Let PS be the median of the triangle with vertices P(2, 2), Q(6, –1) and R(7, 3). -Maths 9th

Let PS be the median of the triangle with vertices P(2, 2), Q(6, –1) and R(7, 3). -Maths 9th
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(b) a = ± √2b Let D be the mid-point of BC. Then D ≡ \(\bigg(rac{a+0}{2},rac{0}{2}\bigg)\)i.e. \(\bigg(rac{a}{2},0\bigg)\)Let E be the mid-point of AC, thenE = \(\bigg(rac{a+0}{2},rac{0+b}{2}\bigg)\) = \(\bigg(rac{a}{2},rac{a}{2}\bigg)\)AD ⊥ BE if slope of AD x Slope of BE = –1       ...(i) Slope of AD = \(rac{0-b}{rac{a}{2}-0}\) = \(rac{-b}{rac{a}{2}}\) = \(rac{-2b}{a}\)Slope of AD = \(\bigg(rac{rac{b}{2}-0}{rac{a}{2}-0}\bigg)\) = \(rac{rac{b}{2}}{rac{a}{2}}\) = \(rac{b}{a}.\)∴ From (i), \(rac{-2b}{a}\) x \(rac{b}{a}\) = -1⇒ 2b2 = a2 ⇒ a = ± √2 .
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Description : Divide Rs 1728 among P, Q, R in such way that 8 times P's share is 12times Q's share and is 6 times R's share. How much does each get? A) Rs. 576,384,768 B) Rs. 588,392,484 C) Rs. 592, 384, 652 D) None of the above

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Description : In the adjoining figure, P and Q have co-ordinates (4, 6)and (0, 3) respectively. Find (i) the co-ordinates of R (ii) Area of quadrilateral OAPQ. -Maths 9th

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The vertices A(4, 5), B(7, 6), C(4, 3) and D(1, 2) from the quadrilateral ABCD whose special name is -Maths 9th

Description : The vertices A(4, 5), B(7, 6), C(4, 3) and D(1, 2) from the quadrilateral ABCD whose special name is -Maths 9th

Last Answer : (a) (5, 2)Let A(8, 6) , B(8, -2) and C(2, -2) be the vertices of the given triangle and P(x, y) be the circum-centre of this triangle. Then, PA2 = PB2 = PC2 Now, PA2 = PB2 ⇒ (x - 8)2 + (y - 6)2 = (x ... 4 = x2 - 4x + 4 + y2 + 4y + 4 ⇒ 12x = 60 ⇒ x = 5. ∴ Co-ordinates of the circumcentre are (5, 2).

Let A = {1, 2, 3, 4}, B = {5, 6, 7, 8}. Then R = {(1, 5), (1, 7), (2, 6)} is a relation from set A to B defined as : -Maths 9th

Description : Let A = {1, 2, 3, 4}, B = {5, 6, 7, 8}. Then R = {(1, 5), (1, 7), (2, 6)} is a relation from set A to B defined as : -Maths 9th

Last Answer : (d) R = {(a, b) : b/a is odd} Since (2, 6) ∈R, the relation a and b are odd does not exist. Since (1, 5) and (1, 7) ∈R, the relation a and b are even does not exist. None of the ... \(rac{7}{1}\) = 7, \(rac{6}{2}\) = 3, quotients being all odd numbers, the relation b/a is odd exists.

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Description : 3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Construction, Join AC and BD. To Prove, PQRS is a rhombus. Proof: In ΔABC P and Q ... (ii), (iii), (iv) and (v), PQ = QR = SR = PS So, PQRS is a rhombus. Hence Proved

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Description : 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. To Prove, PQRS is a rectangle. Construction, Join AC and BD. Proof: In ΔDRS and ... , In PQRS, RS = PQ and RQ = SP from (i) and (ii) ∠Q = 90° , PQRS is a rectangle.

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.