Question
Download Solution PDFखाली दिलेल्या आकृतीमध्ये ABCD हा समांतरभुज चौकोन आहे, DP आणि DQ अनुक्रमे AB आणि BC बाजूंना लंब आहेत, DP = 8 सेमी आणि DQ = 12 सेमी. जर DP ने AB ला 1:2 च्या प्रमाणात भागले आणि DQ ने BC ला 1:1 या प्रमाणात भागले आणि ABCD चे क्षेत्रफळ 72 सेमी2 असेल, तर चतुष्कोण PBQD चे क्षेत्रफळ किती असेल?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFAB = 72/8
AB = 9 सेमी
72 = BC × DQ
BC = 72/12
BC = 6 सेमी
AP = 9 × (1/3)
AP = 3 सेमी
CQ = 6 × (1/2)
CQ = 3 सेमी
∆APD चे क्षेत्रफळ = (1/2) × 3 × 8 = 12 सेमी2
∆CQD चे क्षेत्रफळ = (1/2) × 3 × 12 = 18 सेमी2
चतुष्कोण PBQD चे क्षेत्रफळ = समांतरभुज चौकोनाचे क्षेत्रफळ – ∆APD चे क्षेत्रफळ - ∆CQD चे क्षेत्रफळ
= 72 – 12 – 18 = 42 सेमी2Last updated on Jun 13, 2025
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