The lengths of the sides a, b, c of a ΔABC are connected by the relation a^2 + b^2 = 5c^2. The angle between medians drawn to the sides 'a' and 'b' is -Maths 9th

The lengths of the sides a, b, c of a ΔABC are connected by the relation a^2 + b^2 = 5c^2. The angle between medians drawn to the sides 'a' and 'b' is -Maths 9th
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Let median through C be CF. AF=FB=c2   CF=122(a2+b2)−c2−−−−−−−−−−−−√=3c2   CG=c where G is the centroid and GF=c2   34(a2+b2+c2)=M2a+M2b+M2c   9c22=M2a+M2b+9c24   c2=(23M2a)+(23M2b)   BC2=AG2+BG2   So medians through A and B are perpendicular.
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The lengths of three medians of a triangle are 9 cm, 12 cm and 15 cm. The area (in sq. cm) of this triangle is -Maths 9th

Description : The lengths of three medians of a triangle are 9 cm, 12 cm and 15 cm. The area (in sq. cm) of this triangle is -Maths 9th

Last Answer : (b) 72 cm2Here sm = \(rac{9+12+15}{2}\) = 18 cm, where lengths of medians are m1 = 9 cm, m2 = 12 cm, m3 = 15 cm.∴ Area of triangle = \(rac{4}{3}\sqrt{18(18-9)(18-12)(18-15)}\) cm2= \(rac{4}{3}\sqrt{18 imes9 imes6 imes3}\) cm2 = \(rac{4}{3}\) x 9 x 6 cm2 = 72 cm2.

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Description : The lengths of the perpendiculars drawn from any point in the interior of an equilateral -Maths 9th

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Last Answer : LET EACH SIDE BE X 6X+11X+15X=96 32X=96 X=3 SIDES=6 3=18 11 3=33 15 3=45 AREA OF TRIANGLE BY HERONS FORMULA=S=96/2=48 WHOLE UNDERROOT 48 48-18 48-33 48-45 UNDERROOT=12 4 30 15 3 4 3 15ROOT2 180 ... bh/2 180root2=18 h/2 360root2=18h h=20 root2 But root 2=1.4(approx) h=20 1.4(approx) h=28cm(approx).

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Last Answer : (d) Given, the length of two sides of a triangle are 5 cm and 1.5 cm, respectively. Let sides AB = 5 cm and CA = 1.5 cm We know that, a closed figure formed by three intersecting lines ( ... options (a), (b) and (c) satisfy the above inequality but option (d) does not satisfy the above inequality.

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Description : Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm. -Maths 9th

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Description : The lengths of the sides of a triangle are 7 cm, 13 cm and 12 cm. -Maths 9th

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Description : Let a, b, c be the lengths of the sides of a right angled triangle, the hypotenuse having the length c, then a + b is -Maths 9th

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If a, b, c are the lengths of the sides of a non-equilateral triangle, then -Maths 9th

Description : If a, b, c are the lengths of the sides of a non-equilateral triangle, then -Maths 9th

Last Answer : https://discuss.aiforkids.in/36748/if-are-the-lengths-of-the-sides-non-equilateral-triangle-then

In an equilateral triangle if a, b and c denote the lengths of the perpendicular from A, B and C respectively on the opposite sides, then -Maths 9th

Description : In an equilateral triangle if a, b and c denote the lengths of the perpendicular from A, B and C respectively on the opposite sides, then -Maths 9th

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In a trapezoid ABCD, side BC is parallel to side AD. Also, the lengths of the sides AB, BC, CD and AD are 8, 2, 8 and 10 units respectively -Maths 9th

Description : In a trapezoid ABCD, side BC is parallel to side AD. Also, the lengths of the sides AB, BC, CD and AD are 8, 2, 8 and 10 units respectively -Maths 9th

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Last Answer : . Solution: (i) AB = DE and AB || DE (Given) Two opposite sides of a quadrilateral are equal and parallel to each other. Thus, quadrilateral ABED is a parallelogram (ii) Again BC = EF and BC || EF ... (Given) BC = EF (Given) AC = DF (Opposite sides of a parallelogram) , ΔABC ≅ ΔDEF [SSS congruency]

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Description : G is the centroid of ΔABC with height h units. If a line DE parallel to BC cuts ΔABC at a height h/4 from BC, find the distance GG' in terms of AG -Maths 9th

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In the given figure, ΔABC is an equilateral triangle and ABDC is a cyclic quadrilateral, then find the measure of ∠BDC. -Maths 9th

Description : In the given figure, ΔABC is an equilateral triangle and ABDC is a cyclic quadrilateral, then find the measure of ∠BDC. -Maths 9th

Last Answer : △ABC is an equilateral triangle. ∠BAE=60 ABEC is a cyclic quadrilateral. We know, ∠BAC+∠BEC=180 =>∠BEC=180−60 =120 We know that the angles subtended by an arc on the circumference on the same side are equal. Therefore, ∠ABC=∠BDC =>∠BDC=60

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Description : The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th

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Description : ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle in which AB = AC and BD, CE are its two medians. To show BD = CE.

ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE. -Maths 9th

Description : ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE. -Maths 9th

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Description : The medians AD and BE of the triangle with vertices A(0, b), B(0, 0) and C(a, 0) are mutually perpendicular if -Maths 9th

Last Answer : (c) \(rac{b+k}{f+h}\)Let the slope of the lin passing through the points (-k, h) and (b, - f) be m1. Then m1 = \(rac{-f-h}{b+k}\) = \(-\bigg(rac{f+h}{b+k}\bigg)\)\(\bigg[Slope = rac{y_2-y_1}{x_2-x_1}\bigg]\) ... \(-rac{1}{m_1}\)= \(rac{-1}{-\big(rac{f+h}{b+k}\big)}\) = \(\bigg(rac{b+k}{f+h}\bigg)\)

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

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From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. -Maths 9th

Description : From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. -Maths 9th

Last Answer : Let each side of ㎝ equilateral triangle ABC be ′a′㎝ Now, ar△OAB=21 AB OP=21 a 14=7a㎠→1 ar△OBC= BC OQ =21 a 10=5a㎠→2 ar△OAC=21 AC OR=21 a 6=3a㎠→3 ∴ar△ABC=1+2+3=7a+5a+3a=15a㎠ Also area of equilateral ... ABC=43 a2 Now, 43 a2=15a⇒a=3 15 4 3 3 =3603 =203 ㎝ Now, ar△ABC=43 (203 )2=3003 ㎠

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Description : From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. -Maths 9th

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Description : ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

Last Answer : In △ABE and △ACF, we have ∠BEA=∠CFA (Each 90 0 ) ∠A=∠A (Common angle) AB=AC (Given) ∴△ABE≅△ACF (By SAS congruence criteria) ∴BF=CF [C.P.C.T]

If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lie on the third side. -Maths 9th

Description : If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lie on the third side. -Maths 9th

Last Answer : Solution :- Given: Two circles are drawn on sides AB and AC of a △ABC as diameters. The circles intersects at D. To prove: D lies on BC Construction: Join A and D Proof: ∠ADB = 90° (Angle in the semi-circle ... + 90° => ∠ADB + ∠ADC = 180° => BDC is a straight line. Hence, D lies On third side BC.

How many diagonals can be drawn in a polygon of 15 sides? -Maths 9th

Description : How many diagonals can be drawn in a polygon of 15 sides? -Maths 9th

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From a point O in the interior of a DABC if perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively, then which of the -Maths 9th

Description : From a point O in the interior of a DABC if perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively, then which of the -Maths 9th

Last Answer : (i) In Δ O C E ,D C 2 = D E 2 + E C 2 Δ O B D , D B 2 = O D 2 + B D 2 Δ O A F , O A 2 = O F 2 + A F 2 Adding we get O A 2 + O B 2 + O C 2 = O F 2 + O D 2 + O F 2 + E C 2 + B D 2 + A F 2 A F 2 + B D 2 + C E 2 = O A

Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. The length of each perpendicular is equal to a units. -Maths 9th

Description : Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. The length of each perpendicular is equal to a units. -Maths 9th

Last Answer : answer:

‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Description : ‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Last Answer : No, because in the congruent rule, the two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle i.e., SAS rule.

‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Description : ‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Last Answer : No, because in the congruent rule, the two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle i.e., SAS rule.

Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Description : Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. Draw a line segment BC of length 4.8 cm. From B, point A is at a distance of 3.6 cm. ... at P. Joining BP, we obtain angle bisector of ∠B. Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = ½ x 139° = 19.5°

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Description : The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Last Answer : Given ABCD is a quadrilateral having sides AB=6cm, BC=8cm, CD=12cm and DA=14 cm. Now. Join AC. We have, ABC is a right angled triangle at B. Now, AC2=AB2+BC2 [by Pythagoras theorem]Now, AC2=AB2+BC2 ... =24(1+6-√)cm2=24+246=24(1+6)cm2 Hence, the area of quadrilateral is 241+6-√−−−−−−√cm2241+6cm2 .

Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Description : Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th

Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. 1.Draw a line segment BC of length 4.8 cm. 2.From B, point A is at a distance of 3.6 ... 3.Joining BP, we obtain angle bisector of ∠B. 4.Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = 1/2 x 139° = 19.5°

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Description : The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Last Answer : Given ABCD is a quadrilateral having sides AB = 6 cm, BC = 8 cm, CD = 12 cm and DA = 14 cm. Now, join AC.

In an isosceles triangle, the measure of each of equal sides is 10 cm and the angle between them is 45º. The area of the triangle is: -Maths 9th

Description : In an isosceles triangle, the measure of each of equal sides is 10 cm and the angle between them is 45º. The area of the triangle is: -Maths 9th

Last Answer : (c) 25√2 cm2.ΔABC is an isosceles triangle with AB = AC = 10 cm. ∠A = 45° ∴ Area of ΔABC= \(rac{1}{2}\) x 10 x 10 x sin 45°[Using Δ = \(rac{1}{2}\) bc sin A]= \(rac{50}{\sqrt2}\) = \(rac{50}{\sqrt2}\) x \(rac{\sqrt2}{\sqrt2}\) = 25√2 cm2.

If A is the area of the right angled triangle and b is one of the sides containing the right angle, then what is the length of the -Maths 9th

Description : If A is the area of the right angled triangle and b is one of the sides containing the right angle, then what is the length of the -Maths 9th

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