Question

ral. If
s the
Diagonals AC and BD of quadrilateral ABCD
meet at E. IF AE = 2 cm, BE = 5 cm, CE = 10 cm,
8 सेमी.
मी. में)
DE = 4 cm. BC
2
cm. Find the length of AB
by following figure.
चतुर्भुज ABCD के विकर्ण AC तथा BD, E पर मिलते हैं। यदि
AE = 2 सेमी, BE = 5 सेमी, CE = 10 सेमी, DE = 4 सेमी,
BC =
सेमी. है तो दिये गये चित्र में AB की लम्बाई ज्ञात
कीजिए।​

Answer 1

Given :- Diagonals AC and BD of quadrilateral ABCD meet at E. IF AE = 2 cm, BE = 5 cm, CE = 10 cm, DE = 4 cm. BC = (15/2) cm.

To Find :- The Length of AB ?

Solution :- (Refer to image first.)

In ∆AED and ∆BEC we have,

→ ∠ADE = ∠BCE { Angle at circumference by same chord AB. }

→ ∠AED = ∠BEC { Vertically opposite angles. }

so,

→ ∆AED ~ ∆BEC { By AA similarity. }

then,

→ AE / BE = AD/ BC { By CPCT }

→ 2/5 = AD/(15/2)

→ 2/5 = 2AD/15

→ AD = 3 cm .

Similarly, we get,

→ ∆AEB ~ ∆DEC { By AA similarity. }

then,

→ EB / EC = AB/ DC { By CPCT }

→ 5/10 = AB/DC

→ 1/2 = AB/DC

Let AB = x and DC = 2x.

Now from Ptolemy's theorem we have,

• AB * DC + AD * BC = AC * BD

putting all values in Ptolemy's theorem we get,

→ x * 2x + 3 * (15/2) = (10 + 2) * (5 + 4)

→ 2x² + (45/2) = 12 * 9

→ 2x² = 108 - (45/2)

→ 2x² = (216 - 45)/2

→ 2x² = (171/2)

→ x² = (171/4)

→ x = √171/2

→ x = (1/2)√171 cm (B) (Ans.)

{ Excellent Question. }

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