In the given figure, CAB = CED, CD = 8 cm, CE = 10 cm, BE = 2 cm, AB = 9 cm, AD = b and DE = a, the value of a + b is
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SOLUTION :
GIVEN :∠A = ∠CED. In the figure :
AB = 9 cm , CE= 10 cm , AD = 7cm , DC= 8cm ,BE= 2 cm , DE = x cm
In ∆CAB and ∆CED
∠C = ∠C
[Common]
∠A = ∠CED
[Given]
∴ ∆CAB ~ ∆CED [By AA similarity criterion ]
CA/CE = AB /DE = CB /CD
[Since corresponding sides of two similar triangles are proportional]
AB/DE = CB/CD
9/x = (CE+EB)/CD
9/x = (10+2)/8
9/x = 12/8
12x = 8×9
x = (8×9)/12
x = (2×9)/3
x = 2× 3
x = 6 cm
Hence , the value of x= 6 cm.
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