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`        abc is a right angled triangle with angle b is equal to 90 degree . m is the mid point of ac and bm =under root 117 . sum of the lenght of other two sides ab and bc is 30 degree .the the value of 1\12 (area of triangle abc)is `
2 years ago

```							At first the line bm is perpendicular to AC and let the two sides x and y and therefore  (x^2+y^2)^0.5=AC AM=CM=(1/2)×ACIN TRIANGLE AMB ANGLE AMB IS 90°ThereforeAM^2 +BM^2=AB^2AM=1/2×ACAM ={(x^2+y^2)^0.5}×1/2Am ^2 =(x^2+y^2)×1/4BM^2=117AB=xx^2+y^2+468=4×x^2Similarlyx^2+y^2+468=4×x^2Add this two equationx^2+y^2+936=3×y^2+3×x^2936=2×x^2+2×x^2468=x^2+y^2We know that x+y=30We have to find 1/12×1/2×xyAt first we have to find xy so (X+y)^2-(x^2+y^2)=2xyX+y=30(X+y)^2=900X^2 +y^2=468900-468=2xy432/2=xy216=xyWe have to find 1/12×1/2×xy1/24×216=9Therefore ans is 9
```
2 years ago
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