1.

A triangle if its perimeter is 10.4cm and two angles are 45^(@) and 120^(@) .

Answer»

Solution :Let `ABC` be a triangle. Then, given perimeter = `10.4 CM` i.e., `AB + BC + CA = 10.4cm` and two angles are `45^(@) and 120^(@)` .
say `angleB = 45^(@) and ANGLEC = 120^(@)`
Now, to construct the `Delta ABC` use the FOLLOWING steps.
(i) Draw a line segment say XY and equal to perimeter i.e., `AB + BC + CA = 10.4cm` (ii) Make angle `angleLXY = angleB = 45^(@) and angleMYX = angleC = 120^(@)`.
(III) Bisect `angleLXYand angleMYX` and let these bisectors intersect at a point A (say).

(iv) Draw perpendicular bisectors PQ and RS of AX and AY, respectively.
(v) Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC. Thus , `DELTAABC` is the required triangle.
JustificationSince, B lies on the perpendicular bisector PQ of AX.
`:.`XB = AB
Since, C lies on the perpendicular bisector RS of AY.
`:.`CY = AC
Thus, `AB + BC + CA = XB + CY = XY`
Again, `angleBAX = angleAXB` [`because` in `DeltaAXB`, AB = XB]....(i)
Also,`angleABC = angleBAX + angleAXB` [`because angleABC` is an exterior angle of `DeltaAXB`]
= `angle AXB + angleAXB` [from Eq. (i)]
= `2 angleAXB= angleLXY`[`because` AX is a bisector of `angleLXB`]
Also,`angleCAY = angleAYC` [`because in DeltaAYC`, AC = CY]
`:. angleACB = angleCAY + angleAYC` [`because angle ACB` is an exterior angle of `DeltaAYC`]
= `angleCAY + angleCAY`
= `2 angleCAY = angleMYX` [`because` AY is a bisector of `angleMYX`]
Thus, our construction is justified.


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