If the area of the quadrilateral whose angular points A, B, C, D taken in order are (1, 2), (–5, 6), (7, – 4) and (–2, k) be zero, -Maths 9th

If the area of the quadrilateral whose angular points A, B, C, D taken in order are (1, 2), (–5, 6), (7, – 4) and (–2, k) be zero, -Maths 9th
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1 Answer

Answer :

(d) 1 : 16Gvien \(rac{AD}{AB}=rac{1}{4}\) ⇒ \(rac{AD}{DB}=rac{1}{3}\),i.e., D divides AB internally in the ratio 1 : 3. ∴ Co-odinates of D are\(\bigg(rac{1+12}{1+3},rac{5+18}{1+3}\bigg)\)i.e.,\(\bigg(rac{13}{4},rac{23}{4}\bigg)\)Also, \(rac{AE}{AC}=rac{1}{4}\) ⇒ \(rac{AE}{EC}=rac{1}{3}\), i.e., E divides AC internally in the ratio 1 : 3.Co-ordinates of E are  \(\bigg(rac{7+12}{1+3},rac{2+18}{1+3}\bigg)\)i.e, \(\bigg(rac{19}{4},5\bigg)\)Now, area of Δ ABC = \(rac{1}{2}\) [4(5 – 2) + 1(2 – 6)+ 7(6 – 5)] = \(rac{1}{2}\) [12 – 4 + 7] = \(rac{15}{2}\) sq. unitsArea of Δ ADE, where A(4, 6), D\(\bigg(rac{13}{4},rac{23}{4}\bigg)\), E\(\bigg(rac{19}{4},5\bigg)\) is \(rac{1}{2}\)\(\bigg[4\bigg(rac{23}{4}-5\bigg)+rac{13}{4}(5-6)+rac{19}{4}\bigg(6-rac{23}{4}\bigg)\bigg]\)= \(rac{1}{2}\)\(\bigg[4 imesrac{3}{4}+rac{13}{4} imes-1+rac{19}{4} imesrac{1}{4}\bigg]\)= \(rac{1}{2}\)\(\bigg[3-rac{13}{4}+rac{19}{16}\bigg]\) = \(rac{1}{2}\) \(\bigg[rac{84-5+19}{16}\bigg]\)= \(rac{15}{32}\) sq. units∴ \(rac{ ext{Area of}\,\Delta{ADE}}{ ext{Area of}\,\Delta{ABC}}\) = \(rac{rac{15}{32}}{rac{15}{2}}\) = 1 : 16.
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Last Answer : the answer is 56.55u because height is 8.7 base is 13 cm

5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Description : 5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Last Answer : Solution: Given that, Let ABCD be a quadrilateral and its diagonals AC and BD bisect each other at right angle at O. To prove that, The Quadrilateral ABCD is a square. Proof, In ΔAOB and ΔCOD, AO = ... right angle. Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square. Hence Proved.