Question

answer the above questions please ​

Answer 1

Answer:

1) Option(B) 7√2

2) Option(C) 260°

3) Option(B) 5 : 2

4) Option(D) 3.5cm.

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Question 1:

According to the question, we have a right-angled isosceles triangle, let us name it PQR, let PR be the hypotenuse of the right-angled isosceles triangle.

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Isosceles triangles refer to those triangles where any two sides have the same measure, for a right-angled isosceles triangle, the two sides other than the hypotenuse will have the same measure [as the hypotenuse is the largest side in a right-angles triangle], therefore;

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• PQ = QR = 7cm.

• PR = ?

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According to the Pythagoras theorem;

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$\dashrightarrow \sf \ Hypotenuse^2 = Altitude^2 + Base^2$

$\dashrightarrow \sf \ PR^2 = PQ^2 + QR^2$

$\dashrightarrow \sf \ PR^2 = (7)^2 + (7)^2$

$\dashrightarrow \sf \ PR^2 = 49 + 49$

$\dashrightarrow \sf \ PR^2 = 98$

$\dashrightarrow \sf \ PR = \sqrt{98}$

$\dashrightarrow \sf \ PR = \sqrt{2 \times 7 \times 7}$

$\dashrightarrow \sf \ PR = 7\sqrt{2} \ cm.$

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Therefore the answer is Option (B) 7√2.

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Question 2:

ATQ;

A circle with centre O.

m∠AOB = 100°

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And we're asked to find the angle formed by the major arc AXB. [There's no 'Y' marking in the diagram]

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$\dashrightarrow \sf \ m(arc \ AYB) = 360^\circ - m\angle AOB$

$\dashrightarrow \sf \ m(arc \ AYB) = 360^\circ - 100^\circ$

$\dashrightarrow \sf \ m(arc \ AYB) = 260^\circ$

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Therefore the answer is Option (C) 260°.

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Question 3:

According to the question;

ΔABC ≈ ΔPQR

4A(ΔABC)  = 25A(ΔPQR)

We're asked to find the value of AB : PQ.

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ATQ;

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$\sf \dashrightarrow \ 4A(\Delta ABC) = 25A(\Delta PQR)$

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$\sf \dashrightarrow \ \dfrac{\Delta ABC}{\Delta PQR} = \dfrac{25A}{4A}$

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$\sf \dashrightarrow \ \dfrac{\Delta ABC}{\Delta PQR} = \dfrac{25}{4} \ \ \ \ \longmapsto \ Relation(1)$

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We know that for any two similar triangles, the ratio of the area of the two triangles is equal to the ratio of the squares of the sides of the respective triangles.

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Therefore;

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$\sf \dashrightarrow \ \dfrac{ar(\Delta ABC)}{ar(\Delta PQR)} = \Bigg[\dfrac{AB}{PQ}\Bigg]^2 = \Bigg[\dfrac{BC}{QR}\Bigg]^2 = \Bigg[\dfrac{AC}{PR}\Bigg]^2$

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Substitute the value from Relation (1) in the equation we've got above.

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$\sf \dashrightarrow \ \dfrac{25}{4} = \Bigg[\dfrac{AB}{PQ}\Bigg]^2$

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On square root-ing both sides we get;

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$\sf \dashrightarrow \ \sqrt{\dfrac{25}{4}} = \sqrt{\Bigg[\dfrac{AB}{PQ}\Bigg]^2}$

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$\sf \dashrightarrow \ \dfrac{AB}{PQ} = \sqrt{\dfrac{25}{4}}$

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$\sf \dashrightarrow \ \dfrac{AB}{PQ} = \dfrac{5}{2}$

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Therefore the answer is Option (B) 5:2.

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Question 4:

Let the centre of the first circle be "O", and let the centre of the other circle be "P". These two circles touch each other internally. [Check attachment]

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We're asked to find the distance between their centres. From the figure, we can tell that the distance between the two centres is given by the difference of the radii of the two circles.

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• Radii of the circle with centre O = Diameter/2 = 17/2 cm.

• Radii of the circle with centre P = Diameter/2 = 10/2 = 5 cm.

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Therefore,

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The distance between the centres = Radius(O) - Radius(P)

The distance between the centres = (17/2) - 5

The distance between the centres = (17 - 10/2)

The distance between the centres = 7/2

The distance between the centres = 3.5 cm.

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Therefore the answer is Option (D) 3.5cm.

Answer 2

Answer:

$1) \: .(b)7 \sqrt{2} \\ 2). \: (c) 260$

3).(b) 5:2

4). (d) 3.5cm