Question

Construct a triangle abc if its perimeter is 10.4 cm and two angles are 45 degree and 120 degree

Answer 1

Answer:

Step-by-step explanation:

Our question is: Construct a triangle abc if its perimeter is 10.4 cm and two angles are 45 degree and 120 degree.

Let ∠B = 45° and let ∠C = 120° .

Now, in order to represent the triangle ΔABC we need to follow the next steps.

1. Draw a line segment say XY and equal to perimeter i.e., AB+ BC + CA = 10.4 cm.
2. Draw ∠LXY = ∠B = 45° and ∠MYX = ∠C = 120°.
3. Make the bisector ∠LXY and ∠MYX and let them meet at a know point A.
4. Draw perpendicular bisectors PQ and RS of AX and AY, respectively.
5. Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC.

Hence, we get the triangle which was asked for.

Hope this helped you!!

Answer 2

Answer:

Let ABC be a triangle. Then, given perimeter = 10.4 cm i.e., AB+ BC + CA = 10.4 cm and two angles are 45° and 120°.

say ∠B = 45° and ∠C = 120°

Now, to construct the ΔABC use the following steps.

1.Draw a line segment say XY and equal to perimeter i.e., AB+ BC + CA = 10.4 cm

2.Make angle ∠LXY = ∠B = 45° and ∠MYX = ∠C = 120°.

3.Bisect ∠LXY and ∠MYX and let these bisectors intersect at a point A (say).

4.Draw perpendicular bisectors PQ and RS of AX and AY, respectively.

5.Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC. Thus, ΔABC is the required triangle.

Justification

Since, B lies on the perpendicular bisector PQ of AX.

Thus, AB+ BC + CA = XB+ BC + CY=XY

Again, ∠BAX = ∠AXB [∴ in ΔAXB, AB = XB] …(i)

Also, ∠ABC = ∠BAX + ∠AXB [ ∠ABC is an exterior angle of ΔAXB]

= ∠AXB + ∠AXB [from Eq. (i)]

= 2 ∠AXB= ∠LXY [ AX is a bisector of ∠LXB]

Also, ∠CAY = ∠AYC [∴ in A AYC, AC = CY]

∠ACB=∠CAY + ∠AYC [ ∠ACB is an exterior angle of ΔAYC]

= ∠CAY + ∠CAY

= 2 ∠CAY= ∠MYX [∴ AY is a bisector of ∠MYX]

Thus, our construction is justified.