SCERT AP 7th Class Maths Solutions Pdf Chapter 10 Construction of Triangles Unit Exercise Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 10th Lesson Construction of Triangles Unit Exercise

Construct triangles for the following :

Question 1.

Construct ∆PQR with measurements PQ = 5.8 cm, QR = 6.5 cm and PR = 4.5 cm.

Answer:

Given measurements of ∆PQR are PQ = 5.8 cm, QR = 6.5 cm and PR = 4.5 cm

Steps of Construction:

- Draw a rough sketch of triangle arid label it with given measurements.
- Draw a line segment with PQ = 5.8 cm.,
- Draw ati arc with centre P and radius 4.5 cm.
- Draw another arc with centre Q and radius 6.5 cm. to intersect the previous arc at R.
- Join PR and QR.

Thus, required ∆PQR is constructed with the given measurements.

Question 2.

Construct an isosceles triangle LMN with measurements LM = LN = 6.5 cm and MN=8cm.

Answer:

Given measurements of ∆LMN are LM = LN = 6.5 cm and MN = 8 cm.

Steps of Construction :

- Draw a rough sketch of triangle and label it with given measurements.
- Draw a line segment with MN = 8 cm.
- Draw an arc with centre M and radius 6.5 cm.
- Draw another arc with centre N and same radius (6.5 cm) to intersect the previous are at L
- Join ML and NL.

Hence, required ∆LMN is constructed with the given measurements.

Question 3.

Construct ∆ABC with measurements ∠A = 60°, ∠B = 706 and AB = 7 cm.

Answer:

Given measurements of ∆ABC are ∠A = 60°, ∠B = 70° and AB = 7 cm.

Steps of Construction :

- Draw a rough sketch of triangle and label it with given measurements.
- Draw a line segment AB = 7 cm.
- Draw a ray AX such that ∠BAX = 60°.
- Draw another ray BY such that ∠ABY = 70°.
- Name the intersecting point of AX and BY as C.

Hence, required ∆ABC is constructed with the given measurements.

Question 4.

Construct a right angled triangle XYZ in which ∠Y = 90°, XY = 5 cm and YZ = 7 cm.

Answer:

Given measurements of ∆XYZ are ∠Y = 90°, XY = 5 cm and YZ = 7 Cm.

Steps of Construction:

- Draw a rough sketch of triangle and label it with given measurements.
- Draw a line segment with XY = 5 cm.
- Draw a ray YP such that ∠XYP = 90°
- Draw an arc with centre Y and radius 7 cm to intersect YP at point Z.
- Join XZ.

Hence, required ∆XYZ is constructed with the given measurements.

Question 5.

Construct an equilateral triangle DEF in which DE = EF = FD = 5 cm.

Answer:

Given measurements of ∆DEF are DE = EF = FD = 5 cm.

Steps of Construction:

- Draw a rough sketch of the triangle and label it with the given measurements.
- Draw a line segment DE of length 5 cm.
- Draw an arc with centre D and radius 5 cm.
- Draw another arc with centre E and the same radius (5 cm) to intersect the previous arc at F.
- Join DF and EF.

Thus, the required triangle ∆DEF is constructed with the given measurements.

Question 6.

Construct a triangle with a non-included angle for the sides of ST and SU of lengths 6 v and 7 cm. respectively and ∠T = 80°.

Answer:

Given measurements of ∆STU are ST = 6 cm, SU = 7 cm and ∠T = 80°.

Steps of Construction:

- Draw a rough sketch of triangle and label it with given measurements.
- Draw a line segment with ST = 6 cm.
- Draw a ray TX such that ∠STX = 80°.
- Draw an arc with centre S and radius 7 cm to intersect TX at point U.
- Join SU. Hence, the required triangle ∆STU is constructed with the given measurements.

Question 7.

Can you construct ∆DEF with DE = 7 cm, EF = 14 cm and FD = 5 cm. ? If not give reasons. .

Answer:

No, we can’t construct the triangle. Because the given sides of ∆DEF are DE = 7 cm, EF = 14 cm, FD = 5 cm.

In any triangle sum of any. two sides are always greater than the third side.

DE + FD = 7 + 5 = 12 cm < 14 cm

Sum of DE + FD < EF.

So, with the given measurements construction of ADEF is not possible.