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(iv) If the diagonals of a parallelogram are equal and perpendicular to each other, then quadrilateral will be. (v) Write the name of longest chord of a circle. (vi) Write the Heron's formula to find the area of a triangle.

Mathematics

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here are the solutions to the questions:

Part iv

If the diagonals of a parallelogram are equal and perpendicular to each other, then quadrilateral will be.

  • If the diagonals of a parallelogram are equal, it is a rectangle. If they are perpendicular, it is a rhombus. A quadrilateral that is both a rectangle and a rhombus is a square.

Part v

Write the name of longest chord of a circle.

  • The longest chord of a circle is its diameter.

Part vi· 17 steps

Write the Heron's formula to find the area of a triangle.

  • Heron's formula for the area of a triangle with side lengths a,b,ca, b, c is: Area=s(s−a)(s−b)(s−c)\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} where ss is the semi-perimeter, given by: s=a+b+c2s = \frac{a+b+c}{2}

प्र.6 सरल कीजिए - (Simplify -)
712⋅8127^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}

  1. Use the exponent rule am⋅bm=(ab)ma^m \cdot b^m = (ab)^m.

    712⋅812=(7⋅8)127^{\frac{1}{2}} \cdot 8^{\frac{1}{2}} = (7 \cdot 8)^{\frac{1}{2}}
  2. Multiply the bases.

    (7⋅8)12=5612(7 \cdot 8)^{\frac{1}{2}} = 56^{\frac{1}{2}}
  3. Convert the fractional exponent to a square root and simplify.

    5612=56=4⋅14=21456^{\frac{1}{2}} = \sqrt{56} = \sqrt{4 \cdot 14} = 2\sqrt{14}

    The simplified expression is 214\boxed{2\sqrt{14}}.

    अथवा / OR

    Multiply 656\sqrt{5} and 252\sqrt{5}.

  4. Group the integer parts and the radical parts.

    (65)⋅(25)=(6⋅2)⋅(5⋅5)(6\sqrt{5}) \cdot (2\sqrt{5}) = (6 \cdot 2) \cdot (\sqrt{5} \cdot \sqrt{5})
  5. Perform the multiplications.

    (6⋅2)⋅(5⋅5)=12⋅5(6 \cdot 2) \cdot (\sqrt{5} \cdot \sqrt{5}) = 12 \cdot 5
  6. Calculate the final product.

    12⋅5=6012 \cdot 5 = 60

    The product is 60\boxed{60}.

    प्र.7 17\frac{1}{\sqrt{7}} का हर का परिमेयकरण कीजिए। (Rationalise the denominator of 17\frac{1}{\sqrt{7}}.)

  7. To rationalise the denominator, multiply both the numerator and the denominator by 7\sqrt{7}.

    17=17×77\frac{1}{\sqrt{7}} = \frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}
  8. Perform the multiplication.

    1×77×7=77\frac{1 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \frac{\sqrt{7}}{7}

    The rationalised expression is 77\boxed{\frac{\sqrt{7}}{7}}.

    अथवा / OR

    35\frac{3}{5} और 45\frac{4}{5} के बीच दो परिमेय संख्या ज्ञात कीजिए। (Find two rational numbers between 35\frac{3}{5} and 45\frac{4}{5}.)

  9. To find rational numbers between two fractions, express them with a larger common denominator.

    Multiply both fractions by 1010\frac{10}{10}.

    35=3×105×10=3050\frac{3}{5} = \frac{3 \times 10}{5 \times 10} = \frac{30}{50} 45=4×105×10=4050\frac{4}{5} = \frac{4 \times 10}{5 \times 10} = \frac{40}{50}
  10. Identify two rational numbers between 3050\frac{30}{50} and 4050\frac{40}{50}.

    • Two such numbers could be 3150\frac{31}{50} and 3250\frac{32}{50}. Two rational numbers between 35\frac{3}{5} and 45\frac{4}{5} are 3150,3250\boxed{\frac{31}{50}, \frac{32}{50}}. (Other valid answers exist, e.g., 3350,3450\frac{33}{50}, \frac{34}{50}, etc.)

    प्र.8 गुणनखंड ज्ञात कीजिए - (Factorise -)
    12x2−7x+112x^2 - 7x + 1

  11. Find two numbers that multiply to 12×1=1212 \times 1 = 12 and add up to −7-7.

    These numbers are −3-3 and −4-4.

  12. Rewrite the middle term using these numbers.

    12x2−3x−4x+112x^2 - 3x - 4x + 1
  13. Group the terms and factor out common factors from each group.

    (12x2−3x)−(4x−1)(12x^2 - 3x) - (4x - 1) 3x(4x−1)−1(4x−1)3x(4x - 1) - 1(4x - 1)
  14. Factor out the common binomial factor (4x−1)(4x - 1).

    (4x−1)(3x−1)(4x - 1)(3x - 1)

    The factorised expression is (4x−1)(3x−1)\boxed{(4x - 1)(3x - 1)}.

    अथवा / OR

    सीधे गुणा किये बिना 104×96104 \times 96 का गुणनफल ज्ञात कीजिए। (Evaluate the product of 104×96104 \times 96 without multiplying directly.)

  15. Express the numbers as a sum and difference from a common base (100).

    104=100+4104 = 100 + 4 96=100−496 = 100 - 4
  16. Use the algebraic identity (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2.

    Here, a=100a=100 and b=4b=4.

    104×96=(100+4)(100−4)=1002−42104 \times 96 = (100 + 4)(100 - 4) = 100^2 - 4^2
  17. Calculate the squares and subtract.

    1002−42=10000−16=9984100^2 - 4^2 = 10000 - 16 = 9984

    The product is 9984\boxed{9984}.

    प्र.9 SSS सर्वांगसमता नियम लिखिए। (Write SSS Congruence rule.)

    • SSS (Side-Side-Side) Congruence Rule states that if three sides of one triangle are equal to the three corresponding sides of another triangle, then the two triangles are congruent.
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