The percentage increase in the area of a triangle if each of its side is doubled is -Maths 9th

The percentage increase in the area of a triangle if each of its side is doubled is -Maths 9th
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(c) 300%.Let a, b, c be the sides of the original triangle and s be its semi-perimeter.Then, s = \(rac{1}{2}\)(a+b+c)Let s1 be the semi-perimeter of the new triangle. Then,s1 = \(rac{1}{2}\) (2a + 2b + 2c) = (a + b + c) = 2s∴ Area A of original Δ = \(\sqrt{s(s-a)(s-b)(s-c)}\) and Area A1 of new Δ = \(\sqrt{s_1(s_1-2a)(s_1-2b)(s_1-2c)}\)= \(\sqrt{2s(2s-2a)(2s-2b)(2s-2c)}\)= 4\(\sqrt{s(s-a)(s-b)(s-c)}\) = 4A.∴ Percentage increase in area = \(\bigg(rac{A_1-A}{A} imes100\bigg)\)%= \(rac{(4A-A)}{A} imes100\)% = 300%.
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If each side of a triangle is doubled,... -Maths 9th

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Last Answer : Let a, b, c be the sides of the given triangle and s be its semi-perimeter. Then, s = (a + b + c)/2 ...(i) ∴ Area of the given triangle = root under ( √s(s - a)(s - b)(s - c)) = △, say According to ... - c)) = 4 root under ( √ s(s - a)(s - b)(s - c)) = 4 △ Therefore, the required ratio is 4:1

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Last Answer : We have, Three sides13cm,13cm and 20cm. By using Heron's formula We need to get the semi-perimeter s= 2 a+b+c​ = 2 13+13+20​ = 2 46​ =23 Now, put the heron's formula, s= s(s−a)(s−b)(s−c)​ = 23(23−13)(23−13)(23−20)​ = 23×10×10×3​ =10 23×3​ =83.07cm 2

A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

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A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

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Last Answer : Given = A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined . To Prove : DE || BC and DE = 1 / 2 BC. Const. : Produce the line segment DE to F , such that DE = ... of ||gm are equal and parallel] Also, DE = EF [by construction] Hence, DE || BC and DE = 1 / 2 BC

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

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If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be -Maths 9th

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X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

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Last Answer : Given X and Y are points on the side LN such that LX = XY = YN and XZ || LM To prove ar (ΔLZY) = ar (MZYX) Proof Since, ΔXMZ and ΔXLZ are on the same base XZ and between the same parallel lines LM and XZ. ... get ar (ΔXMZ) + ar (ΔXXZ) = ar (ΔXLZ) + ar (ΔXYZ) => ar (MZYX) = ar (ΔLZY) Hence proved.

Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side, if intersect they will intersect on the circumcircle of the triangle. -Maths 9th

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A right triangle when one side is 3.5 cm and sum of other sides and the hypotenuse is 5.5 cm. -Maths 9th

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Last Answer : Let given right triangle be ABC. Then, given BC = 3.5 cm, ∠B = 90° and sum of other side and hypotenuse i.e., AB + AC = 5.5 cm To construct ΔABC use the following steps 1.Draw the base BC = 3.5 cm 2.Make ... AB = BD - AD = BD - AC [from Eq. (i)] => BD = AB + AC Thus, our construction is justified.

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