1. (i) The median through A meet BC at D. So, D is the mid-point of BC. Therefore Coordinates of D = ((6+1)/2, (5+4)/2) = (7/2, 9/2) (ii) Point P lies on AD, such that AP:PD = 2:1. Therefore (1.4) Coordinates of P = ((2×7/2+1×4)/(2+1), (2×9/2+1×2)/(2+1)) (iii) The median through B meets AC at E. So, ERead more

    (i) The median through A meet BC at D. So, D is the mid-point of BC. Therefore Coordinates of D = ((6+1)/2, (5+4)/2) = (7/2, 9/2)
    (ii) Point P lies on AD, such that AP:PD = 2:1. Therefore (1.4) Coordinates of P = ((2×7/2+1×4)/(2+1), (2×9/2+1×2)/(2+1))
    (iii) The median through B meets AC at E. So, E is the mid-point of AC. Therefore, Coordinates of E = ((4+1)/2, (2+4)/2) = (5/2, 3)
    (iv) Point Q lies on AE, such that AQ:QE = 2:1.Therefore Coordinates of Q = ((2×5/2+1×6)/(2+1), (2×3+1×5)/(2+1)) = (11/3, 11/3)
    The median through C mects AB at F. So, F is the mid-point of AB. Therefore Coordinates of F = ((4+6)/2, (2+5)/2) = (5, 7/2)
    Point R lies on CF, such that CR:RF = 2:1.
    (v) Therefore, the coordinates of R = ((2×5+1×1)/(2+1), (2×7/2+1×4)/(2+1)) = (11/3, 11/3)
    (vi) The coordinates of P, Q and R is same.
    (vii)Points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of triangle ABC. Median through A meets BC at D. So, D is the mid-point of BC. Therefore, The coordinates of D = ((x₂+x₃)/2, (y₂+y₃)/2)
    Let O be the centroid of the triangle. Point O lies on AD such that A0:0D = 2:1.
    Therefore, the coordinates of point O
    ((2×(x₂+x₃)/2 +1×x₁)/(2+1), (2×(y₂+y₃)/2 +1×y₁)/(2+1)) = ((x₁+x₂+x₃)/2, (y₁+y₂+y₃)/2)

    Here you can see the video explanation of this question ✋

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  2. The mid-point of AB is P. Therefore, The coordinates of P = ((-1, 1)/2, (-1+4)/2) = (-1, 3/2) Similarly, the coordinates of points Q, R and S are (2, 4). (5, 3/2) and (2, -1) respectively. Length of PQ = √((-1-2)²+(3/2 -4)²) = √(9+25/4) = √(61/4) Length of QR = √((2-5)²+(4-3/2)²) = √(9+25/4) = √(61/Read more

    The mid-point of AB is P. Therefore,
    The coordinates of P = ((-1, 1)/2, (-1+4)/2) = (-1, 3/2)
    Similarly, the coordinates of points Q, R and S are (2, 4). (5, 3/2) and (2, -1) respectively.
    Length of PQ = √((-1-2)²+(3/2 -4)²) = √(9+25/4) = √(61/4)
    Length of QR = √((2-5)²+(4-3/2)²) = √(9+25/4) = √(61/4)
    Length of PS = √((5-2)²+(3/2+1)²) = √(9+25/4) = √(61/4)
    Length of SP = √((2+1)²+(-1-3/2)²) = √(9+25/4) = √(61/4)
    Length of PR = √((-1-5)²+(3/2-3/2)²) = √(36+0) = 6
    Length of SQ = √((2-2)²+(4+1)²) = √(0+25) = 5
    All the sides of quadrilateral PQRS are equal (PQ = QR = RS = ST) but diagonals are not equal (PR ≠ SQ). So, PQRS is a rhombus.

    For Bette explanation see here👇😃

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  3. In △AABC, by Pythagoras theorem, we have AC² = AB²+BC² = (24 cm)²+(7 cm)² = (576 + 49)cm² = 625 cm² ⇒ AC = √625 = 25 cm (i) sin A = BC/AC = 7/25 and cos A = AB/AC = 24/25 (ii) sin C = AB/AC = 24/25 and cos C = BC/AC = 7/25 See here for video explanation of this above question 😁✌

    In △AABC, by Pythagoras theorem, we have
    AC² = AB²+BC²
    = (24 cm)²+(7 cm)²
    = (576 + 49)cm²
    = 625 cm²
    ⇒ AC = √625 = 25 cm
    (i) sin A = BC/AC = 7/25 and cos A = AB/AC = 24/25
    (ii) sin C = AB/AC = 24/25 and cos C = BC/AC = 7/25

    See here for video explanation of this above question 😁✌

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  4. Where can I get the video solutions of Class 8 Science Chapter 16?

    Where can I get the video solutions of Class 8 Science Chapter 16?

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  5. संज्ञा के बारे में हम पहली कक्षा से ही पढ़ना प्रारंभ कर देते हैं। कक्षा 4 के हिंदी व्याकरण में विद्यार्थी संज्ञा के कुछ और उदाहरणों के साथ अपने ज्ञान को मजबूत करेंगे। कक्षा 4 हिंदी व्याकरण में विभिन्न शब्दों के स्त्रीलिंग और पुल्लिंग शब्दों को भी याद करेंगे। व्याकरण भाषा सीखने का एक महत्वपूर्ण पहलू हRead more

    संज्ञा के बारे में हम पहली कक्षा से ही पढ़ना प्रारंभ कर देते हैं। कक्षा 4 के हिंदी व्याकरण में विद्यार्थी संज्ञा के कुछ और उदाहरणों के साथ अपने ज्ञान को मजबूत करेंगे। कक्षा 4 हिंदी व्याकरण में विभिन्न शब्दों के स्त्रीलिंग और पुल्लिंग शब्दों को भी याद करेंगे।
    व्याकरण भाषा सीखने का एक महत्वपूर्ण पहलू है, और चौथी कक्षा के छात्रों के लिए व्याकरण में एक मजबूत नींव विकसित करना शुरू करना विशेष रूप से महत्वपूर्ण है। व्याकरण सीखने से छात्रों को भाषा की संरचना और तर्क को समझने में मदद मिलती है, जिससे उनके पढ़ने और लिखने के कौशल में सुधार हो सकता है।
    अधिक जानकारी के लिए यहाँ जाएँ => https://hindi.tiwariacademy.com/ncert-solutions/class-4/hindi/grammar/

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