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2019 er junior national er 3 no. Ques er ans. koto?

At first post the problem!

Dustan wrote: ↑

Thu Dec 10, 2020 9:16 pm

At first post the problem!

Here is the question.

To find the value of DI first we need to find the value of IL, which is the height of the parallelogram EHGF. Since the area of a parallelogram is height×base. So, we can find the height of the parallelogram if we find the area of it. Triangle AHE=GCF and EFB=HDG, i dont think i need to explain why. Area of AHE+GCF=2×area of AHE so, area of AHE+GCF=3×4 (1/2 × 2 cancels out). And we do the same with traingle HDG and EBF. So, area of HDG+EBF = 8×6= 48. So, area of the 4 triangles is 12+48=60. Area of the rectangle is (6+3)×(8+4)=108. Area of parallelogram is ABCD-the 4 triangles. So, area of parallelogram is 108-60=48. Now, the base of the parallelogram is HG=√(6^2+8^2)=10. So the length of LI is 48÷10=4.8. now we're (spoiler alert) *halfway* through. We just need to find the value of DL. Now there are 2 ways you can solve this, one is algebra and the other is geometry and im not very good at algebra so ill go with the geometrical one. Now, extend a line HM where M is outisde the rectangle and the line HM be equal to DL and perpendicular to HL. Now extend another line GN thats perpendicular to LG and length is equal to DL. Now connect M and N. We have created a rectangle HMNG and DL=HM=GN. If we find HM or GM we find DL. We already know the length of the long sides which are equal to 10. If we find the area of the rectangle then we can find HM, which is the breadth of the rectangle. Youll notice that triangle DLG is same as triangle GND. And its the same with triangle HLD and HMD. So the area of the rectangle is 2 times the area of the triangle HDG. So, 8×6×1/2×2=48. The length of the rectangle is 10, so the breadth, which is DL is = 48÷10=4.8. now DL=4.8 snd LI=4.8. Which is why i said "we're halfway there".

Your question is quite confusing.
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