = 60°. Measure E.

#### Answer:

Construction of ΔCDE, where CD = 6 cm, .

Steps of construction:

(i) Draw CD = 6 cm.

(ii) Using a protractor at C and D, construct and .

Let them meet at E. Thus, triangle CDE is the required triangle.

On measuring, it is observed that .

#### Question W1:

Construct a triangle LMN, given LM = 5 cm, MN = 6.8 cm and LMN = 70°. Write the steps of construction.

#### Answer:

Construction of ΔLMN, where LM = 5 cm, MN = 6.8 cm and .

Steps of construction:

(i) Draw LM = 5 cm.

(ii) Using a protractor, draw an angle of 70° at M.

(iii) Measure 6.8 cm as radius and cut an arc so that it intersects at point N.

Join LN. Thus, triangle LMN is the required triangle.

#### Question W2:

Construct a triangle ABC, in which AB = 5.6 cm and AB = BC, B = 60°.

#### Answer:

Construction of ΔABC, where AB = 5.6 cm, AB = BC and B = 70°.

Steps of construction:

(i) Draw AB = 5.6 cm.

(ii) Using a protractor, draw an angle of 60° at B.

(iii) Measure 5.6 cm as radius and cut an arc so that it intersects at point C.

Join AC. Thus, triangle ABC is the required triangle.

#### Question W3:

Construct a triangle DEF, in which EF = 4.5 cm, DF = 6.8 cm, F = 110°.

#### Answer:

Construction of ΔDEF, where EF = 4.5 cm, DF = 6.8 cm and F = 110°.

Steps of construction:

(i) Draw EF = 4.5 cm.

(ii) Using a protractor, draw an angle of 110° at F.

(iii) Measure 6.8 cm as radius and cut an arc so that it intersects at point D.

Join ED. Thus, triangle DEF is the required triangle.

#### Question W4:

Construct a triangle PQR in which PQ = 5 cm, PR = 3 cm and P = 30°.

#### Answer:

Construction of ΔPQR, where PQ = 5 cm, PR = 3 cm and P = 30°.

Steps of construction:

(i) Draw PQ = 5 cm.

(ii) Using a protractor, draw an angle of 30° at P.

(iii) Measure 3 cm as radius and cut an arc so that it intersects at point R.

Join QR. Thus, triangle PQR is the required triangle.

#### Question W5:

Construct a triangle XYZ, given XY = 4.2 cm. YZ = 5.3 cm and Y = 55°.

#### Answer:

Construction of ΔXYZ, where XY = 4.2 cm, YZ = 5. 3 cm and Y = 55°.

Steps of construction:

(iv) Draw XY = 4.2 cm.

(v) Using a protractor, draw an angle of 55° at Y.

(vi) Measure 5.3 cm as radius and cut an arc so that it intersects at point Z.

Join XZ. Thus, triangle XYZ is the required triangle.

#### Question O1:

If an angular bisector is drawn to 120° angle, what is the measure of each angle?

#### Answer:

60°

Reason: An angular bisector of an angle bisects the angle.

So, if an angular bisector is drawn to angle, the measure of each angle is .

#### Question W1:

Bisect the given angles and measure the bisected angles in the following:

∠ABC = 60°

#### Answer:

Steps of construction:

(i) Draw ABC = 60°.

(ii) With B as center and a convenient radius, draw arcs which cut AB and AC at F and E respectively.

(iii) With E and F as centers, draw arcs with a convenient radius, cutting each other at P

Join BP. Thus, BP is the required bisector of ABC.

On measuring, it is observed that ABD = DBC = 30.

#### Question O2:

∠PQR is right angle. What is the measure of each angle when an angular bisector is drawn to it?

#### Answer:

45°

Reason: An angular bisector of an angle bisects the angle.

So, if an angular bisector is drawn to angle, the measure of each angle is .

#### Question W2:

Bisect the given angles and measure the bisected angles in the following:

∠PQR = 110°

#### Answer:

Steps of construction:

(i) Draw PQR= 110°.

(ii) With Q as center and a convenient radius, draw arcs which cut PQ and QR at N and M respectively.

(iii) With M and N as centers, draw arcs with a convenient radius, cutting each other at H. Join QH. Thus, QH is the required bisector of PQR.

On measuring, it is observed that PQH = HQR = 55°.

#### Question W3:

Bisect the given angles and measure the bisected angles in the following:

∠XYZ = 80°

#### Answer:

Steps of construction:

(i) Draw XYZ = 80°

(ii) With Y as center and a convenient radius, draw arcs which cut YX and YZ at N and M respectively.

(iv) With M and N as centers, draw arcs with a convenient radius, cutting each other at W. Join YW. Thus, YW is the required bisector of XYZ.

On measuring, it is observed that XYW = WYZ = 40°.

#### Question W4:

Bisect the given angles and measure the bisected angles in the following:

∠LMN = 90°

#### Answer:

(i) Draw LMN = 90°.

(ii) With M as center and a convenient radius, draw arcs which cut ML and MN at Q and R respectively.

(v) With Q and R as centers, draw arcs with a convenient radius, cutting each other at P. Join MP. Thus, MP is the required bisector of LMN.

On measuring, it is observed that LMP = PMN = 45°.

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