Please help me out soon.... it's the 29th question in the image I am in 10th standard....
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You can use Similarity.
In ∆AQR and ∆ADC,
Angle AQR = Angle ADC
Angle ARQ = Angle ACD
Therefore, by AA similarity creteria,
∆AQR~∆ADC.
This Implies,
AR/AC = AQ/AD
4.5/10 = 6/AD
Therefore, On solving, we get AD = 13.33 cm [Approx.]
☺
In ∆AQR and ∆ADC,
Angle AQR = Angle ADC
Angle ARQ = Angle ACD
Therefore, by AA similarity creteria,
∆AQR~∆ADC.
This Implies,
AR/AC = AQ/AD
4.5/10 = 6/AD
Therefore, On solving, we get AD = 13.33 cm [Approx.]
☺
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