Question

Construct a triangle whose perimeter is 10.4 cm and two angles are 45 and 105.

Answer 1

HEY MATE HERE IS YOUR 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 105°.

say ∠B = 45° and ∠C = 105°

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 = 105°

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.

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MATE'S