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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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